Free math worksheets

- An average rate uses a total change over a time interval. - An instantaneous rate describes what is happening at one moment. - Keep the direction as part of each position rate.

1. The average rate of change of position is \(\frac{120\,\text{mi}}{2\,\text{h}}=60\,\text{mph}\) east. 2. The stated velocity \(55\,\text{mph}\) east describes the rate of change of position at one instant. 3. Therefore, \(60\,\text{mph}\) east is the average rate and \(55\,\text{mph}\) east is the instantaneous rate.

Average rate of change of position: \(60\,\text{mph}\) east. Instantaneous rate of change of position: \(55\,\text{mph}\) east.

- Average rate of change is change in output divided by change in input. - Use the later height minus the earlier height. - Use hours for the elapsed time so the rate has units of centimeters per hour.

1. The change in height is \(30-18=12\,\text{cm}\). 2. The elapsed time is \(2\,\text{h}\). 3. The average rate of change is \(\frac{12\,\text{cm}}{2\,\text{h}}=6\,\text{cm/h}\).

\(6\,\text{cm/h}\)

- What does the derivative represent at a point on a graph? - First find the derivative function. - Substitute the given \(x\)-value into the derivative.

1. Differentiate the function: \(f'(x)=10x^4-6x\). 2. The slope at \(x=1\) is \(f'(1)\). 3. Evaluate: \(f'(1)=10(1)^4-6(1)=4\).

\(4\)

- Which quantity gives the slope of a function at a specific input? - Differentiate the function term by term. - Substitute the point's \(x\)-coordinate into the derivative.

1. Differentiate: \(f'(x)=x^3-4x+3\). 2. Evaluate the derivative at the point's \(x\)-coordinate: \(f'(2)=2^3-4(2)+3=3\).

The slope at \(P\) is \(3\).

- The slope of the tangent line at \(x=a\) is the derivative value at \(a\). - Differentiate the function term by term. - Use parentheses when substituting a negative value.

1. Differentiate: \(g'(x)=\frac{1}{2}x^2+\frac{1}{2}x-1\). 2. Evaluate at \(x=-2\): \(g'(-2)=\frac{1}{2}(-2)^2+\frac{1}{2}(-2)-1=0\).

The slope at \(x=-2\) is \(0\).

- What does the first derivative represent at a point? - Differentiate the polynomial term by term. - Substitute the given input into the derivative.

1. Differentiate: \(f'(x)=\frac{1}{2}x^2-6x+10\). 2. Evaluate at \(x=6\): \(f'(6)=\frac{1}{2}(6)^2-6(6)+10=-8\).

The slope at \(x=6\) is \(-8\).

- The slope of a tangent line is a derivative value. - Use parentheses when substituting a negative input. - A constant term has derivative \(0\).

1. Differentiate: \(h'(x)=2x^3-2x^2+1\). 2. Evaluate at \(x=-1\): \(h'(-1)=2(-1)^3-2(-1)^2+1=-3\).

The tangent line has slope \(-3\) at \(x=-1\).

- What does the sign of the derivative tell you about the slope of a graph? - Use parentheses when substituting a negative value. - Find the derivative function before evaluating it at the given input.

1. Differentiate: \(g'(x)=-x^5+8x\). 2. Evaluate at \(x=-1\): \(g'(-1)=-(-1)^5+8(-1)=-7\). 3. Since \(-7<0\), the slope is negative at \(x=-1\).

\(g'(x)=-x^5+8x\); the slope is negative at \(x=-1\).

- How is a specified slope represented using the derivative? - Differentiate the quadratic function. - Set the derivative equal to the given value. - Solve the resulting equation for \(x\).

1. Differentiate: \(f'(x)=3x-7\). 2. Set the derivative equal to the required value: \(3x-7=5\). 3. Solve: \(3x=12\), so \(x=4\).

\(x=4\)

- A smooth connection requires the two graphs to share a point and a slope. - Find the point on the hill at \(x=2\). - Use the derivative to find the hill's slope at that point. - Write the line using the point and slope.

1. Find the connection point: \(f(2)=-0.2(2)^2+2(2)+1=4.2\), so the point is \((2,4.2)\). 2. Differentiate: \(f'(x)=-0.4x+2\). 3. Find the slope at the connection point: \(f'(2)=-0.4(2)+2=1.2\). 4. A smooth connection requires the line to have the same point and slope. Using point-slope form, \(y-4.2=1.2(x-2)\). 5. Simplify: \(y=1.2x+1.8\).

\(y=1.2x+1.8\)

- Evaluate the first derivative at each given value. - A positive derivative means the graph rises; a negative derivative means it falls.

1. Differentiate: \(f'(x)=3x^2-6x-9\). 2. At \(x=-2\), \(f'(-2)=3\cdot4-6(-2)-9=15>0\), so the graph has positive slope. 3. At \(x=2\), \(f'(2)=3\cdot4-6\cdot2-9=-9<0\), so the graph has negative slope.

At \(x=-2\), the slope is positive: \(f'(-2)=15\). At \(x=2\), the slope is negative: \(f'(2)=-9\).

- Use the power rule to find each derivative function. - Rewrite a variable in the denominator using a negative exponent. - A derivative value gives the graph's slope at that input. - Use parentheses when substituting negative values.

1. For part a), \(f'(x)=5x^4\). Therefore, \(f'(1)=5\), \(f'(-2)=80\), and \(f'(0)=0\). 2. For part b), rewrite \(g(x)=2x^{-1}\), so \(g'(x)=-2x^{-2}=-\frac{2}{x^2}\). Thus \(g'(2)=-0.5\) and \(g'(-4)=-0.125\).

a) \(f'(1)=5\), \(f'(-2)=80\), and \(f'(0)=0\) b) \(g'(2)=-0.5\) and \(g'(-4)=-0.125\)

- Differentiate using the sum and constant multiple rules. - Recall the derivatives of sine and cosine. - Use exact unit-circle values at the two inputs. - The derivative value is the graph’s slope.

1. Differentiate: \(f'(x)=3\cos x+2\sin x\). 2. At \(x=0\), \(f'(0)=3\). 3. At \(x=\frac{\pi}{2}\), \(f'\left(\frac{\pi}{2}\right)=2\).

At \(x=0\), the slope is \(3\). At \(x=\frac{\pi}{2}\), the slope is \(2\).

- Differentiate first. - Substitute each candidate input into the derivative. - Use exact unit-circle values. - Select the inputs where the derivative equals \(1\).

1. Differentiate: \(h'(x)=\cos x-\sin x\). 2. Evaluate at the four inputs: \(h'(0)=1\), \(h'\left(\frac{\pi}{2}\right)=-1\), \(h'(\pi)=-1\), and \(h'\left(\frac{3\pi}{2}\right)=1\). Therefore, the tangent slope is \(1\) at \(x=0\) and \(x=\frac{3\pi}{2}\).

\(x=0\) and \(x=\frac{3\pi}{2}\)

- A tangent slope is a value of the derivative. - Find the derivative function first. - Solve one linear equation for each requested slope.

1. Differentiate: \(f'(x)=2x-3\). 2. Set \(2x-3\) equal to each requested slope and solve. 3. For slope \(0\), \(2x-3=0\), so \(x=1.5\). 4. For slope \(1\), \(2x-3=1\), so \(x=2\). 5. For slope \(-1\), \(2x-3=-1\), so \(x=1\). 6. For slope \(\frac{1}{2}\), \(2x-3=\frac{1}{2}\), so \(x=1.75\). 7. For slope \(-\frac{1}{2}\), \(2x-3=-\frac{1}{2}\), so \(x=1.25\).

