Free math worksheets

- Interpret the derivative as a tangent-line slope. - Think about the sign of the slope of a line that rises as \(x\) increases.

1. The derivative at a point is the slope of the tangent line there. 2. A line that rises from left to right has positive slope. 3. Therefore, \(f'(4)>0\).

Positive; \(f'(4)>0\).

- A derivative value equals the slope of the tangent line. - Use two exact points on each tangent to calculate its slope. - Increasing, decreasing, and horizontal behavior correspond to positive, negative, and zero derivative values.

1. The line \(t_1\) passes through \(A(0,2)\) and \((-2,0)\). Its slope is \(\frac{2-0}{0-(-2)}=1\), so \(f'(0)=1\). 2. The line \(t_2\) passes through \(B(4,2)\) and \((6,0)\). Its slope is \(\frac{0-2}{6-4}=-1\), so \(f'(4)=-1\). 3. At \(x=2\), the graph has a local maximum and a horizontal tangent, so \(f'(2)=0\). 4. At \(x=5\), the graph is decreasing, so \(f'(5)<0\). 5. At \(x=-2\), the graph is increasing, so \(f'(-2)>0\).

a) \(f'(0)=1\) and \(f'(4)=-1\) b) 1) \(f'(2)=0\) 2) \(f'(5)<0\) 3) \(f'(-2)>0\)

- The derivative represents the slope of the graph at each input. - Relate increasing and decreasing behavior to the sign of the derivative. - At a smooth local maximum or minimum, the tangent is horizontal. - Compare the derivative's sign pattern with the opening direction of a parabola.

1. A is false. At \(x=0\), the graph of \(f\) is decreasing, so \(f'(0)<0\). Passing through the origin does not determine the derivative sign. 2. B is true. The graph has a local maximum at \(x=-2\) and a local minimum at \(x=2\), so both tangent lines are horizontal. 3. C is true. The function decreases throughout \((-2,2)\), so \(f'(x)<0\) on that interval. 4. D is false. The derivative is positive outside the two zeros and negative between them, which matches an upward-opening parabola.

A) False; \(f'(0)<0\) because \(f\) is decreasing there. B) True; both inputs are smooth extrema. C) True; \(f\) is decreasing on \((-2,2)\). D) False; the sign pattern matches an upward-opening parabola.

- Decide which graph gives function values and which gives slopes. - The derivative value \(f'(x)\) is the slope of \(f\) at \(x\). - For part c), find an \(x\)-coordinate where the graph of \(f\) has height \(1\). - For part d), find where the derivative graph has value \(3\).

1. Read the function value from Figure 1: \(f(2) = -1\). 2. The slope at \(x = -1\) is \(f'(-1)\). Figure 2 gives \(f'(-1) = -1.5\). 3. In Figure 1, the graph passes through \((0, 1)\), so one possible value is \(a = 0\). 4. A slope of \(3\) means \(f'(b) = 3\). Figure 2 shows this at \(b = -2\) and \(b = 2\).

a) \(f(2) = -1\) b) The slope is \(-1.5\). c) \(a = 0\) d) \(b = -2\) and \(b = 2\)

- A tangent line touches the curve at the marked point and has the same local direction there. - Use the slope-intercept form of a line. - Read the y-intercept directly when possible. - Use two grid points to calculate each nonzero slope. - What is the slope of a horizontal line?

1. At \(A(0,2)\), the tangent also passes through \((1,3)\). Its slope is \(\frac{3-2}{1-0}=1\), and its y-intercept is \(2\). Thus, the tangent line is \(y=x+2\). 2. At \(B(2,3)\), the tangent is horizontal. Its slope is \(0\), so its equation is \(y=3\). 3. At \(C(4,2)\), the tangent also passes through \((5,1)\). Its slope is \(\frac{1-2}{5-4}=-1\). Using \((4,2)\), the line is \(y-2=-(x-4)\), or \(y=-x+6\).

At \(A\): \(y=x+2\) At \(B\): \(y=3\) At \(C\): \(y=-x+6\)

- Find the inputs where \(f\) has horizontal tangents. - The derivative is zero at those inputs. - Determine where \(f\) is decreasing. - The derivative must be negative on decreasing intervals.

1. The graph of \(f\) has a local maximum at \(x=-2\) and a local minimum at \(x=2\). Therefore, \(f'(-2)=0\) and \(f'(2)=0\). This eliminates graph \(C\), which has only one zero. 2. The graph of \(f\) decreases between \(-2\) and \(2\), so \(f'(x)<0\) on that interval. 3. Graph \(A\) is below the x-axis between its zeros at \(-2\) and \(2\), while graph \(B\) is above the x-axis there. Therefore, graph \(A\) represents \(f'\).

Graph \(A\) represents \(f'\).

- Locate the horizontal tangents of \(f\). - The derivative has zeros at those x-coordinates. - Determine where \(f\) is increasing and decreasing. - Eliminate candidates using both zeros and sign.

1. The graph of \(f\) has a local maximum at \(x=-1\) and a local minimum at \(x=3\). Thus, \(f'(-1)=0\) and \(f'(3)=0\). This eliminates \(C\), whose zeros are different, and \(D\), which has only one zero. 2. The function decreases between \(-1\) and \(3\), so \(f'(x)<0\) on that interval. 3. Only graph \(A\) has zeros at \(-1\) and \(3\) and is negative between them. Therefore, graph \(A\) represents \(f'\).

Graph \(A\) represents \(f'\).

- Where does the graph of \(f\) have horizontal tangents? - Relate increasing behavior to a positive derivative. - Estimate a curve's slope by imagining a tangent line and a slope triangle. - Use convenient nearby points for the estimate.

1. The graph of \(f\) has a local minimum at \(x=-1\) and a local maximum at \(x=1\). Therefore, the zeros of \(f'\) are \(x=-1\) and \(x=1\). 2. The graph of \(f\) increases between the two extrema, so \(f'(x)>0\) for \(-1<x<1\). 3. At \(x=0\), a reasonable tangent estimate uses \((0,1)\) and approximately \((1,2.5)\), giving \(f'(0)\approx\frac{2.5-1}{1-0}=1.5\). Estimates from about \(1.3\) to \(1.7\) are reasonable from the graph.

a) \(x=-1\) and \(x=1\) b) \((-1,1)\) c) \(f'(0)\approx1.5\)

- Choose two exact grid points on each tangent line. - Use the slope formula with the two points. - The slope and angle of inclination satisfy \(m=\tan(\alpha)\). - Use the inverse tangent function on a calculator to find each angle.

1. The line \(t_1\) passes through \(A(0,-1)\) and \((1,0)\). Its slope is \(m_1=\frac{0-(-1)}{1-0}=1\), so \(t_1\) is \(y=x-1\). 2. The line \(t_2\) passes through \(B(2,3)\) and \((1,0)\). Its slope is \(m_2=\frac{3-0}{2-1}=3\), so \(t_2\) is \(y-3=3(x-2)\), or \(y=3x-3\). 3. Since \(m=\tan(\alpha)\), \(\alpha_1=\arctan(1)=45.0^\circ\). 4. Similarly, \(\alpha_2=\arctan(3)\approx71.6^\circ\).

\(t_1: y=x-1\), with \(\alpha_1=45.0^\circ\) \(t_2: y=3x-3\), with \(\alpha_2\approx71.6^\circ\)

- Relate monotonicity of \(f\) to the sign of \(f'\). - Identify where the monotonicity changes. - The derivative should have zeros at those transition inputs. - Check whether each candidate is above or below the x-axis on each interval.

1. Since the monotonicity changes at \(x=-1\) and \(x=3\), the derivative must be zero there and change sign at each point. 2. Graphs I and II have zeros at \(-1\) and \(3\). Graph III has different zeros, so it cannot represent \(f'\). 3. Because \(f\) increases for \(x<-1\) and \(x>3\), the derivative must be positive in those regions. Because \(f\) decreases for \(-1<x<3\), the derivative must be negative there. 4. Only graph I has this sign pattern. Therefore, graph I could represent \(f'\).

Graph I could represent \(f'\).

- Look for points where the graph has horizontal tangent lines. - The derivative at a point is the slope of the tangent line there. - Use two points on line \(t\) to calculate its slope. - The derivative is negative where the function is decreasing.

1. The zeros of \(f^{\prime}\) occur where the graph of \(f\) has horizontal tangent lines. The graph has a local maximum at \(x=-3\) and a local minimum at \(x=3\), so the zeros are \(x=-3\) and \(x=3\). 2. The value \(f^{\prime}(0)\) is the slope of tangent line \(t\). The line passes through \((0,0)\) and \((2,-3)\), so its slope is \(\frac{-3-0}{2-0}=-1.5\). Therefore, \(f^{\prime}(0)=-1.5\). 3. The graph of \(f\) decreases between the local maximum and local minimum. Therefore, \(f^{\prime}(x)<0\) for \(x\in(-3,3)\).

a) \(x=-3\) and \(x=3\) b) \(f^{\prime}(0)=-1.5\) c) \(x\in(-3,3)\)

- The derivative value gives the tangent slope. - Equal function and derivative values appear as intersections of the two graphs.

1. The tangent slope at \(x = 0\) is \(f'(0) = 0 - 1 = -1\). 2. The function value and derivative value are equal where the two graphs intersect. The intersections occur at \((1, 0)\) and \((3, 2)\), so the required \(x\)-values are \(1\) and \(3\).

a) \(-1\) b) \(x = 1\) and \(x = 3\)

- The derivative at an input is the tangent slope there. - A smooth local extremum has a horizontal tangent line. - Use increasing and decreasing behavior to determine the sign of the derivative. - Use concavity to determine whether derivative values are increasing or decreasing.

1. Statement (A) is true. The graph has a local maximum at \(x=0\), so its tangent line is horizontal and \(f^{\prime}(0)=0\). 2. Statement (B) is true. A tangent-line estimate at \(x=2\) gives a slope of about \(-1\), so \(f^{\prime}(2)\approx-1\). 3. Statement (C) is true. The graph is concave down throughout the displayed interval, so its tangent slopes continually decrease. Therefore, \(f^{\prime}\) is strictly decreasing. 4. Statement (D) is false. For \(x<0\), the graph of \(f\) is increasing, so \(f^{\prime}(x)>0\) and the derivative graph lies above the x-axis.

(A) True (B) True (C) True (D) False

- The derivative is zero where the original graph has a horizontal tangent line. - Determine where \(f\) is increasing and decreasing. - The derivative of a quadratic function is linear.

