Free math worksheets

- Identify the exponent on \(x\). - Recall how the exponent changes when the power rule is applied. - The derivative of a nonconstant monomial has degree one less than the original monomial.

1. Apply the power rule to \(x^6\). 2. The derivative is \(f'(x)=6x^5\).

\(f'(x)=6x^5\)

- Separate the constant coefficient from the power of \(x\). - Check what happens to both the coefficient and the exponent under the power rule. - Compare the student's coefficient with the original exponent.

1. The coefficient \(5\) remains a factor, but the exponent \(3\) must also multiply the coefficient when the power rule is applied. 2. Therefore, \(g'(x)=15x^2\).

The student forgot to multiply by the original exponent \(3\). The correct derivative is \(g'(x)=15x^2\).

- The units of a derivative are the output units divided by the input units. - What physical quantity is distance traveled per unit of time? - Use the power rule to differentiate the function. - Include units after substituting \(t = 4\).

1. Since \(s\) is measured in meters and \(t\) in seconds, the derivative \(s'(t)\) has units \(\frac{\text{m}}{\text{s}}\). 2. The derivative of distance traveled with respect to time is the vehicle’s instantaneous speed at time \(t\). 3. By the power rule, \(s'(t) = t+2\). Therefore, \(s'(4) = 4+2 = 6\,\frac{\text{m}}{\text{s}}\).

1. \(\frac{\text{m}}{\text{s}}\) 2. \(s'(t)\) is the vehicle’s instantaneous speed at time \(t\). 3. \(s'(4) = 6\,\frac{\text{m}}{\text{s}}\)

- The derivative at a point gives the tangent-line slope there. - Differentiate each term of the function. - Substitute \(x = 3\) into the derivative. - Interpret the sign of the result.

1. Differentiate using the power and sum rules: \(f'(x) = x-4\). 2. Evaluate at \(x = 3\): \(f'(3) = 3-4 = -1\). 3. Because \(f'(3) < 0\), the tangent line has negative slope.

The tangent line has negative slope because \(f'(3) = -1\).

- Instantaneous rate of change means the derivative at the point. - First find the derivative function. - Substitute \(x = 4\). - Use the sign of the result, not only its magnitude.

1. Differentiate using the power rule: \(g'(x) = 6x^2 - 18x\). 2. Evaluate at \(x = 4\): \(g'(4) = 6(4)^2-18(4) = 96-72 = 24\). 3. Since \(24 > 0\), the instantaneous rate of change is positive.

The instantaneous rate of change is positive because \(g'(4) = 24\).

- Instantaneous rate of change is the derivative at the given point. - Find a function that gives the slope at any x-value. - Substitute \(x = 2\) after differentiating.

1. Differentiate using the power and sum rules: \(f'(x) = 6x^2-18x+12\). 2. Evaluate at \(x = 2\): \(f'(2) = 6(2)^2-18(2)+12 = 24-36+12 = 0\).

\(f'(2) = 0\)

- A secant slope gives the average rate between two times. - A derivative gives the tangent slope at one time. - Interpret each rate as volume added per second.

1. Evaluate the volumes: \(V(2) = 6\,\text{L}\) and \(V(5) = 22.5\,\text{L}\). 2. The average inflow rate is \(\frac{V(5)-V(2)}{5-2} = \frac{22.5-6}{3} = 5.5\,\frac{\text{L}}{\text{s}}\). 3. Differentiate: \(V'(t) = t+2\). Thus, \(V'(4) = 6\,\frac{\text{L}}{\text{s}}\). 4. The average rate describes the mean volume increase per second throughout \([2, 5]\). The instantaneous rate describes the inflow at the single time \(t = 4\).

a) \(5.5\,\frac{\text{L}}{\text{s}}\) b) \(6\,\frac{\text{L}}{\text{s}}\) c) The first is an average over an interval; the second is the rate at one exact time.

- The slope of a position-time graph represents velocity. - Differentiate the position function. - The derivative’s units are meters per second.