Slope \(0\): \(x=1.5\) Slope \(1\): \(x=2\) Slope \(-1\): \(x=1\) Slope \(\frac{1}{2}\): \(x=1.75\) Slope \(-\frac{1}{2}\): \(x=1.25\)

- The derivative gives the graph's slope at each input. - Substitute a specified input into the derivative to find the slope there. - Rewrite the reciprocal power using a negative exponent before differentiating. - Use parentheses when substituting negative values.

1. Rewrite the function as \(f(x)=x^4-x^{-2}\). Then \(f'(x)=4x^3+2x^{-3}=4x^3+\frac{2}{x^3}\). 2. Evaluate: \(f'(1)=4+2=6\). 3. At \(x=-1\), \(f'(-1)=4(-1)^3+\frac{2}{(-1)^3}=-6\). Since \(-6<0\), the slope is negative.

a) \(f'(x)=4x^3+\frac{2}{x^3}\) b) \(6\) c) \(f'(-1)=-6<0\), so the slope is negative.

- The derivative gives the slope at each input. - Use the power rule for a function with a negative exponent. - Set each derivative equal to the specified slope. - A quadratic equation can have two real solutions.

1. Differentiate: \(f'(x)=x-3\) and \(g'(x)=-\frac{2}{x^2}\). 2. For part a), solve \(x-3=2\), giving \(x=5\). 3. For part b), solve \(-\frac{2}{x^2}=-0.5\). This gives \(x^2=4\), so \(x=-2\) or \(x=2\).

a) \(x=5\) b) \(x=-2\) and \(x=2\)

- The derivative value at a point is the graph's slope there. - Rewrite the radical and reciprocal using exponents before differentiating. - After finding the derivative function, substitute the given input.

1. Rewrite the function as \(f(x)=4x^{1/2}-9x^{-1}+2\). 2. Differentiate: \(f'(x)=2x^{-1/2}+9x^{-2}=\frac{2}{\sqrt{x}}+\frac{9}{x^2}\). 3. Evaluate at \(x=1\): \(f'(1)=2+9=11\).

a) \(f'(x)=\frac{2}{\sqrt{x}}+\frac{9}{x^2}\), where \(x>0\) b) \(11\)

- Differentiate using the power rule. - Compare each derivative directly with the corresponding perimeter or surface-area formula. - Find the constant factor that multiplies the boundary measure to produce the derivative. - Check whether the factor is the same in both cases.

1. For the square, \(A'(x)=2x\). Since \(P(x)=4x\), \(A'(x)=\frac{1}{2}P(x)\). 2. For the cube, \(V'(x)=3x^2\). Since \(S(x)=6x^2\), \(V'(x)=\frac{1}{2}S(x)\). 3. In both cases, the ratio of the derivative to the boundary measure is \(\frac{1}{2}\).

a) \(A'(x)=2x=\frac{1}{2}P(x)\) b) \(V'(x)=3x^2=\frac{1}{2}S(x)\) c) The ratio is \(\frac{1}{2}\), or \(1:2\).

- Rewrite the expression so that the term containing \(k\) is isolated. - Find the x-value that makes the coefficient of \(k\) equal to zero. - Evaluate the derivative at that x-value.

1. Rewrite the function as \(f_k(x)=k(x-2)^2+2x-3\). The parameter term is zero when \(x=2\). 2. Then \(f_k(2)=1\), so every graph passes through \(P(2, 1)\). 3. Differentiate: \(f_k'(x)=2kx+2-4k\). At \(x=2\), \(f_k'(2)=4k+2-4k=2\), independent of \(k\). Therefore, every graph has slope \(2\) at \(P\).

a) \(P(2, 1)\) b) Every graph has slope \(2\) at \(P\).

- Apply the power rule term by term. - Substitute \(x=0\) into the derivative. - Use two points on the tangent line to compute rise over run.

1. Differentiate: \(f'(x)=0.5\cdot2x-2=x-2\). 2. Evaluate at \(x=0\): \(f'(0)=0-2=-2\). 3. The tangent line passes through \((0,0)\). Moving \(1\) unit to the right along the line corresponds to moving \(2\) units down, so its slope is \(\frac{-2}{1}=-2\). This agrees with the derivative.

a) \(f'(x)=x-2\) b) \(f'(0)=-2\) c) The tangent line has slope \(-2\), confirming the calculated value.

- Differentiate the polynomial term by term. - Evaluate the derivative at the specified input. - Compare the result with the slope of a horizontal line.

1. Differentiate term by term: \(f'(x)=3x^2-6x+3\). 2. Evaluate at \(x=1\): \(f'(1)=0\). 3. Since a horizontal tangent has slope \(0\), the graph has a horizontal tangent at \(x=1\).

\(f'(x)=3x^2-6x+3\) and \(f'(1)=0\), so the tangent is horizontal at \(x=1\).

- Apply the Power Rule to the quadratic term. - Keep the constant factor attached to each differentiated term. - Evaluate only after finding the derivative function.

1. Differentiate term by term: \(f'(x)=4x-3\). 2. Evaluate the derivative: \(f'(1)=1\). 3. The instantaneous rate of change at \(x=1\) is \(1\).

\(f'(1)=1\)

- Differentiate each term separately. - The derivative of the constant term is \(0\). - Substitute \(x=2\) only after differentiating.

1. Differentiate term by term: \(g'(x)=3x^2-8x\). 2. Evaluate at \(x=2\): \(g'(2)=12-16=-4\). 3. The graph has slope \(-4\) at \(x=2\).

\(g'(2)=-4\)

- Rearrange the average-rate relationship to recover total change. - The interval length is the difference between the two times. - Interpret the sign using the direction of the volume change.

1. The time interval has length \(7-4=3\,\text{min}\). 2. Total change equals average rate times elapsed time: \((-2.5\,\text{gal/min})(3\,\text{min})=-7.5\,\text{gal}\). 3. The negative sign means the tank contains less water at \(t=7\) than at \(t=4\); the volume decreased by \(7.5\,\text{gal}\).

a) \(-7.5\,\text{gal}\) b) The tank's volume decreased by \(7.5\,\text{gal}\) over the interval.

- Compute the hour-long average from total distance and total time. - Ask which information an interval average ignores about what happens inside the interval. - Distinguish a whole-interval quantity from a one-moment quantity.

1. Each cyclist's average speed over the hour is \(\frac{12\,\text{mi}}{1\,\text{h}}=12\,\text{mph}\). 2. An average speed depends only on the total distance and total elapsed time over the interval. 3. An instantaneous speed describes one moment, so the cyclists can have different values at \(1{:}30\) p.m. even though their one-hour averages are equal.

Both average \(12\,\text{mph}\) over the hour, but average speed does not determine the speed at a particular instant; the instantaneous speeds can therefore be \(8\,\text{mph}\) and \(16\,\text{mph}\).

- Compare how each line meets the function graph rather than using its label as a clue. - The average rate over an interval comes from a line through the two endpoint values. - The instantaneous rate at a point comes from the tangent slope there.