1. The derivative is zero where the graph of \(f\) has a horizontal tangent line. The vertex occurs at \(x=0\), so \(f^{\prime}(0)=0\). 2. The graph of \(f\) is a downward-opening parabola. It increases for \(x<0\), so \(f^{\prime}(x)>0\), and decreases for \(x>0\), so \(f^{\prime}(x)<0\). From the vertex \((0,2)\) and the point \((2,0)\), write \(f(x)=ax^2+2\) and solve \(0=4a+2\), giving \(a=-0.5\). Thus, \(f(x)=-0.5x^2+2\) and \(f^{\prime}(x)=-x\). The derivative graph is a decreasing line through the origin, crossing the x-axis from positive values to negative values.

1) The zero of \(f^{\prime}\) is \(x=0\). 2) The graph is the line \(y=-x\). The derivative is positive for \(x<0\) and negative for \(x>0\).

- At the vertex of a smooth parabola, the tangent line is horizontal. - The derivative is positive where the function is increasing.

1. The graph of \(f\) has its minimum at \(x=2\). The tangent line is horizontal there, so \(f^{\prime}(2)=0\). Thus, the derivative graph crosses the x-axis at \(x=2\). 2. For \(x>2\), the graph of \(f\) is increasing. Therefore, its tangent slopes are positive and \(f^{\prime}(x)>0\).

1) \(x=2\) 2) \(f^{\prime}(x)>0\) for \(x>2\).

- Locate the local maximum and local minimum. - The derivative is zero where the tangent line is horizontal. - The derivative is negative where the function is decreasing.

1. The derivative is zero at the local extrema of \(f\). The graph has a local maximum at \(x=-2\) and a local minimum at \(x=2\). Therefore, the derivative graph crosses the x-axis at \(x=-2\) and \(x=2\). 2. The graph of \(f\) is decreasing between the local maximum and local minimum. Therefore, \(f^{\prime}(x)<0\) for \(-2<x<2\).

1) \(x=-2\) and \(x=2\) 2) \(f^{\prime}(x)<0\) on \((-2,2)\).

- Use the tangent lines' intercepts or other exact grid points to find their slopes and y-intercepts. - For a shallow line, choose grid points that are farther apart to reduce reading error.

1. At \(A(1,4)\), the tangent passes through \((0,8)\) and \((2,0)\). Its slope is \(\frac{0-8}{2-0}=-4\), and its y-intercept is \(8\). Thus, the line is \(y=-4x+8\). 2. At \(B(2,2)\), the tangent passes through \((0,4)\) and \((4,0)\). Its slope is \(\frac{0-4}{4-0}=-1\), and its y-intercept is \(4\). Thus, the line is \(y=-x+4\). 3. At \(C(4,1)\), the tangent passes through \((0,2)\) and \((8,0)\). Its slope is \(\frac{0-2}{8-0}=-0.25\), and its y-intercept is \(2\). Thus, the line is \(y=-0.25x+2\).

At \(A\): \(y=-4x+8\) At \(B\): \(y=-x+4\) At \(C\): \(y=-0.25x+2\)

- A horizontal tangent has an equation of the form \(y=\text{constant}\). - At \(B\), use grid points to estimate how quickly the tangent falls.

1. At \(A(-2,4)\), the graph has a local maximum, so the tangent is horizontal. Its equation is \(y=4\). 2. At \(B(0,0)\), the tangent passes through the origin and falls \(3\) units for each \(1\)-unit increase in \(x\). Its slope is \(-3\), so its equation is \(y=-3x\). 3. At \(C(2,-4)\), the graph has a local minimum, so the tangent is horizontal. Its equation is \(y=-4\).

At \(A\): \(y=4\) At \(B\): \(y=-3x\) At \(C\): \(y=-4\)

- Find the horizontal tangent line on the parabola. - Estimate tangent slopes at one or two additional inputs. - The derivative of a quadratic function is linear. - Use the increasing and decreasing intervals to check the sign of the derivative graph.

1. The graph of \(f\) has a minimum at \(x=2\), so \(f^{\prime}(2)=0\). 2. From the graph, the tangent slope is about \(-2\) at \(x=0\) and about \(2\) at \(x=4\). 3. Since \(f\) is quadratic, \(f^{\prime}\) is linear. The line through \((0,-2)\), \((2,0)\), and \((4,2)\) is \(f^{\prime}(x)=x-2\).

The derivative graph is the line \(f^{\prime}(x)=x-2\). It crosses the x-axis at \(x=2\) and passes through \((0,-2)\).

- What value does a derivative have where the original function has a horizontal tangent? - Compare the minimum of one graph with the x-intercept of the other. - The derivative graph gives the tangent slope at each input. - A tangent equation requires both a point on \(f\) and the derivative value there.

1. The red graph represents \(f\), and the blue graph represents \(f'\). 2. The red graph has a minimum at \(x=0\), where its tangent is horizontal. The blue graph has value \(0\) at \(x=0\), as a derivative must at that critical point. 3. From the red graph, \(f(2)=0\), so the point of tangency is \((2,0)\). From the blue graph, \(f'(2)=1\), so the tangent slope is \(1\). 4. The line through \((2,0)\) with slope \(1\) is \(y=x-2\).

a) The red graph is \(f\), and the blue graph is \(f'\). The derivative is zero at the x-coordinate of the minimum of \(f\). b) \(y=x-2\)

- Compare the x-intercepts of one graph with the maxima and minima of the other. - The derivative is zero at a smooth local maximum or minimum. - Read both \(f(0)\) and \(f'(0)\) from the graphs. - Use the point and slope to write the tangent line.

1. The green graph represents \(f\), and the orange graph represents \(f'\). 2. At the maxima and minimum of the green graph, the orange graph crosses the x-axis. This matches the fact that \(f'(x)=0\) at critical points of \(f\). 3. At \(x=0\), the green graph gives \(f(0)=0\), and the orange graph gives \(f'(0)=2\). Thus, the tangent line passes through \((0,0)\) with slope \(2\). 4. The tangent line is \(y=2x\).

a) The green graph is \(f\), and the orange graph is \(f'\). The zeros of the orange graph occur at the extrema of the green graph. b) \(y=2x\)

- Determine the sign of every tangent slope on the graph of \(f\). - Check whether the graph ever has a horizontal tangent line. - Observe how the steepness changes from left to right.

1. The exponential function \(f(x)=1.25^x\) is positive and strictly increasing for every real \(x\). Therefore, \(f^{\prime}(x)>0\) for all \(x\), so the derivative has no zeros. 2. The graph of \(f\) becomes steeper from left to right. Therefore, \(f^{\prime}\) is also strictly increasing. 3. As \(x\to-\infty\), the graph of \(f\) becomes nearly horizontal, so \(f^{\prime}(x)\to0\) from above. The derivative graph is qualitatively another increasing exponential curve above the x-axis.

The graph of \(f^{\prime}\) is an increasing exponential curve entirely above the x-axis. It approaches \(0\) from above as \(x\to-\infty\) and has no zeros.

- Determine whether each branch is increasing or decreasing. - A decreasing function has negative tangent slopes. - Remember that \(x=0\) is not in the domain.

1. For \(x<0\), the left branch of \(f\) is strictly decreasing. Therefore, every tangent slope is negative and \(f^{\prime}(x)<0\). 2. For \(x>0\), the right branch is also strictly decreasing. Therefore, every tangent slope is negative and \(f^{\prime}(x)<0\). 3. Thus, the derivative graph lies below the x-axis at every input in the domain, \(x\ne0\).

For both \(x<0\) and \(x>0\), \(f^{\prime}(x)<0\).

- The derivative describes the tangent slope of the original function. - Look for local maxima and minima. - A smooth local extremum has a horizontal tangent line.

1. Zeros of \(f^{\prime}\) occur where the graph of \(f\) has horizontal tangent lines. 2. The graph has a local maximum at \(x=-2\) and a local minimum at \(x=2\). 3. Therefore, \(f^{\prime}(-2)=0\) and \(f^{\prime}(2)=0\).

The zeros of \(f^{\prime}\) are \(x=-2\) and \(x=2\).

- Calculate each line slope as change in \(y\) divided by change in \(x\). - The tangent line at the vertex of a smooth parabola is horizontal. - Compare a negative slope with a positive slope.

1. The tangent line at \(x=0\) passes through \((0,-1)\) and \((1,-2)\). Its slope is \(\frac{-2-(-1)}{1-0}=-1\), so \(g^{\prime}(0)=-1\). 2. The tangent line at \(x=4\) passes through \((4,-1)\) and \((5,0)\). Its slope is \(\frac{0-(-1)}{5-4}=1\), so \(g^{\prime}(4)=1\). 3. At the vertex \(S(2,-2)\), the tangent line is horizontal. Thus, \(g^{\prime}(2)=0\). Point \(S\) is the vertex and a local minimum. 4. At \(x=1\), the graph is decreasing, so the slope is negative. At \(x=3\), it is increasing, so the slope is positive. Therefore, \(g^{\prime}(3)>g^{\prime}(1)\).

a) \(g^{\prime}(0)=-1\) and \(g^{\prime}(4)=1\) b) \(g^{\prime}(2)=0\); \(S\) is the vertex and a local minimum. c) The slope at \(x=3\) is greater.

- Determine where the original function is increasing and decreasing. - The derivative is zero where the original graph has a horizontal tangent line. - Check the sign of each candidate on both sides of the zero.

1. The graph of \(f\) is an upward-opening parabola with a local minimum at \(x=0\). 2. Therefore, \(f^{\prime}(0)=0\). Both candidate graphs have a zero at \(x=0\). 3. For \(x<0\), the graph of \(f\) is decreasing, so \(f^{\prime}(x)<0\). For \(x>0\), it is increasing, so \(f^{\prime}(x)>0\). 4. Graph 1 has this sign pattern, while graph 2 has the opposite sign pattern. Therefore, graph 1 represents \(f^{\prime}\).

Graph 1 represents \(f^{\prime}\).

- The tangent line at the highest point of a smooth parabola is horizontal. - Use symmetry to compare slopes at opposite inputs. - The derivative of a quadratic function is linear.

1. At \(x=0\), the graph has its maximum and the tangent line is horizontal. Therefore, \(f^{\prime}(0)=0\). 2. Because \(f\) is a downward-opening parabola, its tangent slopes decrease steadily as \(x\) increases. By symmetry, the slopes are positive to the left of the vertex and negative to the right. 3. Therefore, the graph of \(f^{\prime}\) is a decreasing line through \((0,0)\). Its vertical scale can be estimated from the slopes of \(f\). 4. The x-coordinate of the vertex of \(f\) is the zero of \(f^{\prime}\).