1. Instantaneous velocity is the derivative of the position function. 2. Differentiate: \(s'(t) = 0.8t+3\). 3. Evaluate at \(t = 5\): \(s'(5) = 0.8(5)+3 = 7\,\frac{\text{m}}{\text{s}}\). 4. In context, \(s'(t)\) gives the car’s instantaneous velocity at time \(t\).

a) \(7\,\frac{\text{m}}{\text{s}}\) b) \(s'(t)\) is the car’s instantaneous velocity at time \(t\).

- Average growth uses the change over a time interval. - Instantaneous growth uses the derivative at one time. - Keep track of the population unit, which is thousands of bacteria.

1. Evaluate the population values: \(B(2) = 122\) and \(B(4) = 168\), in thousands of bacteria. 2. The average growth rate is \(\frac{B(4)-B(2)}{4-2} = \frac{168-122}{2} = 23\) thousand bacteria per hour. 3. Differentiate: \(B'(t) = 6t+5\). Then \(B'(3) = 6(3)+5 = 23\) thousand bacteria per hour.

a) \(23\) thousand bacteria per hour b) \(23\) thousand bacteria per hour

- Use the slope formula for the secant line. - The derivative gives the tangent-line slope. - Compare the two numerical slopes.

1. Evaluate the function: \(f(-2) = \frac{1}{2}(-2)^2-2 = 0\), and \(f(2) = \frac{1}{2}(2)^2+2 = 4\). 2. The secant slope is \(\frac{f(2)-f(-2)}{2-(-2)} = \frac{4-0}{4} = 1\). 3. Differentiate: \(f'(x) = x+1\). The tangent slope at \(x = 0\) is \(f'(0) = 1\). 4. Both slopes are \(1\), so the secant and tangent lines are parallel.

1. The secant slope is \(1\). 2. The tangent slope is \(f'(0) = 1\), so the slopes are equal.

- Instantaneous rate of change is the derivative at a point. - Apply the power rule. - Substitute each x-value into the derivative.

1. Differentiate using the power rule: \(f'(x) = -2x\). 2. Evaluate at \(x = 2\): \(f'(2) = -2(2) = -4\). 3. Evaluate at \(x = -1\): \(f'(-1) = -2(-1) = 2\).

At \(x = 2\): \(-4\) At \(x = -1\): \(2\)

- The derivative gives tangent-line slope. - Expanding the square first may make differentiation easier. - Evaluate the derivative at each requested point.

1. Expand the function: \(g(x) = x^2+2x+1\). 2. Differentiate: \(g'(x) = 2x+2\). 3. At \(x = 1\), \(g'(1) = 2(1)+2 = 4\). 4. At \(x = -3\), \(g'(-3) = 2(-3)+2 = -4\).

At \(x = 1\): \(4\) At \(x = -3\): \(-4\)

- The derivative gives the slope of the graph at each x-value. - Set the derivative equal to \(5\). - Solve the resulting linear equation.

1. Differentiate using the power and sum rules: \(f'(x) = 4x-7\). 2. Set the derivative equal to the required slope: \(4x-7 = 5\). 3. Solve: \(4x = 12\), so \(x = 3\).

\(x = 3\)

- Apply the power rule to the quadratic term. - A marginal rate is an instantaneous rate of change. - Convert the decimal rate to a percentage for interpretation.

1. Differentiate using the power and sum rules: \(S'(x) = 0.000004x+0.15\). 2. Evaluate at \(x = 30{,}000\): \(S'(30{,}000) = 0.000004(30{,}000)+0.15 = 0.27\). 3. The value \(0.27\) is equivalent to \(27\%\). Near an income of \(\$30{,}000\), the model predicts that each additional dollar of taxable income increases the tax owed by approximately \(\$0.27\).

1. \(S'(x) = 0.000004x+0.15\) 2. \(S'(30{,}000) = 0.27 = 27\%\) 3. Near \(\$30{,}000\), one additional dollar of taxable income increases the tax owed by approximately \(\$0.27\).

- What physical quantity is the rate of change of distance traveled with respect to time? - Differentiate with respect to \(t\), treating the angle factor as a constant. - Substitute \(t = 2\) after differentiating. - Make sure your calculator is in degree mode.