1. Line \(p\) passes through the two function points \((0,0)\) and \((2,4)\), so it is the secant for the interval and represents the average rate of change. 2. Its slope is \(\frac{4-0}{2-0}=2\). 3. Line \(q\) touches the graph at \((2,4)\) with the same local direction as the curve, so it is the tangent and represents the instantaneous rate at \(x=2\). 4. Using \((1,0)\) and \((2,4)\) on \(q\), its slope is \(\frac{4-0}{2-1}=4\).

a) \(p\), because it is the secant through the two interval endpoints. b) \(q\), because it is tangent to the graph at \(x=2\). c) Average rate: \(2\); instantaneous rate at \(x=2\): \(4\)

- Write the average-rate formula with the unknown endpoint value still in place. - Use the full input change from \(2\) to \(5\). - Solve the resulting linear equation for the unknown output.

1. Use the average-rate equation \(\frac{R(5)-R(2)}{5-2}=-3\). 2. Substitute \(R(2)=14\): \(\frac{R(5)-14}{3}=-3\). 3. Thus \(R(5)-14=-9\), so \(R(5)=5\).

\(R(5)=5\)

- Find the y-intercept by setting \(x=0\). - Find the x-intercepts by solving \(f(x)=0\). - The slope at an intercept is the derivative evaluated at that intercept's \(x\)-coordinate. - Be sure to evaluate the derivative at every distinct intercept input.

1. Differentiate: \(f'(x)=4x^3-15x^2+8x\). 2. The graph intersects the y-axis at \((0,0)\). 3. Factor to find the x-intercepts: \(f(x)=x^2(x^2-5x+4)=x^2(x-1)(x-4)\). Thus the distinct intercept inputs are \(x=0\), \(x=1\), and \(x=4\). 4. Evaluate the derivative: \(f'(0)=0\), \(f'(1)=-3\), and \(f'(4)=48\).

At \((0,0)\), the slope is \(0\). At \((1,0)\), the slope is \(-3\). At \((4,0)\), the slope is \(48\).

- First find the coordinates of all intercepts. - Differentiate the function to obtain its slope function. - Check the signs carefully when solving the quadratic equation. - Substitute each intercept's \(x\)-coordinate into the derivative.

1. Differentiate: \(g'(x)=x-2\). 2. The y-intercept is \((0,-6)\), and \(g'(0)=-2\). 3. To find the x-intercepts, solve \(0.5x^2-2x-6=0\). Multiplying by \(2\) gives \(x^2-4x-12=0=(x-6)(x+2)\), so \(x=6\) or \(x=-2\). 4. Evaluate the derivative: \(g'(6)=4\) and \(g'(-2)=-4\).

At the y-intercept \((0,-6)\), the slope is \(-2\). At the x-intercepts \((6,0)\) and \((-2,0)\), the slopes are \(4\) and \(-4\), respectively.

- Parallel lines have equal slopes. - Which coefficient in slope-intercept form gives the slope? - The derivative gives the tangent slope at each input. - After setting the derivative equal to the line's slope, solve the resulting quadratic equation.

1. The line \(y=8x-1\) has slope \(8\). 2. Differentiate: \(f'(x)=x^2+x-4\). 3. Parallel tangent lines must have slope \(8\), so solve \(x^2+x-4=8\). 4. This gives \(x^2+x-12=0=(x+4)(x-3)\). 5. Therefore, \(x=-4\) or \(x=3\).

\(x=-4\) and \(x=3\)

- The derivative gives the slope at a specific input. - Set a derivative equal to a required slope to form an equation. - Equal slopes at the same input mean the derivative values are equal. - Set the two derivative functions equal and solve for \(x\).

1. Differentiate: \(f'(x)=2x-4\) and \(g'(x)=x^2-4x+1\). 2. For part a), solve \(2x-4=6\), giving \(x=5\). 3. For part b), \(g'(5)=25-20+1=6\), so both graphs have slope \(6\) at \(x=5\). 4. For part c), set the derivatives equal: \(2x-4=x^2-4x+1\). 5. Rearranging gives \(x^2-6x+5=0=(x-1)(x-5)\). The solutions are \(x=1\) and \(x=5\), so the additional value is \(x=1\).

a) \(x=5\) b) Yes. Both graphs have slope \(6\) at \(x=5\). c) The other value is \(x=1\).

- What is the slope of a horizontal line? - After finding an \(x\)-coordinate, substitute it into the original function to find the point. - If two graphs have the same slope at the same input, their derivatives are equal there. - Solve the resulting quadratic equations by factoring or another valid method.

1. Differentiate: \(h'(x)=3x^2-12x+9\) and \(k'(x)=-3x+3\). 2. For part a), horizontal tangent lines have slope \(0\), so solve \(3x^2-12x+9=0\). 3. Dividing by \(3\) gives \(x^2-4x+3=0=(x-1)(x-3)\), so \(x=1\) or \(x=3\). 4. Evaluate the function: \(h(1)=6\) and \(h(3)=2\). The points are \((1,6)\) and \((3,2)\). 5. For part b), set the derivatives equal: \(3x^2-12x+9=-3x+3\). 6. Simplifying gives \(x^2-3x+2=0=(x-1)(x-2)\), so \(x=1\) or \(x=2\).

a) \((1,6)\) and \((3,2)\) b) \(x=1\) and \(x=2\)

- What equation identifies the point where the object reaches the ground? - The derivative at the impact point gives the tangent slope. - Use the magnitude of the slope when finding the acute angle with the horizontal. - Round the angle as requested.

1. Find where the shot put reaches the ground by solving \(h(x)=0\). The solutions are \(x=-2\) and \(x=10\), so the physically relevant impact point is \(x=10\). 2. Differentiate: \(h'(x)=-0.2x+0.8\). 3. Find the slope at impact: \(h'(10)=-0.2(10)+0.8=-1.2\). 4. The acute angle \(\alpha\) with the horizontal satisfies \(\tan(\alpha)=|-1.2|=1.2\). Thus, \(\alpha=\arctan(1.2)\approx 50.2^\circ\).

\(50.2^\circ\)

- What does the derivative tell you about the graph at a point? - For a smooth transition to the ground, which two quantities must match? - Use the relationship between slope and angle.

1. Differentiate: \(f'(x)=\frac{1}{4}x-1\). 2. At \(x=0\), the slope is \(f'(0)=-1\). The acute angle \(\alpha\) satisfies \(\tan(\alpha)=|-1|=1\), so \(\alpha=45^\circ\). 3. At \(x=4\), \(f(4)=\frac{1}{8}\cdot 4^2-4+2=0\), so the ramp reaches the ground. 4. Also, \(f'(4)=\frac{1}{4}\cdot 4-1=0\), which matches the slope of the horizontal line \(y=0\). Since both the height and slope agree, the ramp joins the ground smoothly at \((4,0)\).

a) \(45^\circ\) b) \(f(4)=0\) and \(f'(4)=0\), so the ramp and the ground line \(y=0\) have the same point and slope at \((4,0)\).

- A tangent continuation has the same value and slope as the curve at the transition point. - Differentiate the given function. - Use point-slope form for the tangent line. - Verify that the line passes through the point on the curve.

1. Find the transition point: \(g(2)=2^3-4\cdot 2^2+5\cdot 2=2\), so the point is \((2,2)\). 2. Differentiate: \(g'(x)=3x^2-8x+5\). 3. Find the slope at \(x=2\): \(g'(2)=3\cdot 2^2-8\cdot 2+5=1\). 4. Use point-slope form: \(y-2=1(x-2)\). 5. Simplify: \(y=x\).

\(y=x\)

- Find the point where the straight continuation begins. - A smooth connection requires the line to have the same slope as the curve. - Write the tangent line at \(x=10\). - Set the line's output equal to the target height.