The slope at \(x=0\) is \(0\). The derivative graph is a decreasing line through \((0,0)\). The vertex of \(f\) occurs at the same input as the zero of \(f^{\prime}\).

- Use rise over run to find each tangent-line slope. - The y-coordinate on the derivative graph equals the tangent slope at that input. - Plot the three derivative points mentally and identify their pattern. - Recall the degree of the derivative of a quadratic polynomial.

1. At \(x=-2\), tangent line \(t_1\) falls \(2\) units for each unit to the right, so \(m_1=-2\). At \(x=0\), tangent line \(t_2\) is horizontal, so \(m_2=0\). At \(x=2\), tangent line \(t_3\) rises \(2\) units for each unit to the right, so \(m_3=2\). 2. The corresponding derivative points are \(Q_1(-2,-2)\), \(Q_2(0,0)\), and \(Q_3(2,2)\). 3. These three points lie on a line through the origin. Also, the derivative of a quadratic function is linear. Therefore, \(f^{\prime}\) is a linear function.

a) \(m_1=-2\), \(m_2=0\), and \(m_3=2\) b) \(Q_1(-2,-2)\), \(Q_2(0,0)\), and \(Q_3(2,2)\) c) \(f^{\prime}\) is linear.

- Choose two exact grid points on the tangent line. - Use the slope formula. - The slope and angle of inclination satisfy \(m=\tan(\alpha)\). - Measure the angle from the positive x-axis.

1. The tangent passes through \(A(2,0)\) and \((3,2)\), so its slope is \(m=\frac{2-0}{3-2}=2\). 2. Using \(A\), \(0=2(2)+b\), so \(b=-4\). Therefore, \(t\) is \(y=2x-4\). 3. Since \(\tan(\alpha)=2\), \(\alpha=\arctan(2)\approx63.43^\circ\).

1. \(t: y=2x-4\) 2. \(\alpha\approx63.43^\circ\)

- Choose two exact points on the orange line. - Relate slope to the angle of inclination. - A line with negative slope has an inclination angle between \(90^\circ\) and \(180^\circ\).

1. The tangent passes through \(P(1,0)\) and \((3,-1)\), so \(m=\frac{-1-0}{3-1}=-0.5\). 2. Using \(P\), \(0=-0.5(1)+b\), so \(b=0.5\). Therefore, \(t\) is \(y=-0.5x+0.5\). 3. Since \(\arctan(-0.5)\approx-26.57^\circ\), add \(180^\circ\) to obtain the angle of inclination: \(\alpha\approx153.43^\circ\).

1. \(t: y=-0.5x+0.5\) 2. \(\alpha\approx153.43^\circ\)

- Use the origin as one of the two points for calculating the slope. - Read the horizontal and vertical changes to another grid point. - A negative slope corresponds to an inclination angle between \(90^\circ\) and \(180^\circ\).

1. The tangent passes through \(W(0,0)\) and the marked point \((2,-1.5)\), so its slope is \(m=\frac{-1.5-0}{2-0}=-0.75\). 2. Because the line passes through the origin, its equation is \(y=-0.75x\). 3. Since \(\arctan(-0.75)\approx-36.87^\circ\), its angle of inclination is \(\alpha\approx143.13^\circ\).

1. \(t: y=-0.75x\) 2. \(\alpha\approx143.13^\circ\)

- Use vertical position to determine the sign of the function value. - Use increasing, decreasing, or horizontal behavior to determine the sign of the first derivative. - Use concavity to determine the sign of the second derivative. - Examine the tangent line at each marked point.

1. At \(A(1,0)\), the point lies on the x-axis, so \(f(1)=0\). The graph is decreasing, so \(f^{\prime}(1)<0\). The graph is concave up, so \(f^{\prime\prime}(1)>0\). 2. At \(B(2,-0.5)\), the point lies below the x-axis, so \(f(2)<0\). The tangent line is horizontal at the vertex, so \(f^{\prime}(2)=0\). The graph is concave up, so \(f^{\prime\prime}(2)>0\). 3. At \(C(3,0)\), the point lies on the x-axis, so \(f(3)=0\). The graph is increasing, so \(f^{\prime}(3)>0\). The graph is concave up, so \(f^{\prime\prime}(3)>0\).

The completed table is: <table> <tr><th>Point</th><th>\(f(x)\)</th><th>\(f^{\prime}(x)\)</th><th>\(f^{\prime\prime}(x)\)</th></tr> <tr><td>\(A\)</td><td>\(0\)</td><td>\(-\)</td><td>\(+\)</td></tr> <tr><td>\(B\)</td><td>\(-\)</td><td>\(0\)</td><td>\(+\)</td></tr> <tr><td>\(C\)</td><td>\(0\)</td><td>\(+\)</td><td>\(+\)</td></tr> </table>

- An increasing graph has a positive first derivative. - A decreasing graph has a negative first derivative. - Concave down means the second derivative is negative. - Locate the point with a horizontal tangent line.

1. At point \(A\), the graph is increasing, so \(f^{\prime}(x)>0\), and it is concave down, so \(f^{\prime\prime}(x)<0\). Thus, \(A\) matches the first description. 2. Point \(B\) is the local maximum. The tangent line is horizontal, so \(f^{\prime}(x)=0\). Thus, \(B\) matches the second description. 3. At point \(C\), the graph is decreasing, so \(f^{\prime}(x)<0\), and it remains concave down, so \(f^{\prime\prime}(x)<0\). Thus, \(C\) matches the third description.

a) \(A\); b) \(B\); c) \(C\)

- Use increasing or decreasing behavior to determine the sign of the first derivative. - Use concavity to determine the sign of the second derivative.

1. At point \(A\), the graph is increasing, so \(f^{\prime}(x)>0\). The graph is concave down, so \(f^{\prime\prime}(x)<0\). 2. At point \(B\), the graph is decreasing, so \(f^{\prime}(x)<0\). The graph is concave up, so \(f^{\prime\prime}(x)>0\).

Point \(A\): \(f^{\prime}(x)>0\) and \(f^{\prime\prime}(x)<0\) Point \(B\): \(f^{\prime}(x)<0\) and \(f^{\prime\prime}(x)>0\)

- Determine the tangent slope at a horizontal inflection point. - Check whether the original function increases or decreases on each side. - A zero without a sign change appears as a touch rather than a crossing.

1. At a horizontal inflection point, the tangent line is horizontal, so \(h^{\prime}(1)=0\). 2. The function \(h\) is increasing on both sides of \(x=1\). Therefore, \(h^{\prime}(x)>0\) for \(x\ne1\). 3. The derivative graph touches the x-axis at \(x=1\) without crossing it. Because the original graph has point symmetry about \((1,1)\), the derivative graph is symmetric about the vertical line \(x=1\). 4. Thus, the derivative graph is a smooth curve symmetric about \(x=1\), with a minimum at \((1,0)\) and values above the x-axis elsewhere.

The derivative graph is symmetric about \(x=1\), touches the x-axis at \((1,0)\), and remains positive on both sides, so the zero has no sign change.

- A graph approaching a horizontal line becomes nearly flat. - Determine whether the original function is increasing or decreasing. - Nearly horizontal tangent lines have slopes near zero.

1. The graph of \(f\) is strictly increasing for \(x>0\), so \(f^{\prime}(x)>0\) throughout that interval. 2. As the graph approaches the horizontal asymptote, it becomes flatter and its tangent slopes decrease. 3. Therefore, \(f^{\prime}(x)\to0\) as \(x\to\infty\). The derivative graph stays above the x-axis and approaches it from above.

For \(x>0\), \(f^{\prime}(x)>0\), and \(f^{\prime}(x)\to0\) as \(x\to\infty\).

- Zeros of the first derivative occur at horizontal tangent lines. - Zeros of the second derivative can occur at inflection points. - Analyze each continuous branch separately.

1. On each interval of its domain, the graph is strictly decreasing and has no horizontal tangent lines. Therefore, \(f^{\prime}\) has \(0\) zeros. 2. Each branch keeps the same concavity and has no inflection point. The change across the vertical asymptote does not count because the function is undefined there. Therefore, \(f^{\prime\prime}\) has \(0\) zeros.

\(f^{\prime}\) has \(0\) zeros, and \(f^{\prime\prime}\) has \(0\) zeros.

- The derivative graph gives the tangent slope of the original function. - Read the derivative value at each specified x-value. - The two functions have the same value where their graphs intersect.

1. The slope of the tangent line at \(x\) equals \(f^{\prime}(x)\). 2. From the derivative graph, \(f^{\prime}(1)=0.5\) and \(f^{\prime}(4)=0.25\). 3. The function and derivative have the same value at an intersection of their graphs. The intersection occurs at \(x=0.5\). 4. Algebraically, \(\sqrt{0.5}=\frac{1}{2\sqrt{0.5}}\approx0.707\), confirming the intersection.

a) At \(x=1\), the slope is \(0.5\); at \(x=4\), the slope is \(0.25\). b) \(x=0.5\)

- Choose two points on the tangent line. - The tangent slope equals the derivative value at the point of tangency. - Use vertical change divided by horizontal change.

1. The derivative value \(f'(0)\) equals the slope of the tangent line at \(P\). 2. The tangent passes through \(P(0,2)\) and the grid point \((4,3)\). 3. Its slope is \(m=\frac{3-2}{4-0}=\frac{1}{4}=0.25\). 4. Therefore, \(f'(0)=0.25\).

The tangent slope is \(0.25\), so \(f'(0)=0.25\).

- Identify the local maximum and local minimum. - What is the slope of the tangent line at a local extremum? - Determine whether the graph is increasing or decreasing at \(x=0\).

1. The graph has a local maximum at \(x=-2\) and a local minimum at \(x=2\). 2. The tangent line is horizontal at each local extremum, so \(f^{\prime}(-2)=0\) and \(f^{\prime}(2)=0\). 3. Near \(x=0\), the graph is decreasing. Therefore, its tangent slope is negative and \(f^{\prime}(0)<0\).

a) \(f^{\prime}(x)=0\) at \(x=-2\) and \(x=2\), where the graph has horizontal tangent lines. b) \(f^{\prime}(0)\) is negative because the graph is decreasing at \(x=0\).

- Look for the number the listed average velocities are approaching. - Think about what happens to the time interval as its right endpoint moves closer to \(4\). - Connect the limiting average velocity to the idea of velocity at a single instant.

1. The average velocities \(11.5\), \(11.1\), and \(11.01\) approach \(11\) as the right endpoint moves toward \(4\). 2. The instantaneous velocity is modeled by the limiting value of these average velocities. 3. Therefore the velocity at \(t=4\) is \(11\,\text{ft/s}\).