1. Find the transition point: \(f(10)=0.01(10)^3-0.2(10)^2+1.5(10)+2=7\), so the point is \((10,7)\). 2. Differentiate: \(f'(x)=0.03x^2-0.4x+1.5\). 3. Find the slope at the transition: \(f'(10)=0.03(100)-0.4(10)+1.5=0.5\). 4. A smooth straight continuation is the tangent line: \(y-7=0.5(x-10)\), so \(y=0.5x+2\). 5. Set the height equal to \(12\): \(12=0.5x+2\). Solving gives \(x=20\).

\(x=20\,\text{m}\)

- A smooth transition means the linear model has the same value and slope at the joining time. - Find the temperature and rate of change at \(t=10\). - Write the linear model for times after \(10\) minutes. - Convert the final time to the requested unit.

1. Find the temperature at the transition: \(T(10)=0.1(10)^2-4(10)+80=50\). 2. Differentiate: \(T'(t)=0.2t-4\). The rate at \(t=10\) is \(T'(10)=-2\) degrees Celsius per minute. 3. A smooth linear continuation has the same value and rate at \(t=10\): \(L(t)-50=-2(t-10)\), so \(L(t)=-2t+70\). 4. Solve \(25=-2t+70\). This gives \(t=22.5\) minutes. 5. Convert to seconds: \(22.5\cdot 60=1350\) seconds.

\(1350\,\text{s}\)

- The derivative gives the local slope of the trail. - How are slope, percent grade, and angle of inclination related? - A tangent line requires both the point and the slope. - Use point-slope form for the tangent line.

1. Differentiate: \(h'(x)=-0.006x^2+0.12x\). 2. At \(x=5\), \(h'(5)=-0.006(25)+0.12(5)=0.45\). The percent grade is \(0.45\cdot 100\%=45\%\). 3. At \(x=15\), \(h'(15)=0.45\). Therefore, \(\alpha=\arctan(0.45)\approx 24.23^\circ\). 4. At \(x=25\), the slope is \(h'(25)=-0.75\), and \(h(25)=16.25\). The tangent line is \(y-16.25=-0.75(x-25)\), so \(y=-0.75x+35\).

a) \(45\%\) b) \(24.23^\circ\), approximately c) \(y=-0.75x+35\)

- A line perpendicular to the curve at a point is the normal line. - Use the derivative to find the curve's tangent slope. - Tangent and normal slopes are negative reciprocals. - Intersect the normal line with \(y=7\).

1. Differentiate: \(f'(x)=0.4x\). At \(x=5\), the tangent slope is \(f'(5)=2\). 2. The support follows the normal line, whose slope is \(-\frac{1}{2}\). 3. Through \(P(5,3)\), the normal line is \(y-3=-0.5(x-5)\), or \(y=-0.5x+5.5\). 4. Set \(y=7\): \(7=-0.5x+5.5\), so \(x=-3\). 5. Therefore, \(Q=(-3,7)\).

\(Q=(-3,7)\)

- Use the point-slope form of a tangent line. - Find both the function value and derivative value at \(x=0\). - A point on the tangent line must satisfy its equation. - Express the tangent slope and y-intercept in terms of \(k\).

1. Differentiate: \(f_k'(x)=ke^x-1\). 2. At \(x=0\), \(f_k(0)=k\) and \(f_k'(0)=k-1\). 3. The tangent line is \(y=(k-1)x+k\). 4. Substitute \((2, 5)\): \(5=2(k-1)+k\). 5. Solve \(5=3k-2\) to obtain \(k=\frac{7}{3}\).

a) \(f_k'(x)=ke^x-1\) b) \(k=\frac{7}{3}\)

- Find both the function value and slope at \(x=1\). - Simplify the tangent equation before applying the conditions. - Positive slope determines the sign of \(a\). - Set the tangent equation equal to \(0\) to find its x-intercept. - Translate the x-intercept condition into an inequality for \(a\).

1. At \(x=1\), \(g_a(1)=a+2\). 2. Differentiate: \(g_a'(x)=ae^{x-1}\), so the tangent slope at \(x=1\) is \(a\). 3. The tangent line is \(y-(a+2)=a(x-1)\), which simplifies to \(y=ax+2\). 4. Positive slope requires \(a>0\). 5. The x-intercept is \(x=-\frac{2}{a}\). Require \(-\frac{2}{a}>-1\). 6. Since \(a>0\), multiplication by \(a\) preserves the inequality: \(-2>-a\), so \(a>2\).

a) \(y=ax+2\) b) \(a>2\)

- A horizontal tangent has slope \(0\). - Treat the parameter \(k\) as a constant when differentiating with respect to \(x\). - Substitute the specified input into the derivative before solving for \(k\).

1. Treat \(k\) as a constant and differentiate: \(f_k'(x)=3kx^2-6x\). 2. A horizontal tangent has slope \(0\), so require \(f_k'(1)=0\). 3. Substitute \(x=1\): \(3k-6=0\). 4. Solving gives \(k=2\).

a) \(f_k'(x)=3kx^2-6x\) b) \(k=2\)

- The derivative gives the slope at each input. - What value must the derivative equal at the requested points? - Solve the resulting quadratic equation. - Check that you found every solution.

1. Differentiate: \(f'(x)=x^2-2x-5\). 2. Set the derivative equal to the required slope: \(x^2-2x-5=3\). 3. Rearrange: \(x^2-2x-8=0\). 4. Factor: \((x-4)(x+2)=0\). 5. Therefore, \(x=4\) or \(x=-2\).

\(x=-2\) and \(x=4\)

- Rewrite the square root as a rational power. - Interpret a derivative of speed with respect to distance using quotient units. - To show a rate decreases, examine the sign of its derivative. - Simplify the units after dividing speed units by distance units.

1. a) Rewrite \(v(s)=\sqrt{2g}\,s^{\frac{1}{2}}\). Then \(v'(s)=\frac{\sqrt{2g}}{2\sqrt{s}}=\sqrt{\frac{g}{2s}}\), for \(s>0\). 2. b) \(v'(2)=\sqrt{\frac{9.81}{4}}\approx1.566\,\text{s}^{-1}\), and \(v'(8)=\sqrt{\frac{9.81}{16}}\approx0.783\,\text{s}^{-1}\). 3. The derivative measures the increase in speed, in meters per second, per additional meter fallen. Its units simplify to \(\text{s}^{-1}\). 4. c) Differentiate again: \(v''(s)=-\frac{\sqrt{2g}}{4s^{\frac{3}{2}}}\). 5. Since \(v''(s)<0\) for every \(s>0\), \(v'(s)\) is strictly decreasing. Therefore, the speed gained per additional meter becomes smaller as the object falls farther.

a) \(v'(s)=\sqrt{\frac{g}{2s}}\) b) \(v'(2)\approx1.566\,\text{s}^{-1}\); \(v'(8)\approx0.783\,\text{s}^{-1}\). It is the increase in speed per additional meter fallen. c) \(v''(s)=-\frac{\sqrt{2g}}{4s^{\frac{3}{2}}}<0\), so \(v'(s)\) decreases as \(s\) increases.

- Rewrite the function before differentiating. - Write the tangent line at a general x-coordinate \(x_0\). - Use the fact that the origin lies on the tangent line. - Substitute the resulting x-coordinate into the original function.