\(11\,\text{ft/s}\). It is the value approached by the average velocity over intervals whose endpoints get closer to \(t=4\).

- Compare the left and right entries in the row closest to \(t=2\). - Look for a common value between the two sequences. - An instantaneous rate is supported when shrinking-interval averages from both sides agree.

1. The left averages increase toward \(6.2^\circ/\text{s}\). 2. The right averages decrease toward \(6.2^\circ/\text{s}\). 3. Because the average rates from both sides approach the same value, the instantaneous angular velocity is \(6.2^\circ/\text{s}\).

\(6.2^\circ/\text{s}\). The left- and right-side average angular velocities converge to the same value.

- Compare the values for small positive and negative \(h\). - Look for a number between \(11.41\) and \(12.61\) that both sides appear to approach. - The quotient represents an average rate over an interval whose width is \(|h|\).

1. For negative \(h\) close to \(0\), \(Q(h)\) rises toward \(12\). 2. For positive \(h\) close to \(0\), \(Q(h)\) falls toward \(12\). 3. Therefore the instantaneous rate of change at \(x=2\) is \(12\).

\(12\).

- Use both sides of the instant, not just one measurement. - Ask which number the two nearby averages appear to bracket. - Distinguish a finite-interval average from the limiting instantaneous estimate.

1. The two closest averages lie nearly symmetrically around \(8.000\,\text{L/min}\). 2. Their behavior supports a common limiting value of \(8.000\,\text{L/min}\). 3. The value \(8.002\,\text{L/min}\) is one nearby average rate, not the common value suggested by both sides.

Yes. \(8.000\,\text{L/min}\) is consistent with the two-sided data; \(8.002\,\text{L/min}\) is only one finite-interval average.

- Check whether each set settles toward a finite value. - When two sides are listed, compare their limiting behavior. - Eliminate patterns that disagree, oscillate, or grow without bound.

1. In A, the left-side rates approach \(2\) from below and the right-side rates approach \(2\) from above. 2. In B, the two sides approach different values. 3. C oscillates rather than settling, and D grows without bound. 4. Therefore A gives the strongest evidence for an instantaneous rate of \(2\).

A. Both sides approach \(2\).

- The derivative at a point is the slope of the tangent line there. - Choose two well-separated x-values on the tangent and estimate the corresponding y-values from the grid. - Use rise divided by run; small differences in your visual readings are expected.

1. The derivative value \(f'(1)\) is the slope of the tangent line at \(P\). 2. From the grid, the tangent is approximately at \((0,0.3)\) and \((2,3.7)\). These readings are estimates because the line does not pass through exact grid intersections there. 3. The estimated slope is \(\frac{3.7-0.3}{2-0}\approx1.7\). Therefore, \(f'(1)\approx1.7\).

\(f'(1)\approx1.7\)

- Substitute each base into the given difference quotient. - Keep enough calculator precision before rounding. - Isolate \(b^h\) before taking the \(h\)th root. - Rewrite an \(h\)th root as a power of \(1/h\).

1. For \(b=2\), \(\frac{2^{0.0001}-1}{0.0001}\approx 0.6932\). For \(b=4\), \(\frac{4^{0.0001}-1}{0.0001}\approx 1.3864\). 2. Solve the equation algebraically: \(\frac{b^h-1}{h}=1\) implies \(b^h=1+h\). Therefore, \(b=(1+h)^{1/h}\). With \(h=0.0001\), \(b=(1.0001)^{10000}\approx 2.7181\).

a) For \(b=2\), \(f'(0)\approx 0.6932\). For \(b=4\), \(f'(0)\approx 1.3864\). b) \(b\approx 2.7181\)

- Locate the local maximum and minimum of \(f\). - What is the derivative value at a smooth extremum? - Determine where \(f\) is increasing and decreasing. - Match those intervals to the sign of each candidate graph.

1. The graph of \(f\) has a local maximum at \(x=-1\) and a local minimum at \(x=1\). Therefore, \(f'(-1)=0\) and \(f'(1)=0\). This eliminates graph \(C\), which has only one zero. 2. The function increases for \(x<-1\) and \(x>1\), so \(f'(x)>0\) there. It decreases for \(-1<x<1\), so \(f'(x)<0\) there. 3. Only graph \(A\) has zeros at \(-1\) and \(1\), is negative between them, and is positive outside them. Therefore, graph \(A\) represents \(f'\).

Graph \(A\) represents \(f'\).

- Identify where \(f\) is increasing and decreasing. - Relate the sign of \(f'\) to the direction of change of \(f\). - At a smooth extremum, the tangent slope is zero. - Where does \(f\) have its most negative slope?

1. Statement 1 is true. The function \(f\) is decreasing throughout \([-1,1]\), so \(f'(x)<0\) there. 2. Statement 2 is true. The graph of \(f\) has a local maximum at \(x=-2\), so its tangent is horizontal and \(f'(-2)=0\). 3. Statement 3 is false. At \(x=0\), the graph of \(f\) has its steepest negative slope. Therefore, \(f'\) has a local minimum there, not a local maximum. 4. Statement 4 is false. To the right of the local minimum at \(x=2\), the graph of \(f\) is increasing, so \(f'(3)>0\).

1) True 2) True 3) False 4) False

- The derivative gives the tangent slope at each input. - At a smooth local maximum or minimum, the tangent is horizontal. - Use a slope triangle on the red tangent line. - Identify the interval where the graph moves downward from left to right.

1. The graph has a local maximum at \(x=-2\) and a local minimum at \(x=2\). The tangents are horizontal there, so \(f'(-2)=0\) and \(f'(2)=0\). 2. The tangent \(t\) passes through \((0,0)\) and \((1,-3)\). Its slope is \(\frac{-3-0}{1-0}=-3\), so \(f'(0)=-3\). 3. The graph decreases from the local maximum to the local minimum. Thus, the largest interval is \([-2,2]\).

a) \(x=-2\) and \(x=2\) b) \(f'(0)=-3\) c) \([-2,2]\)

- Compare the coordinates of the two vertices. - Use two readable points on \(t\) to calculate its slope. - A translation does not change the slope at corresponding points. - Use the translated point and the slope in point-slope form.

1. The vertices are \(S_f(1,0.5)\) and \(S_g(4,2.5)\). Therefore, the graph is translated \(3\) units right and \(2\) units up, so \(g(x)=f(x-3)+2\). 2. The line \(t\) passes through \((1,0)\) and \((2,1)\), so its slope is \(\frac{1-0}{2-1}=1\). Therefore, \(f'(2)=1\). 3. A translation preserves slopes at corresponding points. Since the graph moved \(3\) units right, \(x=5\) on \(g\) corresponds to \(x=2\) on \(f\). Thus, \(g'(5)=f'(2)=1\). 4. The translated point is \((5,3)\). A line of slope \(1\) through that point is \(y-3=x-5\), or \(y=x-2\).

a) The graph is translated \(3\) units right and \(2\) units up, so \(g(x)=f(x-3)+2\). b) The slope is \(1\), so \(f'(2)=1\). c) The slope is \(1\), and the tangent line is \(y=x-2\).

- The derivative value equals the tangent slope. - Use two points on \(t\) to calculate its slope. - The derivative is zero at smooth local extrema. - Match the sign of each candidate to the increasing and decreasing intervals of \(f\).

1. The tangent passes through \(P(2,2)\) and \((4,0)\). Its slope is \(\frac{0-2}{4-2}=-1\), so \(f'(2)=-1\). 2. All three candidates have value \(-1\) at \(x=2\), so that value alone does not determine the answer. 3. The graph of \(f\) has a local maximum near \(x=0.85\) and a local minimum near \(x=3.15\). Therefore, \(f'\) must have zeros near those inputs. 4. Only \(g_1\) has zeros near \(0.85\) and \(3.15\). It is also positive where \(f\) increases and negative where \(f\) decreases. Thus, \(g_1\) represents \(f'\).

a) The slope is \(-1\), so \(f'(2)=-1\). b) \(g_1\) represents \(f'\).

- Zeros of the derivative occur at smooth extrema of the original function. - Use increasing and decreasing behavior to determine the derivative's sign. - Look for where the graph changes concavity. - Consider the slope of the graph far to the left.

1. Statement 1 is true. The graph of \(f\) has a local maximum at \(x=-2\) and a local minimum at \(x=2\), so \(f'\) has zeros at both inputs. 2. Statement 2 is false. The graph of \(f\) decreases between the extrema, so \(f'(x)<0\) for \(-2<x<2\), with zeros at the endpoints. 3. Statement 3 is true. The change in concavity occurs midway between the symmetric extrema, at \(x=0\), and the graph gives \(f(0)=1\). 4. Statement 4 is false. Far to the left, the graph is increasing ever more steeply, so \(f'(x)\to+\infty\) as \(x\to-\infty\).

1) True 2) False 3) True 4) False

- Find the x-coordinates of the local maximum and minimum. - The derivative must be zero at those inputs. - Use the increasing and decreasing intervals to determine the derivative's sign pattern. - Estimate one slope to distinguish candidates with the same zeros.

1. The graph of \(f\) has a local maximum at \(x=-1\) and a local minimum at \(x=3\). Therefore, \(f'\) must have zeros at \(-1\) and \(3\). This eliminates option D. 2. The function increases for \(x<-1\) and \(x>3\), so \(f'\) must be positive there. It decreases between the extrema, so \(f'\) must be negative there. Thus, its graph opens upward, eliminating option B. 3. Options A and C have the correct zeros and sign pattern. Estimate the slope of \(f\) near the inflection point \(x=1\). The graph suggests a slope near \(-2\). 4. Option A gives \(f'(1)=0.5-1-1.5=-2\), while option C gives \(f'(1)=1-2-3=-4\). Therefore, option A matches the graph.

A: \(f'(x)=0.5x^2-x-1.5\)

- The derivative is the graph's tangent slope. - Relate increasing and decreasing behavior to the sign of the derivative. - At a smooth extremum, the derivative is zero. - Do not confuse a function value with a derivative value. - Use symmetry about the inflection point.

1. Statement 1 is true. The graph of \(f\) increases throughout \((0,4)\) and has no horizontal tangent there, so \(f'(x)>0\) on that interval. 2. Statement 2 is false. At \(x=2\), the graph has an inflection point with positive slope; the slope is approximately \(3\), not \(0\). 3. Statement 3 is true. The graph is point-symmetric about its inflection point \((2,2)\), so the tangent slopes at inputs equally spaced from \(2\), such as \(1\) and \(3\), are equal. 4. Statement 4 is false. At \(x=0\), the graph has a local minimum, so \(f'(0)=0\). The value \(-2\) is \(f(0)\), not the derivative value.

1) True 2) False 3) True 4) False

- Zeros of the derivative correspond to horizontal tangents on the function graph. - Use two readable points on tangent line \(t\) to calculate its slope. - The derivative is positive where the function graph is increasing.

1. The zeros of \(f'\) occur where the graph of \(f\) has horizontal tangent lines. The graph has a local maximum at \(x=-2\) and a local minimum at \(x=2\), so the zeros are \(x=-2\) and \(x=2\). 2. The tangent line passes through \((0,0)\) and \((1,-3)\). Its slope is \(\frac{-3-0}{1-0}=-3\), so \(f'(0)=-3\). 3. On the interior of the displayed interval, the graph of \(f\) is increasing for \(-5<x<-2\) and for \(2<x<5\). Therefore, \(f'(x)>0\) on \((-5,-2)\) and \((2,5)\).

a) \(x=-2\) and \(x=2\) b) \(f'(0)=-3\) c) \((-5,-2)\) and \((2,5)\)

- Zeros of the first derivative occur where the graph has horizontal tangent lines. - The sign of the first derivative indicates whether the function is increasing or decreasing. - Zeros of the second derivative can occur where the graph changes concavity. - Recall how the derivative of an even function behaves.

1. Statement (1) is true. The graph of \(f\) has horizontal tangent lines at its local maxima \(x=-2\) and \(x=2\) and at its local minimum \(x=0\). Thus, \(f^{\prime}\) has exactly three zeros. 2. Statement (2) is false. The function is decreasing for \(x>2\), so \(f^{\prime}(x)<0\) there. 3. Statement (3) is true. The graph changes concavity twice, near \(x=-1.15\) and \(x=1.15\), so \(f^{\prime\prime}\) has exactly two zeros on the interval. 4. Statement (4) is false. The graph of \(f\) is symmetric about the y-axis, so \(f\) is even and \(f^{\prime}\) is odd. Therefore, the graph of \(f^{\prime}\) is symmetric about the origin. 5. Statement (5) is true. The graph of \(f\) is increasing at \(x=1\), so \(f^{\prime}(1)>0\).

(1) True (2) False (3) True (4) False (5) True

- At a smooth local maximum or minimum, the tangent line is horizontal. - Use a rise-over-run calculation on the tangent line at the origin. - Concavity describes whether tangent slopes are decreasing or increasing. - Relate concavity to the sign of the second derivative.

1. The graph has a local maximum at \((-2,4)\) and a local minimum at \((2,-4)\). The tangent line is horizontal at each point, so \(f^{\prime}(-2)=0\) and \(f^{\prime}(2)=0\). 2. The tangent line at the origin falls \(3\) units for every increase of \(1\) unit in \(x\). Its slope is \(-3\), so \(f^{\prime}(0)=-3\). 3. For \(x<0\), the graph is concave down, so \(f^{\prime\prime}(x)<0\). For \(x>0\), the graph is concave up, so \(f^{\prime\prime}(x)>0\).

a) Local maximum: \((-2,4)\); local minimum: \((2,-4)\). At both inputs, \(f^{\prime}(x)=0\). b) \(f^{\prime}(0)=-3\) c) Concave down with \(f^{\prime\prime}(x)<0\) for \(x<0\); concave up with \(f^{\prime\prime}(x)>0\) for \(x>0\).

- Locate the horizontal tangent lines on the graph of \(f\). - The derivative must be zero at those inputs. - Determine where \(f\) is increasing and decreasing, then match the derivative sign. - A cubic polynomial has a quadratic derivative.

1. The graph of \(f\) has a local maximum near \(x=-0.15\) and a local minimum near \(x=2.15\). Therefore, \(f^{\prime}\) must have zeros near those inputs. Both \(g\) and \(h\) have the required zeros. 2. Between the two extrema, \(f\) is decreasing. Therefore, \(f^{\prime}\) must be negative on that interval. 3. Graph \(g\) lies below the x-axis between the two zeros, while graph \(h\) lies above it. Thus, \(g\) has the correct sign pattern. 4. Graph \(k\) has only one zero and is linear, so it cannot be the derivative of the displayed cubic function. Therefore, graph \(g\) represents \(f^{\prime}\).

The red graph \(g\) represents \(f^{\prime}\).

- Read function values from the graph of \(g\) and slopes from the graph of \(g'\). - Horizontal tangents occur where the derivative is zero. - A requested slope is a requested value of the derivative.

1. From Figure 1, \(g(2) = -3\). 2. The slope at \(x = 2\) is \(g'(2) = 0\). 3. From Figure 2, \(g'(1) = -3\). 4. Figure 1 shows \(g(0) = 1\), so \(x = 0\) is one solution. 5. Horizontal tangents occur where \(g'(x) = 0\), at \(x = -2\), \(x = 0\), and \(x = 2\). 6. Figure 2 shows \(g'(-1) = 3\), so \(x = -1\) works.

a) \(g(2) = -3\) b) \(0\) c) \(-3\) d) \(x = 0\) e) \(x = -2\), \(x = 0\), and \(x = 2\) f) \(x = -1\)

- Examine the tangent line at \((-1,-1)\). - Determine whether the function increases or decreases on each side of \(x=-1\). - A derivative can equal zero without changing sign.

1. Yes. At \(x=-1\), the graph of \(f\) has a horizontal inflection point, so its tangent line is horizontal and \(f^{\prime}(-1)=0\). 2. The function \(f\) is increasing on both sides of \(x=-1\). Thus, \(f^{\prime}(x)>0\) near \(-1\), except that \(f^{\prime}(-1)=0\). The derivative graph touches the x-axis at \(x=-1\) and does not cross it, so there is no sign change.

1) Yes, \(f^{\prime}(-1)=0\). 2) No. The derivative is positive on both sides of \(x=-1\), so its graph touches the x-axis without changing sign.

- Count the local maxima and minima of \(f\). - A horizontal tangent line gives a zero of \(f^{\prime}\). - An inflection point of \(f\) corresponds to a local extremum of \(f^{\prime}\).

1. The derivative is zero where the graph of \(f\) has horizontal tangent lines. The graph has local minima at \(x=-2\) and \(x=2\) and a local maximum at \(x=0\). Therefore, \(f^{\prime}(x)=0\) at \(x=-2\), \(x=0\), and \(x=2\). 2. The graph of \(f\) changes concavity once between \(-2\) and \(0\) and once between \(0\) and \(2\). At each inflection point of \(f\), the slope function \(f^{\prime}\) has a local extremum. Thus, \(f^{\prime}\) has at least two local extrema.

1) \(x=-2\), \(x=0\), and \(x=2\) 2) At least \(2\) local extrema

- Locate the local maximum and local minimum of \(f\). - The derivative is zero at horizontal tangent lines. - Look for the point where the graph rises most steeply.

1. The derivative is zero at the local extrema of \(f\). The graph has a local minimum at \(x=-\frac{\pi}{2}\) and a local maximum at \(x=\frac{\pi}{2}\). Therefore, the graph of \(f^{\prime}\) crosses the x-axis at those two inputs. 2. The derivative is greatest where the graph of \(f\) rises most steeply. This occurs at the origin, so \(f^{\prime}\) reaches its greatest value at \(x=0\).

1) \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{2}\) 2) \(x=0\)

- The derivative gives the tangent slope of \(f\). - Local extrema of \(f\) produce zeros of \(f^{\prime}\). - Increasing and decreasing intervals determine the sign of the derivative. - Use the symmetry of the original graph to infer the symmetry of its derivative.

1. Statement (1) is true. The graph of \(f\) is increasing at \(x=0\), so \(f^{\prime}(0)>0\). 2. Statement (2) is true. The graph has two local extrema, near \(x=-1.4\) and \(x=1.4\). The tangent line is horizontal at each point, so \(f^{\prime}\) has exactly two zeros. 3. Statement (3) is true. The graph of \(f\) is decreasing on \([1.5,3]\), so \(f^{\prime}(x)<0\) throughout that interval. 4. Statement (4) is true. The graph of \(f\) has point symmetry about \((0,1)\), so its derivative is symmetric about the y-axis. Therefore, \(f^{\prime}(2)=f^{\prime}(-2)\). 5. Statement (5) is false. Passing through the origin would require \(f^{\prime}(0)=0\), but the graph of \(f\) has positive slope at \(x=0\).

(1) True (2) True (3) True (4) True (5) False

- Find all horizontal tangent lines on the graph. - The derivative is positive where the function is increasing and negative where it is decreasing. - Recall the derivative symmetry associated with an even function.

1. Statement (1) is true. The graph of \(g\) has a local maximum at \(x=0\), so the tangent line is horizontal and \(g^{\prime}(0)=0\). 2. Statement (2) is true. The function is increasing for every \(x<0\), so \(g^{\prime}(x)>0\) there. 3. Statement (3) is false. The only horizontal tangent line in \([-4,4]\) occurs at \(x=0\), so \(g^{\prime}\) has exactly one zero. 4. Statement (4) is true. The graph is decreasing for \(x>0\), so \(g^{\prime}(1)<0\). 5. Statement (5) is true. The graph of \(g\) is symmetric about the y-axis, so \(g\) is even and \(g^{\prime}\) is odd. Therefore, the derivative graph is symmetric about the origin.

(1) True (2) True (3) False (4) True (5) True

- Horizontal tangent lines on \(f\) give zeros of \(f^{\prime}\). - Determine whether \(f\) increases or decreases on each interval between its extrema. - The derivative of a fourth-degree polynomial is cubic. - Use the symmetry of \(f\) to check the symmetry of \(f^{\prime}\).

1. The graph of \(f\) has local maxima at \(x=-2\) and \(x=2\) and a local minimum at \(x=0\). Therefore, \(f^{\prime}(-2)=0\), \(f^{\prime}(0)=0\), and \(f^{\prime}(2)=0\). 2. The function increases for \(x<-2\), decreases for \(-2<x<0\), increases for \(0<x<2\), and decreases for \(x>2\). Thus, \(f^{\prime}\) has the sign pattern positive, negative, positive, negative across those intervals. 3. Since \(f\) is a fourth-degree polynomial, \(f^{\prime}\) is cubic. It is an origin-symmetric cubic that crosses the x-axis at \(x=-2\), \(x=0\), and \(x=2\) with the sign pattern positive, negative, positive, negative from left to right.

a) \(x=-2\), \(x=0\), and \(x=2\) b) An origin-symmetric cubic crossing the x-axis at \(x=-2\), \(x=0\), and \(x=2\), with signs \(+,-,+,-\) from left to right.