1. Rewrite the function as \(f_k(x)=1+\frac{k}{x}\), so \(f_k'(x)=-\frac{k}{x^2}\). 2. Let the point of tangency have x-coordinate \(x_0\). The tangent line is \(y=f_k'(x_0)(x-x_0)+f_k(x_0)\). 3. Since the line passes through \((0, 0)\), \(0=-\frac{k}{x_0^2}(0-x_0)+1+\frac{k}{x_0}\). 4. Simplify: \(0=1+\frac{2k}{x_0}\), so \(x_0=-2k\). Since \(k>0\), this is a valid domain value. 5. The y-coordinate is \(f_k(-2k)=\frac{-2k+k}{-2k}=\frac{1}{2}\). Therefore, \(B=\left(-2k, \frac{1}{2}\right)\).

\(B=\left(-2k, \frac{1}{2}\right)\)

- Tangent slopes are the values taken by the first derivative. - After differentiating, determine the smallest value the derivative can have without substituting trial inputs. - Connect the sign and minimum value of the derivative to the directions of possible tangent lines.

1. Differentiate: \(f'(x)=3x^2+3\). 2. Since \(x^2\ge0\), \(f'(x)\ge3\) for every real \(x\). Therefore, the possible slopes are \([3,\infty)\), and every slope \(m<3\) is impossible. 3. Geometrically, the graph is strictly increasing everywhere. It has no horizontal or negative-slope tangent, and its smallest slope is \(3\) at \(x=0\).

1) \(f'(x)=3x^2+3\) 2) Every slope \(m<3\) is impossible. 3) The graph rises everywhere, and its minimum tangent slope is \(3\) at \(x=0\).

- Rewrite the function with a negative exponent. - Determine the sign of \(x^3\) on each side of zero. - A fraction with a nonzero constant numerator cannot equal zero.

1. Rewrite and differentiate: \(g(x)=-x^{-2}\), so \(g'(x)=2x^{-3}=\frac{2}{x^3}\). 2. For \(x<0\), \(x^3<0\), so \(g'(x)<0\). The left branch is strictly decreasing. 3. For \(x>0\), \(x^3>0\), so \(g'(x)>0\). The right branch is strictly increasing. 4. The derivative can never equal zero because its numerator is the nonzero constant \(2\). Every nonzero real slope occurs, but \(m=0\) is excluded. 5. Geometrically, neither branch has a horizontal tangent.

For \(x<0\), slopes are negative. For \(x>0\), slopes are positive. The only excluded slope is \(m=0\).

- First find the derivative function. - Set the derivative equal to each required value. - Solve each resulting equation for \(x\). - An equation of the form \(x^2=a\) can have two real solutions.

1. Differentiate: \(f'(x)=x^2-\frac{1}{4}\). 2. For \(f'(x)=0\), solve \(x^2=\frac{1}{4}\), giving \(x=\pm\frac{1}{2}\). 3. For \(f'(x)=\frac{3}{4}\), solve \(x^2=1\), giving \(x=\pm1\). 4. For \(f'(x)=2\), solve \(x^2=\frac{9}{4}\), giving \(x=\pm\frac{3}{2}\).

For \(f'(x)=0\): \(x\in\left\{-\frac{1}{2},\frac{1}{2}\right\}\) For \(f'(x)=\frac{3}{4}\): \(x\in\{-1,1\}\) For \(f'(x)=2\): \(x\in\left\{-\frac{3}{2},\frac{3}{2}\right\}\)

- Which function gives the tangent slope at each input? - Set the derivative equal to the proposed slope. - What values can the square of a real number take?

1. Differentiate: \(f'(x)=3x^2+3\). 2. A tangent line with slope \(2\) would require \(3x^2+3=2\). 3. This equation simplifies to \(x^2=-\frac{1}{3}\), which has no real solution. 4. Equivalently, since \(x^2\geq0\), \(f'(x)=3x^2+3\geq3\) for every real \(x\).

\(f'(x)=3x^2+3\geq3\) for every \(x\in\mathbb{R}\), so the derivative never equals \(2\).

- Find the derivative. - Use the range of sine. - Account for the constant factor and negative sign. - Report the values outside the derivative’s range.

1. Differentiate: \(g'(x)=-4\sin x\). 2. Since \(-1\le\sin x\le 1\), the derivative satisfies \(-4\le g'(x)\le 4\). Therefore, tangent slopes outside this interval cannot occur.

The impossible slopes are \(m<-4\) or \(m>4\).

- Use symmetry to locate the right edge from the total width. - Convert a percent grade to a decimal slope. - The derivative at the edge gives a condition on \(a\). - Evaluate the model at the edge to find the depth.

1. Because the bowl is \(12\,\text{m}\) wide and symmetric, the right edge is at \(x=6\). 2. A grade of \(120\%\) corresponds to slope \(1.2\), so \(f'(6)=1.2\). 3. Since \(f'(x)=2ax\), the condition gives \(12a=1.2\), so \(a=0.1\). Thus \(f(x)=0.1x^2\). 4. The depth is the height of the edge above the origin: \(f(6)=0.1(6)^2=3.6\,\text{m}\).

a) \(f(x)=0.1x^2\) b) \(3.6\,\text{m}\)

- The tangent slope at \(x=-1\) is \(g'(-1)\). - Use the given slope to form an equation containing only \(a\). - Find the point of tangency after determining \(a\). - Use the point-slope form of a line.

1. Differentiate: \(g'(x)=2ax+4\). 2. Use the given slope: \(g'(-1)=2\). Therefore, \(-2a+4=2\), so \(a=1\). 3. With \(a=1\), the point of tangency is \((-1,g(-1))=(-1,-5)\). 4. A line with slope \(2\) through \((-1,-5)\) satisfies \(y+5=2(x+1)\), so \(y=2x-3\).

a) \(a=1\) b) \(y=2x-3\)

- Use only odd powers for origin symmetry. - Translate the slope condition into an equation involving the derivative. - Leave one coefficient as a parameter when finding all possible functions. - Use the additional point to determine the parameter.

1. A cubic polynomial symmetric about the origin has the form \(f(x)=ax^3+cx\), where \(a\ne0\). 2. Since \(f'(x)=3ax^2+c\), the slope condition \(f'(1)=6\) gives \(3a+c=6\), so \(c=6-3a\). 3. Therefore, all such cubic polynomials are \(f_a(x)=ax^3+(6-3a)x\), where \(a\in\mathbb{R}\setminus\{0\}\). 4. To pass through \(Q(2, 20)\), the polynomial must satisfy \(8a+2(6-3a)=20\). Thus, \(2a+12=20\), so \(a=4\) and \(c=-6\). 5. The required polynomial is \(f(x)=4x^3-6x\).

All such polynomials: \(f_a(x)=ax^3+(6-3a)x\), where \(a\in\mathbb{R}\setminus\{0\}\) Polynomial through \(Q(2, 20)\): \(f(x)=4x^3-6x\)

- Tangency to the x-axis gives both a zero and a zero slope. - Use the origin to determine the constant term. - Translate each remaining condition into an equation. - Solve the resulting linear system.

1. Let \(f(x)=ax^3+bx^2+cx+d\). Passing through the origin gives \(d=0\). 2. The point \(P(2, 4)\) gives \(8a+4b+2c=4\). Tangency to the x-axis at \(x=4\) gives \(f(4)=0\) and \(f'(4)=0\), so \(64a+16b+4c=0\) and \(48a+8b+c=0\). 3. Solving the system gives \(a=\frac{1}{2}\), \(b=-4\), and \(c=8\). 4. Therefore, \(f(x)=\frac{1}{2}x^3-4x^2+8x\).