- Determine the tangent slope at a horizontal inflection point. - Check whether the original function is increasing or decreasing on each side. - A zero without a sign change appears as a touch rather than a crossing of the x-axis.

1. At the horizontal inflection point, the tangent line is horizontal, so \(f^{\prime}(1)=0\). 2. The graph of \(f\) is increasing on both sides of \(x=1\), so \(f^{\prime}(x)>0\) for \(x\ne1\). 3. Because the displayed function is cubic, its derivative is quadratic. The derivative graph is an upward-opening parabola with vertex \((1,0)\); the visible slopes of \(f\) determine its vertical scale. 4. The horizontal inflection point of \(f\) appears as a zero of \(f^{\prime}\) where the derivative graph touches the x-axis without changing sign.

The derivative graph is an upward-opening parabola with vertex \((1,0)\). It touches the x-axis there without changing sign.

- Use the sign of \(f^{\prime}\) to decide whether the graph is increasing or decreasing. - Use the sign of \(f^{\prime\prime}\) to decide whether the graph is concave up or concave down. - The minimum value of \(f^{\prime}\) occurs where the graph has its steepest downward slope.

1. At \(x_A=-1\), the graph is increasing, so \(f^{\prime}(-1)>0\), and it is concave down, so \(f^{\prime\prime}(-1)<0\). Thus, \(x_A\) matches description 2. 2. At \(x_B=1\), the graph has an inflection point where the tangent slope is most negative. Thus, \(f^{\prime}\) has its minimum value there, so \(x_B\) matches description 3. 3. At \(x_C=2\), the graph is decreasing, so \(f^{\prime}(2)<0\), and it is concave up, so \(f^{\prime\prime}(2)>0\). Thus, \(x_C\) matches description 1. 4. At \(x_D=4\), the graph is increasing and concave up, so it does not match any listed description.

1) \(x_C=2\); 2) \(x_A=-1\); 3) \(x_B=1\)

- Examine how the graph approaches its horizontal asymptote. - Tangent slopes approach zero when a graph becomes nearly horizontal. - Determine whether the right branch is increasing or decreasing. - Observe whether the tangent slopes become larger or smaller.

1. As \(x\to\infty\), the graph of \(f\) approaches the horizontal asymptote \(y=3\). Its tangent lines become nearly horizontal, so \(\lim_{x\to\infty}f^{\prime}(x)=0\). 2. For \(x>0\), the function is increasing, so \(f^{\prime}(x)>0\). 3. The tangent lines become less steep as \(x\) increases. Therefore, \(f^{\prime}\) is strictly decreasing for \(x>0\) and approaches the x-axis from above. The line \(y=0\) is a horizontal asymptote of the derivative graph.

For \(x>0\), \(f^{\prime}(x)>0\) and \(f^{\prime}\) is strictly decreasing. As \(x\to\infty\), \(f^{\prime}(x)\to0\) from above.

- Locate the local maximum and local minimum of \(g\). - The derivative must be zero at those inputs. - Determine the sign of the derivative between the extrema. - Select the candidate with the matching sign pattern.

1. The graph of \(g\) has a local maximum near \(x=-1.73\) and a local minimum near \(x=1.73\). 2. Therefore, \(g^{\prime}\) must have zeros near those inputs. Both candidate graphs have the required zeros. 3. Between the two extrema, \(g\) is decreasing, so \(g^{\prime}(x)<0\) there. 4. Graph 3 lies below the x-axis between its zeros, while graph 4 lies above it. Therefore, graph 3 represents \(g^{\prime}\).

Graph 3 represents \(g^{\prime}\).

- Mark every input where the graph has a horizontal tangent line. - Determine whether \(f\) increases or decreases on each interval. - The derivative of a fourth-degree polynomial is cubic. - A steeper graph corresponds to a derivative value farther from zero.

1. Zeros of \(f^{\prime}\) occur where \(f\) has horizontal tangent lines. The graph has local minima at \(x=-2\) and \(x=2\) and a local maximum at \(x=0\). Thus, the zeros are \(x=-2\), \(x=0\), and \(x=2\). 2. For \(x<-2\), \(f\) is decreasing, so \(f^{\prime}(x)<0\). For \(-2<x<0\), \(f\) is increasing, so \(f^{\prime}(x)>0\). For \(0<x<2\), \(f\) is decreasing, so \(f^{\prime}(x)<0\). For \(x>2\), \(f\) is increasing, so \(f^{\prime}(x)>0\). 3. Since \(f\) is a fourth-degree polynomial, \(f^{\prime}\) is cubic. Its graph crosses the x-axis at \(-2\), \(0\), and \(2\) with the sign pattern negative, positive, negative, positive.

a) \(x=-2\), \(x=0\), and \(x=2\) b) The derivative graph is cubic. It lies below the x-axis where \(f\) is decreasing and above the x-axis where \(f\) is increasing.

- A local extremum of \(f^{\prime}\) occurs where the tangent slope of \(f\) is largest or smallest. - Look for the inflection point of \(f\). - Use the local extrema of \(f\) to locate zeros of \(f^{\prime}\). - The derivative of a cubic polynomial is quadratic.

1. The graph of \(f\) has a local maximum at \(x=-2\) and a local minimum at \(x=2\). Therefore, \(f^{\prime}(-2)=0\) and \(f^{\prime}(2)=0\). 2. The graph has an inflection point at the origin, where the slope is most negative. The tangent slope there is approximately \(-3\), so \(f^{\prime}\) has a local minimum at \(x=0\). 3. Since \(f\) is cubic, \(f^{\prime}\) is quadratic. Its graph is an upward-opening parabola with zeros at \(x=-2\) and \(x=2\) and a vertex approximately at \((0,-3)\).

The derivative has a local minimum at \(x=0\). Its graph is an upward-opening parabola with zeros at \(x=-2\) and \(x=2\) and a vertex approximately at \((0,-3)\).

- Find where the original function changes from increasing to decreasing. - A nearly horizontal graph has derivative values near zero. - Locate where the original graph rises and falls most steeply.

1. The graph of \(f\) has a local maximum at \(x=0\). Therefore, \(f^{\prime}(0)=0\). 2. For \(x<0\), the graph is increasing, so \(f^{\prime}(x)>0\). For \(x>0\), the graph is decreasing, so \(f^{\prime}(x)<0\). 3. Near \(x=-1\), the graph rises most steeply, so \(f^{\prime}\) has a local maximum. Near \(x=1\), the graph falls most steeply, so \(f^{\prime}\) has a local minimum. 4. As \(x\to\pm\infty\), the graph of \(f\) becomes nearly horizontal. Therefore, \(f^{\prime}(x)\to0\).

The derivative has a zero at \(x=0\). It is positive for \(x<0\), negative for \(x>0\), and approaches \(0\) as \(x\to\pm\infty\). Its graph has a local maximum at a negative input and a local minimum at the corresponding positive input.

- Use the monotonicity of \(f\) to determine the sign of \(f^{\prime}\). - Horizontal tangent lines give zeros of the derivative. - Compare a negative slope with a positive slope. - An inflection point of \(f\) corresponds to a local extremum of \(f^{\prime}\).

1. Statement (1) is false. The function \(f\) is decreasing on \((-2,2)\), so \(f^{\prime}(x)<0\) and its graph lies below the x-axis. 2. Statement (2) is true. The graph has horizontal tangent lines at the local maximum \(x=-2\) and the local minimum \(x=2\). Therefore, \(f^{\prime}(-2)=0=f^{\prime}(2)\). 3. Statement (3) is true. At \(x=0\), the graph is decreasing, so \(f^{\prime}(0)<0\). At \(x=4\), it is increasing, so \(f^{\prime}(4)>0\). Thus, \(f^{\prime}(0)<f^{\prime}(4)\). 4. Statement (4) is true. The graph of \(f\) has an inflection point at \(x=0\). Therefore, its slope function \(f^{\prime}\) has a local extremum there.

(1) False (2) True (3) True (4) True

- Use increasing and decreasing behavior to determine the derivative sign. - Look for horizontal tangent lines. - Estimate the tangent slope at the origin. - When comparing negative slopes, the value closer to zero is greater.

1. Statement (1) is true. The graph of \(f\) is decreasing for \(x\le-1\) and for \(x\ge1\). Thus, \(f^{\prime}(x)\le0\) whenever \(|x|\ge1\). 2. Statement (2) is false. From the graph, the tangent slope at the origin is approximately \(2\), not \(1\). 3. Statement (3) is true. For \(x>1\), the graph is decreasing but becomes flatter as \(x\) increases. Thus, the negative slope at \(x=3\) is closer to zero than the negative slope at \(x=2\), so \(f^{\prime}(2)<f^{\prime}(3)\). 4. Statement (4) is true. The graph has a local maximum at \(x=1\), so the tangent line is horizontal and \(f^{\prime}(1)=0\).

(1) True (2) False (3) True (4) True

- Local extrema of \(f\) give zeros of \(f^{\prime}\). - Use increasing and decreasing intervals to determine the derivative sign. - An inflection point of \(f\) corresponds to a local extremum of \(f^{\prime}\). - Consider the degree and leading coefficient of the derivative.

1. Statement (1) is true. The graph has a local maximum at \(x=-1\) and a local minimum at \(x=3\), so \(f^{\prime}\) has zeros at those two inputs. 2. Statement (2) is false. The graph of \(f\) is decreasing on \((-1,3)\), so \(f^{\prime}(x)<0\) and the derivative graph lies below the x-axis. 3. Statement (3) is true. The graph of \(f\) has an inflection point at \(x=1\), so \(f^{\prime}\) has a local extremum there. 4. Statement (4) is true. The cubic has a positive leading coefficient, so its derivative is an upward-opening quadratic. Therefore, \(f^{\prime}(x)\to\infty\) as \(x\to\infty\).

(1) True (2) False (3) True (4) True

- Find the local extrema of \(f\). - The derivative must be zero at those inputs. - Use the increasing and decreasing intervals to determine the sign of the derivative. - The derivative of a cubic polynomial is quadratic.

1. The graph has a local minimum at \(x=-1\) and a local maximum at \(x=3\). Therefore, \(f^{\prime}\) must have zeros at \(x=-1\) and \(x=3\). 2. Options A and B have those two zeros. Option C is linear and has only one zero, and option D has different zeros. 3. Between \(-1\) and \(3\), the graph of \(f\) is increasing, so \(f^{\prime}(x)>0\) there. 4. At \(x=0\), option A gives \(0.9>0\), while option B gives \(-0.9<0\). Therefore, option A has the correct sign pattern.