\(f(x)=\frac{1}{2}x^3-4x^2+8x\)

- Recall the derivative of \(\log_bx\). - Substitute the given input and slope. - Solve the resulting equation for \(\ln b\). - Exponentiate to solve for \(b\).

1. Differentiate: \(h'(x)=\frac{1}{x\ln b}\). 2. Use the given slope: \(\frac{1}{2\ln b}=0.5=\frac{1}{2}\). Therefore, \(\ln b=1\), so \(b=e\).

\(b=e\)

- Use the given point to obtain one equation in the parameters. - A horizontal tangent gives a condition on the first derivative. - Apply the quotient rule before substituting the tangent input.

1. The point condition gives \(f(4)=8\): \(\frac{16a+4b}{2}=8\), so \(4a+b=4\). 2. By the quotient rule, \(f'(x)=\frac{(2ax+b)(x-2)-(ax^2+bx)}{(x-2)^2}=\frac{ax^2-4ax-2b}{(x-2)^2}\). 3. A horizontal tangent at \(x=0\) requires \(f'(0)=0\), so \(b=0\). 4. Substituting \(b=0\) into \(4a+b=4\) gives \(a=1\).

\(a=1\) and \(b=0\)

- Convert each graph condition into an equation involving \(f\) or \(f'\). - Differentiate the general function before using the slope condition. - Use exact sine and cosine values at \(0\), \(\frac{\pi}{2}\), and \(\pi\). - Solve the resulting linear system for the three parameters.

1. Translate the conditions into equations: \(f(0)=4\), \(f\left(\frac{\pi}{2}\right)=1\), and \(f'(0)=2\). 2. Differentiate: \(f'(x)=a\cos x-2b\sin(2x)\). From \(f'(0)=2\), \(a=2\). 3. From \(f(0)=4\), \(b+c=4\). 4. From \(f\left(\frac{\pi}{2}\right)=1\), \(2-b+c=1\), so \(-b+c=-1\). 5. Solve the system \(b+c=4\) and \(-b+c=-1\). This gives \(b=\frac{5}{2}\) and \(c=\frac{3}{2}\).

\(a=2\), \(b=\frac{5}{2}\), and \(c=\frac{3}{2}\) Therefore, \(f(x)=2\sin x+\frac{5}{2}\cos(2x)+\frac{3}{2}\).

- Write the tangent-line equation at a general input \(x_0\). - Use the derivative for the tangent slope. - Substitute the sensor coordinates into the tangent-line equation. - Solve the resulting quadratic equation. - Check each solution against the stated domain.

1. Differentiate: \(f'(x)=-0.5x\). 2. The tangent line at \(x=x_0\) is \(t(x)=f'(x_0)(x-x_0)+f(x_0)\). Substitution and simplification give \(t(x)=-0.5x_0x+0.25x_0^2+4\). 3. Since the line must pass through \(S(4,1)\), set \(t(4)=1\): \(1=-2x_0+0.25x_0^2+4\). 4. Rearranging gives \(x_0^2-8x_0+12=0\), so \((x_0-2)(x_0-6)=0\). Thus, \(x_0=2\) or \(x_0=6\). 5. Only \(x_0=2\) lies in the rail's domain \([0,4]\).

The laser must be at \(x_0=2\).

- Find the vertex of the parabola. - Translate “stay at or above the roadway” into an inequality for the minimum value. - Differentiate to find the slope at the tower. - Use the tangent relationship between slope and angle.

1. The parabola's lowest point occurs halfway between the towers, at \(x=50\). Its height is \(f_a(50)=30-2500a\). 2. To stay at or above the roadway, the minimum height must satisfy \(30-2500a\ge0\). Therefore, \(a\le0.012\). Together with \(a>0\), this gives \(0<a\le0.012\). 3. If the minimum height is \(5\,\text{m}\), then \(30-2500a=5\), so \(a=0.01\). 4. For \(a=0.01\), \(f_a'(x)=0.02x-1\). At \(x=0\), the slope is \(-1\). The acute angle relative to the horizontal is \(\arctan(1)=45^\circ\).

a) \(0<a\le0.012\) b) \(a=0.01\) c) \(45^\circ\)

- Perpendicular lines have slopes that are negative reciprocals. - Use the derivative to find the walking path's tangent slope at \(A\). - Write the connector line through \(A\). - Set the connector and creek equations equal to find their intersection.

1. Differentiate: \(f'(x)=-x+2\). At \(x=3\), the tangent slope is \(f'(3)=-1\). 2. The perpendicular connector has slope \(1\). 3. Through \(A(3,3)\), its equation is \(y-3=x-3\), or \(y=x\). 4. Intersect this line with the creek: \(x=-2x+6\). Thus, \(3x=6\), so \(x=2\) and \(y=2\). 5. Therefore, the bridge location is \(B=(2,2)\).

The connector path is \(y=x\), and the bridge is at \(B=(2,2)\).

- A logarithm requires a positive argument. - Apply the chain rule to \(\ln(u(x))\). - Use the monotonicity and zero of \(f\) to determine where \(f(x)>0\). - Express \(g'\) in terms of \(f\) and \(f'\), then simplify the equation. - Read the required function value from the graph.

1. For part a, require \(3x-5>0\), so \(D_h=\left(\frac{5}{3}, \infty\right)\). The zero satisfies \(3x-5=1\), giving \(x=2\). By the chain rule, \(h'(x)=\frac{3}{3x-5}\). 2. Since \(f\) is strictly increasing and has its only zero at \(-2\), \(f(x)>0\) exactly when \(x>-2\). Therefore, \(D_g=(-2, \infty)\). Also, \(g'(x)=\frac{f'(x)}{f(x)}\). The equation \(\frac{f'(x)}{f(x)}=\frac{1}{2}f'(x)\) can be divided by \(f'(x)>0\), giving \(f(x)=2\). From the graph, \(f(x)=2\) at \(x=2\).

a) \(D_h=\left(\frac{5}{3}, \infty\right)\), zero at \(x=2\), and \(h'(x)=\frac{3}{3x-5}\) b) 1. \(D_g=(-2, \infty)\) 2. \(x=2\)

- First find the exact takeoff point on the ramp. - Use the derivative to find the ramp's slope at takeoff. - Model the initial path with the tangent line. - Substitute the wall's x-coordinate into the tangent-line equation.

1. Find the takeoff point: \(f(2)=-0.1\cdot2^3+0.6\cdot2^2=1.6\), so the point is \((2,1.6)\). 2. Differentiate: \(f'(x)=-0.3x^2+1.2x\). 3. Find the slope at takeoff: \(f'(2)=-0.3\cdot2^2+1.2\cdot2=1.2\). 4. The tangent line is \(y-1.6=1.2(x-2)\), or \(y=1.2x-0.8\). 5. At the wall, \(y=1.2\cdot6-0.8=6.4\). Therefore, \(h=6.4\,\text{m}\).

\(h=6.4\,\text{m}\)

- Set the two function expressions equal to find all intersections. - Differentiate both functions to obtain the tangent slopes at each common input. - Use the acute angle-between-lines relation once the slopes are known.

1. Set \(\frac{1}{4}x^3=x\). Then \(x(x^2-4)=0\), so the intersection points are \((-2,-2)\), \((0,0)\), and \((2,2)\). 2. The tangent slopes are given by \(f'(x)=\frac{3}{4}x^2\) and \(g'(x)=1\). 3. At \((0,0)\), the slopes are \(0\) and \(1\), so the acute angle is \(45^\circ\). 4. At \((-2,-2)\) and \((2,2)\), the slopes are \(3\) and \(1\). Thus, \(\tan\theta=\frac{1}{2}\), so \(\theta=\arctan\!\left(\frac{1}{2}\right)\approx26.6^\circ\).

a) \((-2,-2)\), \((0,0)\), and \((2,2)\) b) At \((0,0)\), the acute angle is \(45^\circ\). At the other two points, it is \(\arctan\!\left(\frac{1}{2}\right)\approx26.6^\circ\).

- Use the midpoint coordinates in the cable model. - The derivative gives the slope of the cable at the tower. - Relate slope to angle using the tangent function. - For an upward-opening parabola, the minimum occurs at the vertex.

1. At the midpoint, \(x=0\). Set the height equal to \(30\): \(50-2500k=30\). Thus, \(k=\frac{20}{2500}=\frac1{125}\). 2. The right tower is at \(x=50\). Differentiate: \(f_k'(x)=2kx\). For \(k=\frac1{125}\), the slope at the tower is \(f_k'(50)=\frac45\). 3. If \(\theta\) is the angle of inclination, then \(\tan(\theta)=\frac45\), so \(\theta=\arctan\left(\frac45\right)\approx38.66^\circ\). 4. Because \(k>0\), the lowest point is at \(x=0\). The cable touches or passes below the roadway when \(50-2500k\le0\). Therefore, \(k\ge\frac1{50}\). It first touches the roadway when \(k=\frac1{50}\).

a) \(k=\frac1{125}\) b) \(\theta\approx38.66^\circ\) c) \(k\ge\frac1{50}\); it first touches the roadway at \(k=\frac1{50}\).

- Landing at the same height means the function value is zero at the basin. - Use the vertex of the downward-opening parabola. - Convert slope to angle with the inverse tangent.

1. The basin condition is \(h_c(4)=0\): \(-0.5(4)^2+4c=0\). Thus \(-8+4c=0\), so \(c=2\). 2. For \(c=2.5\), the vertex occurs at \(x=2.5\). The maximum height is \(h_{2.5}(2.5)=3.125\,\text{m}\). 3. The initial slope is \(h_c'(0)=c\). For \(c=1.5\), \(\tan(\alpha)=1.5\), so \(\alpha=\arctan(1.5)\approx56.31^\circ\).

a) \(c=2\) b) \(3.125\,\text{m}\) c) \(56.31^\circ\)

- Use the power rule to find the derivative. - The derivative value is the tangent slope. - Parallel lines have equal slopes. - Substitute each resulting input into the original function to find the full point.

1. Differentiate: \(f'(x)=2x-2\). 2. For slope \(2\), solve \(2x-2=2\), giving \(x=2\). Then \(f(2)=-1\), so \(P=(2,-1)\). 3. A line parallel to \(y=-4x+5\) has slope \(-4\). Solve \(2x-2=-4\), giving \(x=-1\). Then \(f(-1)=2\), so \(Q=(-1,2)\).

a) \(f'(x)=2x-2\) b) \(P=(2,-1)\) c) \(Q=(-1,2)\)

- Differentiate the polynomial term by term. - A horizontal tangent has slope \(0\). - A tangent parallel to \(y=x\) must have slope \(1\). - Substitute each solved input into \(f\) to recover the point.

1. Differentiate term by term: \(f'(x)=-x+3\). 2. For a horizontal tangent, solve \(-x+3=0\), giving \(x=3\). Since \(f(3)=4.5\), \(S=(3,4.5)\). 3. The line \(y=x\) has slope \(1\). Solve \(-x+3=1\), giving \(x=2\). Since \(f(2)=4\), \(R=(2,4)\).

a) \(f'(x)=-x+3\) b) \(S=(3,4.5)\) c) \(x=2\), and \(R=(2,4)\)

- Add corresponding y-values of the two component functions. - Parallel lines have equal slopes. - Use the bound \(-1\le\sin x\le1\) for the final positivity argument.

1. Since \(f(x)=g(x)+h(x)\), the graph of \(f\) is obtained by adding the corresponding y-values, so it oscillates around \(y=x\). 2. The line \(g\) has slope \(1\). Since \(f'(x)=1+1.5\cos x\), parallel tangents satisfy \(\cos x=0\), giving \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\) on the stated interval. 3. For \(0<x\le\pi\), \(\sin x\ge0\), so \(f(x)>0\). For \(x>\pi\), \(f(x)\ge x-1.5>\pi-1.5>0\).

a) The graph of \(f\) is the pointwise sum of \(g\) and \(h\), so it oscillates around \(y=x\). b) \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\) c) For \(0<x\le\pi\), both terms are nonnegative. For \(x>\pi\), \(f(x)\ge x-1.5>0\).

- Compare the shapes and formulas of the three functions. - Set the derivative equal to the desired slope in part b. - A horizontal tangent requires the derivative to equal \(0\). - Use the unit circle for the resulting sine equations.

1. Graph B is the line through the origin, so it represents \(g\). Graph C is the reflected cosine curve, so it represents \(h\). Their pointwise sum is graph A, which represents \(f\). 2. Since \(f'(x)=0.5+\sin x\), slope \(0.5\) occurs when \(\sin x=0\). On \([0,2\pi]\), the inputs are \(0\), \(\pi\), and \(2\pi\), giving points \((0,-1)\), \(\left(\pi,\frac{\pi}{2}+1\right)\), and \((2\pi,\pi-1)\). 3. Horizontal tangents satisfy \(\sin x=-\frac{1}{2}\), so \(x=\frac{7\pi}{6}+2\pi n\) or \(x=\frac{11\pi}{6}+2\pi n\), where \(n\in\mathbb{Z}\).

a) A: \(f\); B: \(g\); C: \(h\) b) \((0,-1)\), \(\left(\pi,\frac{\pi}{2}+1\right)\), and \((2\pi,\pi-1)\) c) \(x=\frac{7\pi}{6}+2\pi n\) or \(x=\frac{11\pi}{6}+2\pi n\), where \(n\in\mathbb{Z}\)

- Interpret a function value as height and a derivative value as slope. - Use vertex form because the highest point is given. - The water reaches the surface at a zero of the function. - The tangent of the launch angle equals the slope at the nozzle.

1. The condition \(f(0)=2\) means the water leaves the nozzle \(2\,\text{ft}\) above the surface. The condition \(f'(4)=0\) means the path has a horizontal tangent at a horizontal distance of \(4\,\text{ft}\), where it reaches its maximum height. 2. Using vertex form with vertex \((4, 10)\), write \(f(x)=a(x-4)^2+10\). Since \(f(0)=2\), \(16a+10=2\), so \(a=-\frac{1}{2}\). Thus, \(f(x)=-\frac{1}{2}(x-4)^2+10=-\frac{1}{2}x^2+4x+2\). 3. Set \(f(x)=0\): \(-\frac{1}{2}x^2+4x+2=0\), or \(x^2-8x-4=0\). The positive solution is \(x=4+2\sqrt{5}\approx8.47\). The stream hits the water about \(8.47\,\text{ft}\) from the nozzle. 4. The initial slope is \(f'(0)\). Since \(f'(x)=-x+4\), \(f'(0)=4\). Therefore, \(\tan(\theta)=4\), so \(\theta=\arctan(4)\approx76.0^\circ\).

a) \(f(0)=2\) gives the nozzle's height, and \(f'(4)=0\) identifies the horizontal tangent at the maximum height. b) \(f(x)=-\frac{1}{2}x^2+4x+2\) c) \(4+2\sqrt{5}\,\text{ft}\approx8.47\,\text{ft}\) d) \(\theta\approx76.0^\circ\)

- 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.