A) \(f^{\prime}(x)=-0.3x^2+0.6x+0.9\)

- The derivative sign indicates whether the original graph is increasing or decreasing. - Horizontal tangent lines give zeros of the derivative. - Check the entire stated interval, including its endpoints. - Count all local extrema of the original graph.

1. Statement (1) is true. The graph is increasing at \(x=-1\), so \(f^{\prime}(-1)>0\). 2. Statement (2) is true. The graph has a local maximum at \(x=0\), so \(f^{\prime}(0)=0\). 3. Statement (3) is false. At \(x=0\), the derivative equals zero, and for \(0<x\le1.5\), the graph of \(f\) is decreasing, so \(f^{\prime}(x)<0\). 4. Statement (4) is true. The graph has a local minimum near \(x=1.73\), so \(f^{\prime}\) has a zero there. 5. Statement (5) is false. The graph is increasing at \(x=2\), so \(f^{\prime}(2)>0\). 6. Statement (6) is true. The graph has three horizontal tangent lines, so \(f^{\prime}\) has three zeros.

(1) True (2) True (3) False (4) True (5) False (6) True

- Local extrema of \(g\) give zeros of \(g^{\prime}\). - The derivative is negative where the function is decreasing. - Compare the signs of the tangent slopes at \(x=-1\) and \(x=1\).

1. The derivative is zero where the graph has horizontal tangent lines. The graph has a local maximum at \(x=0\) and a local minimum at \(x=2\). Thus, \(g^{\prime}(0)=0\) and \(g^{\prime}(2)=0\). 2. The graph of \(g\) is decreasing on \((0,2)\), so \(g^{\prime}(x)<0\) throughout that interval. 3. At \(x=-1\), the graph is increasing, so \(g^{\prime}(-1)>0\). At \(x=1\), it is decreasing, so \(g^{\prime}(1)<0\). Therefore, \(g^{\prime}(-1)>g^{\prime}(1)\).

a) \(x=0\) and \(x=2\) b) Negative: \(g^{\prime}(x)<0\) on \((0,2)\). c) \(g^{\prime}(-1)>g^{\prime}(1)\).

- The derivative is the tangent slope of the original graph. - Use increasing and decreasing intervals to determine the derivative sign. - Horizontal tangent lines produce zeros of the derivative. - Count all local extrema in the displayed interval.

1. Statement (1) is true. The graph has a local maximum at \(x=-2\), so the tangent line is horizontal and \(h^{\prime}(-2)=0\). 2. Statement (2) is false. The graph is decreasing at \(x=0\), so \(h^{\prime}(0)<0\). 3. Statement (3) is true. The graph is strictly decreasing throughout \([-1,1]\), so \(h^{\prime}(x)<0\) there and the derivative graph lies below the x-axis. 4. Statement (4) is true. The graph has exactly two horizontal tangent lines, at the local maximum \(x=-2\) and the local minimum \(x=2\). Therefore, \(h^{\prime}\) has exactly two zeros.

(1) True (2) False (3) True (4) True

- Count all horizontal tangent lines on the graph of \(g\). - Match their x-coordinates to zeros of the candidates. - Check whether \(g\) is increasing or decreasing on each interval. - Eliminate any candidate with the wrong sign.

1. The fourth-degree graph of \(g\) has local minima at \(x=-2\) and \(x=2\) and a local maximum at \(x=0\). Therefore, \(g^{\prime}\) must have zeros at \(x=-2\), \(x=0\), and \(x=2\). 2. Graph A has only two zeros, so it cannot represent \(g^{\prime}\). 3. Graph C has three zeros, but they occur at different inputs. Its sign pattern also does not match: for \(x<-2\), the graph of \(g\) is decreasing, so \(g^{\prime}(x)\) must be negative. 4. Graph B has the required zeros and sign pattern. Therefore, graph B represents \(g^{\prime}\).

Graph B represents \(g^{\prime}\). Graph A has too few zeros, and graph C has zeros at the wrong inputs and the wrong sign pattern.

- Function values are compared by the vertical positions of the marked points. - Derivative values are compared by tangent-line steepness and direction. - A steep downward tangent has a large negative derivative. - Concave down means tangent slopes decrease as \(x\) increases.

1. Compare the heights of the marked points. The greatest marked function values occur at \(x_1\) and \(x_7\). The least marked values occur at the local minima \(x_2\) and \(x_6\). 2. The derivative is greatest where the graph rises most steeply, at \(x_7\). It is least where the graph falls most steeply, at \(x_1\). 3. The graph is concave down near its local maximum, which is the marked point \(x_4\). The points \(x_3\) and \(x_5\) are approximately inflection points, while the other marked points are in concave-up regions.

a) Greatest at \(x_1\) and \(x_7\); least at \(x_2\) and \(x_6\) b) Greatest at \(x_7\); least at \(x_1\) c) Concave down at \(x_4\)

- The maximum occurs at the highest marked point. - The smallest derivative occurs where the graph has its steepest downward tangent. - An inflection point is where the graph changes concavity.

1. The highest marked point is \(x_3\), so \(g(x)\) reaches its maximum there. 2. The derivative is smallest where the graph falls most steeply. Among the marked points on the decreasing part of the graph, this occurs at \(x_5\). 3. At \(x_5\), the graph changes from concave down to concave up. Therefore, \(x_5\) is an inflection point and \(g^{\prime\prime}(x_5)=0\).

a) \(x_3\) b) \(x_5\) c) \(x_5\)

- A local extremum of \(f\) corresponds to a zero of \(f^{\prime}\). - At a local minimum, the derivative changes from negative to positive. - At a local maximum, the derivative changes from positive to negative. - Check both the number of zeros and the direction of each sign change.

1. At the local minimum \(x=-2\), the derivative must be zero and change sign from negative to positive. 2. At the local maximum \(x=2\), the derivative must be zero and change sign from positive to negative. 3. Graph C has only one zero, so it cannot represent the derivative. 4. Graph B has zeros at \(-2\) and \(2\), but its sign changes correspond to a maximum at \(-2\) and a minimum at \(2\), the reverse of the given information. 5. Graph A has zeros at \(-2\) and \(2\), with a negative-to-positive sign change at \(-2\) and a positive-to-negative sign change at \(2\). Therefore, graph A represents \(f^{\prime}\).

Graph A represents \(f^{\prime}\).

- Horizontal tangent lines on \(k\) give zeros of \(k^{\prime}\). - Determine where \(k\) is increasing or decreasing. - The derivative of a cubic polynomial is quadratic. - Check the sign of each candidate between the two zeros.

1. The graph of \(k\) has a local maximum at \(x=0\) and a local minimum at \(x=2\). Therefore, \(k^{\prime}\) must have zeros at \(x=0\) and \(x=2\). 2. Panel d) is a line with only one zero. Also, the derivative of a cubic function must be quadratic, so panel d) cannot be correct. 3. Panels b) and c) are parabolas with zeros at \(x=0\) and \(x=2\). 4. The graph of \(k\) is decreasing for \(0<x<2\), so \(k^{\prime}(x)<0\) on that interval. 5. Panel b) lies below the x-axis between the zeros, while panel c) lies above it. Therefore, panel b) represents \(k^{\prime}\).

Panel b) represents \(k^{\prime}\).

- Local extrema of \(f\) give zeros of \(f^{\prime}\). - Inflection points occur where the graph changes concavity. - The first derivative gives tangent slope. - The derivative of an odd function is even. - Concave down corresponds to a negative second derivative.

1. Statement (1) is true. The graph of \(f\) has a local minimum at \(x=-2\) and a local maximum at \(x=2\), so \(f^{\prime}\) has exactly two zeros. 2. Statement (2) is true. The graph changes concavity near \(x=-3.5\), at \(x=0\), and near \(x=3.5\), giving three inflection points in the displayed interval. 3. Statement (3) is true. The graph is increasing at the origin, so \(f^{\prime}(0)>0\). 4. Statement (4) is false. The graph of \(f\) is symmetric about the origin, so \(f\) is odd and \(f^{\prime}\) is even. Therefore, the derivative graph is symmetric about the y-axis. 5. Statement (5) is false. At \(x=1\), the graph is concave down, so \(f^{\prime\prime}(1)<0\).

(1) True (2) True (3) True (4) False (5) False

- Find the horizontal tangent lines on the graph of \(f\). - Use increasing and decreasing intervals to determine the derivative sign. - Identify where \(f\) has its most negative slope. - The derivative of a cubic polynomial is quadratic.

1. The derivative is zero at the local extrema of \(f\). From the graph, these occur near \(x=-1.4\) and \(x=1.4\). 2. The graph of \(f\) is increasing for \(x<-1.4\), decreasing for \(-1.4<x<1.4\), and increasing for \(x>1.4\). Therefore, \(f^{\prime}\) is positive, then negative, then positive across those intervals. 3. Since \(f\) is cubic, \(f^{\prime}\) is an upward-opening parabola. Its minimum occurs at \(x=0\), where the graph of \(f\) has its most negative slope.

1) The zeros are approximately \(x=-1.4\) and \(x=1.4\). 2) The derivative graph is an upward-opening parabola, positive outside the two zeros and negative between them, with a minimum at \(x=0\).

- The derivative is zero at the maximum of the bell curve. - Determine the derivative sign from increasing and decreasing behavior. - The extrema of the derivative occur where the original graph is steepest. - Use symmetry to check your description.

1. The graph of \(f\) is symmetric about the y-axis and has a maximum at \(x=0\). Therefore, \(f^{\prime}(0)=0\). The function increases for \(x<0\), so \(f^{\prime}(x)>0\), and decreases for \(x>0\), so \(f^{\prime}(x)<0\). 2. The local extrema of \(f^{\prime}\) occur where the graph of \(f\) is steepest. These are the inflection points of \(f\), at approximately \(x=-1\) and \(x=1\). 3. Thus, the derivative graph rises from values near zero to a local maximum near \(x=-1\), crosses the x-axis at \(x=0\), falls to a local minimum near \(x=1\), and approaches zero from below.

1) The derivative is positive for \(x<0\), zero at \(x=0\), and negative for \(x>0\), approaching \(0\) at both ends. 2) The local extrema of \(f^{\prime}\) occur at approximately \(x=-1\) and \(x=1\). These are the inflection points of \(f\).

- Count all horizontal tangent lines. - Count all changes in concavity.

1. The graph has one local maximum at \(x=0\). The tangent line is horizontal there, so \(h^{\prime}\) has exactly \(1\) zero. 2. The graph changes concavity once on each side of the maximum. Therefore, it has two inflection points and \(h^{\prime\prime}\) has exactly \(2\) zeros.

\(h^{\prime}\) has \(1\) zero, and \(h^{\prime\prime}\) has \(2\) zeros.

- Count points with horizontal tangent lines. - Check for a concavity change within each continuous branch. - Do not count a vertical asymptote as an inflection point.

1. The graph has one local maximum on the left branch and one local minimum on the right branch. Therefore, \(p^{\prime}\) has \(2\) zeros. 2. Neither branch changes concavity. The function is not defined at the vertical asymptote, so no change across it is an inflection point. Therefore, \(p^{\prime\prime}\) has \(0\) zeros.

\(p^{\prime}\) has \(2\) zeros, and \(p^{\prime\prime}\) has \(0\) zeros.

- Look for horizontal tangent lines. - Identify any point where a continuous branch changes concavity. - Analyze the separate branches around the vertical asymptotes.

1. The function is strictly decreasing on every interval of its domain and has no horizontal tangent lines. Therefore, \(r^{\prime}\) has \(0\) zeros. 2. On the middle branch, the graph changes concavity at the origin. Thus, \((0,0)\) is an inflection point and \(r^{\prime\prime}\) has one zero. The outer branches have no additional concavity changes.

\(r^{\prime}\) has \(0\) zeros, and \(r^{\prime\prime}\) has \(1\) zero.

- Consider the degree of the derivative of a quadratic function. - A local extremum of the original function occurs where its derivative is zero. - Read the derivative value at \(x=3\) to find the tangent slope. - The function and derivative have the same value where their graphs intersect.

1. Curve \(k\) is a parabola and curve \(m\) is a line. Since the derivative of a quadratic function is linear, \(k\) represents \(g\) and \(m\) represents \(g^{\prime}\). This also agrees with the local minimum of \(k\) and the zero of \(m\) at \(x=0\). 2. The slope of \(g\) at \(x=3\) is \(g^{\prime}(3)\). From curve \(m\), \(g^{\prime}(3)=1.5\). 3. The curves intersect for \(x>0\) at \(x=2\). At that point, \(g(2)=g^{\prime}(2)=1\).

a) Curve \(k\) represents \(g\), and curve \(m\) represents \(g^{\prime}\). b) \(1.5\) c) \(x=2\)

- Where does the graph of \(f\) have horizontal tangent lines? - How are the inflection points of \(f\) related to the local extrema of \(f^{\prime}\)? - On the interval where \(f\) is decreasing, where is its slope most negative? - Use the degree and symmetry of \(f\) to describe \(f^{\prime}\).

1. The zeros of \(f^{\prime}\) occur at the local extrema of \(f\). Thus, \(f^{\prime}(x)=0\) at \(x\approx -1.73\), \(x=0\), and \(x\approx 1.73\). 2. The zeros of \(f^{\prime\prime}\) occur at the inflection points of \(f\). Thus, \(f^{\prime\prime}(x)=0\) at \(x=-1\) and \(x=1\). 3. On \(0\leq x\leq 1.73\), the graph is decreasing, so the steepest decrease occurs where \(f^{\prime}\) is smallest. The local minimum of \(f^{\prime}\) occurs where \(f^{\prime\prime}=0\), at \(x=1\). 4. Because \(f\) is an even quartic polynomial, \(f^{\prime}\) is an odd cubic polynomial. Its graph is symmetric about the origin, has zeros at \(x\approx -1.73\), \(x=0\), and \(x\approx 1.73\), has a local maximum at \(x=-1\), and has a local minimum at \(x=1\).

a) Zeros of \(f^{\prime}\): \(x\approx -1.73\), \(x=0\), and \(x\approx 1.73\). Zeros of \(f^{\prime\prime}\): \(x=-1\) and \(x=1\). b) The graph decreases most steeply at \(x=1\), where \(f^{\prime}\) has a local minimum. c) The graph of \(f^{\prime}\) is an odd cubic graph with zeros at \(x\approx -1.73\), \(x=0\), and \(x\approx 1.73\), a local maximum at \(x=-1\), and a local minimum at \(x=1\).

- Track what happens to both interval endpoints from row to row. - Decide whether symmetry is essential to the idea of shrinking an interval. - Use the trend in the average rates to identify the estimate.

1. Each interval contains \(t=1\), and both endpoints move closer to \(1\) across the table. 2. The corresponding average rates \(1.7\), \(1.95\), and \(1.991\) approach \(2\). 3. Centering is not required; shrinking the interval around the instant is the key idea. The instantaneous rate is \(2\).

\(2\). The intervals need not be symmetric as long as their endpoints approach \(1\) and the average rates approach a common value.

- Focus on the rates from right-side intervals that shrink toward \(t=3\). - Compare the limiting patterns for the two functions separately. - A whole-interval average summarizes net change over that interval, not the detailed behavior inside it.

1. The shrinking right-side interval rates for \(f\) approach \(6\). 2. The shrinking right-side interval rates for \(g\) approach \(4\). 3. Thus the right-hand instantaneous rates are \(6\) for \(f\) and \(4\) for \(g\). 4. Equal average rates on the larger interval \([2, 4]\) do not force equal local behavior at its midpoint.

At \(t=3\), \(f\) has right-hand instantaneous rate \(6\) and \(g\) has right-hand instantaneous rate \(4\). Their equal average rate on \([2, 4]\) does not determine these local rates.

- Do not treat opposite signs as evidence that the limiting values are different. - Compare the magnitudes of the left and right averages as the distance shrinks. - Interpret what a negative rate before the instant and a positive rate after it mean.

1. Both columns approach \(0\) as the intervals shrink toward \(t=2\). 2. The instantaneous rate is \(0\). 3. Negative averages just before \(t=2\) and positive averages just after \(t=2\) suggest that the output is changing from decreasing to increasing near that time.

\(0\). The nearby averages suggest a change from decreasing before \(t=2\) to increasing after \(t=2\).

- Zeros of the first derivative occur where the graph has horizontal tangent lines. - Zeros of the second derivative can occur at inflection points. - A change across a discontinuity is not an inflection point. - Count each qualifying point within the function domain.

1. For \(f\), zeros of \(f^{\prime}\) occur at horizontal tangent lines. The graph has one local minimum at \(x=-1\) and one local maximum at \(x=1\), so \(f^{\prime}\) has \(2\) zeros. Zeros of \(f^{\prime\prime}\) correspond to inflection points. The graph changes concavity three times: once on the left side, once at the origin, and once on the right side. Therefore, \(f^{\prime\prime}\) has \(3\) zeros. 2. For \(g\), the graph has one local minimum at \(x=0\), so \(g^{\prime}\) has \(1\) zero. The graph does not change concavity within any interval of its domain. Changes across the vertical asymptotes \(x=-1\) and \(x=1\) are not inflection points because the function is not defined there. Therefore, \(g^{\prime\prime}\) has \(0\) zeros.

a) \(f^{\prime}\) has \(2\) zeros, and \(f^{\prime\prime}\) has \(3\) zeros. b) \(g^{\prime}\) has \(1\) zero, and \(g^{\prime\prime}\) has \(0\) zeros.

- Zeros of the first derivative occur at horizontal tangent lines. - Zeros of the second derivative can occur where the graph changes concavity. - Do not count a change across a vertical asymptote as an inflection point. - Track each branch from left to right.

1. In panel a), the graph of \(f\) has one local minimum at \(x=0\), so \(f^{\prime}\) has \(1\) zero. Within each continuous interval of its domain, the graph does not change concavity. The changes across the vertical asymptotes are not inflection points because \(f\) is undefined there. Thus, \(f^{\prime\prime}\) has \(0\) zeros. 2. In panel b), the graph of \(g\) has one local minimum on the left branch and one local maximum on the right branch, so \(g^{\prime}\) has \(2\) zeros. The graph changes concavity once on each branch, so \(g^{\prime\prime}\) has \(2\) zeros.

a) \(f^{\prime}\) has \(1\) zero, and \(f^{\prime\prime}\) has \(0\) zeros. b) \(g^{\prime}\) has \(2\) zeros, and \(g^{\prime\prime}\) has \(2\) zeros.

- Find each point where the graph of \(f\) has a horizontal tangent line. - Distinguish a local extremum from a horizontal inflection point. - Determine the sign of the derivative from the increasing and decreasing intervals. - A sign change produces a crossing; no sign change produces a touch.

1. The notable points give the following table: <table> <tr><th>Input \(x\)</th><th>Type of point on \(f\)</th><th>Value of \(f^{\prime}(x)\)</th></tr> <tr><td>\(0\)</td><td>Local minimum</td><td>\(0\)</td></tr> <tr><td>\(3\)</td><td>Horizontal inflection point</td><td>\(0\)</td></tr> </table> 2. For \(x<0\), the function is decreasing, so \(f^{\prime}(x)<0\). For \(0<x<3\), the function is increasing, so \(f^{\prime}(x)>0\). For \(x>3\), the function continues increasing, so \(f^{\prime}(x)>0\). 3. The derivative graph crosses the x-axis at \(x=0\), changing from negative to positive. At \(x=3\), it touches the x-axis from above without changing sign. One additional reference value is \(f^{\prime}(1)\approx0.8\).

The derivative has zeros at \(x=0\) and \(x=3\). At \(x=0\), it changes sign from negative to positive and crosses the x-axis. At \(x=3\), it remains positive on both sides and touches the x-axis.

- Zeros of the first derivative occur where the graph has a horizontal tangent line. - Zeros of the second derivative can occur where the graph changes concavity. - Analyze each continuous branch separately; a vertical asymptote is not an extremum or an inflection point.

1. For \(h\), the graph has a local minimum at \(x=-1\) and a local maximum at \(x=1\). These are the two points with horizontal tangent lines, so \(h^{\prime}\) has \(2\) zeros. 2. The graph of \(h\) changes concavity at \(x=-\sqrt{3}\), \(x=0\), and \(x=\sqrt{3}\). Therefore, \(h^{\prime\prime}\) has \(3\) zeros. 3. For \(k\), the graph has one local maximum at \(x=2\). Therefore, \(k^{\prime}\) has \(1\) zero. 4. The right branch of \(k\) changes concavity once, at \(x=3\). The left branch has no concavity change, and the vertical asymptote is not an inflection point. Therefore, \(k^{\prime\prime}\) has \(1\) zero.

Graph \(h\): \(h^{\prime}\) has \(2\) zeros, and \(h^{\prime\prime}\) has \(3\) zeros. Graph \(k\): \(k^{\prime}\) has \(1\) zero, and \(k^{\prime\prime}\) has \(1\) zero.