- Use endpoint position change over endpoint time change for part a). - For part b), use the two times that bracket \(t=2\) most closely. - Keep the time interval used for each rate clear when comparing them.

1. The average velocity from \(t=1\) to \(t=3\) is \(\frac{27-1}{3-1}=13\,\text{m/s}\). 2. A local secant estimate around \(t=2\) is \(\frac{9.261-6.859}{2.1-1.9}=12.01\,\text{m/s}\). 3. The local estimate is slightly smaller than the longer-interval average: \(12.01\,\text{m/s}\) versus \(13\,\text{m/s}\).

a) \(13\,\text{m/s}\) b) Approximately \(12.01\,\text{m/s}\) c) The estimated instantaneous velocity is slightly smaller than the average velocity over \([1,3]\).

- Treat radius as the input and volume as the output. - Compute output change before dividing by input change. - The units should be volume units per radius unit.

1. The change in volume is \(\frac{32\pi}{3}-\frac{4\pi}{3}=\frac{28\pi}{3}\,\text{in}^3\). 2. The change in radius is \(2-1=1\,\text{in}\). 3. The average rate is \(\frac{28\pi}{3}\,\text{in}^3/\text{in}\).

\(\frac{28\pi}{3}\,\text{in}^3/\text{in}\)

- Represent each average rate as a secant slope. - Equal average rates correspond to equal secant slopes. - Keep the input differences in the same order as the output differences.

1. The slope of \(\overline{AB}\) is \(\frac{10-2}{5-1}=2\). 2. The slope of \(\overline{CD}\) is \(\frac{y-3}{2-(-2)}=\frac{y-3}{4}\). 3. Set the average rates equal: \(\frac{y-3}{4}=2\). 4. Solving gives \(y=11\).

\(y=11\)

- Use the y-intercept condition before differentiating. - Relate the stated angle with the y-axis to a line's angle of inclination. - Evaluate the derivative at the y-intercept and compare that slope with the possible line directions. - Remember to enforce the restriction on \(k\).

1. The y-intercept gives \(f_{a,k}(0)=a=3\). 2. Differentiate using the product and chain rules: \(f_{a,k}'(x)=(1+kx+ak)e^{kx}\). 3. At \(x=0\), the tangent slope is \(1+3k\). 4. A line making a \(45^\circ\) angle with the y-axis also makes a \(45^\circ\) or \(135^\circ\) angle of inclination with the x-axis, so its slope is \(1\) or \(-1\). 5. The equation \(1+3k=1\) gives \(k=0\), which is excluded. The equation \(1+3k=-1\) gives \(k=-\frac{2}{3}\). 6. Therefore, \(a=3\) and \(k=-\frac{2}{3}\).

\(a=3\) and \(k=-\frac{2}{3}\)

- Differentiate the general function \(f_n(x)=x^n\). - How many real solutions can \(x^k=c\), with \(c>0\), have when \(k\) is even or odd? - Connect the parity of the exponent in the derivative to the number of solutions. - Check \(n=1\) separately.

1. Differentiate: \(f_n'(x)=nx^{n-1}\). The slope condition is \(nx^{n-1}=10\), or \(x^{n-1}=\frac{10}{n}\). 2. Since \(\frac{10}{n}>0\), the equation has exactly two real solutions when the positive exponent \(n-1\) is even. 3. The case \(n=1\) must be checked separately: \(f_1'(x)=1\), so there are no solutions. 4. For \(n=3\), \(x^2=\frac{10}{3}\) has two solutions. For \(n=5\), \(x^4=2\) has two solutions. 5. For \(n=2,4,6\), the exponent \(n-1\) is odd, so each equation has exactly one real solution.

\(n=3\) and \(n=5\)

- The upper edges lie on the x-axis, so use their coordinates in the model. - A \(50\%\) grade corresponds to derivative value \(0.5\). - The parabola's lowest point occurs at its vertex. - Report depth as a positive distance.

1. The top edges are at \(x=\pm10\), and \(f(10)=0\). 2. Since \(f'(x)=2ax\) and \(f'(10)=0.8\), \(20a=0.8\), so \(a=0.04\). 3. Use \(f(10)=0\): \(0.04(10)^2+c=0\), giving \(c=-4\). Thus \(f(x)=0.04x^2-4\). 4. The lowest point is \(f(0)=-4\), so the maximum depth is \(4\,\text{m}\). 5. A \(50\%\) slope is \(0.5\). Solve \(f'(x)=0.08x=0.5\), giving \(x=6.25\).

a) \(f(x)=0.04x^2-4\) b) \(4\,\text{m}\) c) \(6.25\,\text{m}\) from the center

- A tangential pulse follows the tangent line at the release point. - Use the derivative to find the tangent slope at each candidate input. - Write each tangent line using its release point and slope. - Evaluate each tangent line at the sensor's x-coordinate.

1. Differentiate: \(g'(x)=0.5\cos(0.5x)\). 2. At \(x_1=\frac{3\pi}{2}\approx4.712\), \(g(x_1)\approx2.707\) and \(g'(x_1)\approx-0.354\). The tangent gives \(y\approx-0.354(10-4.712)+2.707\approx0.835\), which does not match the sensor height. 3. At \(x_2=2\pi\approx6.283\), \(g(x_2)\approx2.000\) and \(g'(x_2)\approx-0.500\). The tangent gives \(y\approx-0.500(10-6.283)+2.000\approx0.142\), which matches the sensor height. 4. At \(x_3=\frac{5\pi}{2}\approx7.854\), \(g(x_3)\approx1.293\) and \(g'(x_3)\approx-0.354\). The tangent gives \(y\approx-0.354(10-7.854)+1.293\approx0.533\), which does not match the sensor height. 5. Therefore, the pulse should be released at \(x_2=2\pi\).

\(x_2=2\pi\)

- A point on the path gives a condition on the function value. - At the highest point, the tangent is horizontal. - Start with a general quadratic and solve for its coefficients. - The tangent slope and the angle with the horizontal are related by tangent.

1. The takeoff condition is \(g(0)=0\). A horizontal tangent at the highest point gives \(g'(6)=0\). The remaining given point gives \(g(10)=5\). 2. Let \(g(x)=ax^2+bx+c\). From \(g(0)=0\), \(c=0\). Since \(g'(x)=2ax+b\), the condition \(g'(6)=0\) gives \(12a+b=0\), so \(b=-12a\). Using \(g(10)=5\), \(100a+10b=5\). Substitution gives \(-20a=5\), so \(a=-\frac{1}{4}\) and \(b=3\). Therefore, \(g(x)=-\frac{1}{4}x^2+3x\). 3. The maximum occurs at \(x=6\): \(g(6)=-\frac{1}{4}(36)+18=9\). The maximum height is \(9\,\text{m}\). 4. The slope at landing is \(g'(12)=-\frac{1}{2}(12)+3=-3\). If \(\theta\) is the magnitude of the angle with the horizontal ground, then \(\tan(\theta)=3\). Thus \(\theta=\arctan(3)\approx71.6^\circ\).

a) \(g(0)=0\) and \(g'(6)=0\). b) \(g(x)=-\frac{1}{4}x^2+3x\). c) \(9\,\text{m}\). d) \(\arctan(3)\approx71.6^\circ\).