1. The instantaneous rate of change \(s'(t)\) is the sled’s instantaneous speed, measured in meters per second. 2. For a fixed angle \(\theta\), the factor \(4.9\sin(\theta)\) is constant. By the power rule, \(s'(t) = 9.8\sin(\theta)t\). 3. For \(\theta = 30^\circ\) and \(t = 2\), \(s'(2) = 9.8\sin(30^\circ)(2) = 9.8\,\frac{\text{m}}{\text{s}}\). 4. For \(\theta = 50^\circ\) and \(t = 2\), \(s'(2) = 9.8\sin(50^\circ)(2) \approx 15.01\,\frac{\text{m}}{\text{s}}\).

The instantaneous rate of change is the sled’s instantaneous speed. For \(\theta = 30^\circ\): \(9.8\,\frac{\text{m}}{\text{s}}\) For \(\theta = 50^\circ\): \(\approx 15.01\,\frac{\text{m}}{\text{s}}\)

- What does the derivative of an area function represent? - Differentiate with respect to time, treating \(v\) as a constant. - Apply the power rule to \(t^2\). - The output unit is area divided by time.

1. The derivative \(A'(t)\) gives the instantaneous rate at which the oil slick’s area is growing. Its units are square meters per hour. 2. Treating \(v\) as a constant and applying the power rule gives \(A'(t) = \sqrt{v}\,t\). 3. For \(v = 10\) and \(t = 3\), \(A'(3) = 3\sqrt{10} \approx 9.49\,\frac{\text{m}^2}{\text{h}}\). 4. For \(v = 25\) and \(t = 3\), \(A'(3) = 3\sqrt{25} = 15\,\frac{\text{m}^2}{\text{h}}\).

The instantaneous rate is the rate at which the slick’s area is growing, in square meters per hour. For \(v = 10\,\text{mph}\): \(\approx 9.49\,\frac{\text{m}^2}{\text{h}}\) For \(v = 25\,\text{mph}\): \(15\,\frac{\text{m}^2}{\text{h}}\)

- Distinguish between average velocity over an interval and velocity at one instant. - Use a difference quotient for part a. - Differentiate the height function to obtain the velocity function. - Positive and negative vertical velocity indicate opposite directions.

1. Since \(s(0) = 4\) and \(s(2) = 4\), the average velocity is \(\frac{s(2)-s(0)}{2-0} = 0\,\frac{\text{m}}{\text{s}}\). 2. Differentiate to obtain velocity: \(v(t) = s'(t) = -10t+10\). Then \(v(1.5) = -10(1.5)+10 = -5\,\frac{\text{m}}{\text{s}}\). The negative sign means the ball is moving downward. 3. Set the velocity equal to \(2\): \(-10t+10 = 2\). Solving gives \(t = 0.8\,\text{s}\). Because the velocity is positive, the ball is rising at that time.

a) \(0\,\frac{\text{m}}{\text{s}}\) b) \(-5\,\frac{\text{m}}{\text{s}}\); the ball is falling. c) \(t = 0.8\,\text{s}\); the ball is rising.

- Differentiate the position function to obtain velocity. - The tangent-line slope has a physical meaning in a position-time model. - Substitute the given time into both the position and velocity functions. - For part b, set the velocity function equal to the target value and solve.

1. Evaluate the position: \(s(2.5) = 0.4(2.5)^3 = 6.25\,\text{m}\). 2. Differentiate to obtain the velocity function: \(v(t) = s'(t) = 1.2t^2\). 3. Evaluate the velocity: \(v(2.5) = 1.2(2.5)^2 = 7.5\,\frac{\text{m}}{\text{s}}\). 4. Set \(v(t) = 30\): \(1.2t^2 = 30\), so \(t^2 = 25\). In the relevant time interval, \(t = 5\,\text{s}\).

a) Position: \(6.25\,\text{m}\); instantaneous velocity: \(7.5\,\frac{\text{m}}{\text{s}}\) b) \(t = 5\,\text{s}\)

- A function value gives the covered area at a specific time. - The derivative gives the instantaneous growth rate. - Apply the power rule to the cubic term. - To find a time from a given rate, set the derivative equal to that rate and solve.

1. Evaluate the area: \(A(4) = 0.5(4)^3+10 = 42\,\text{mm}^2\). 2. Differentiate to find the growth-rate function: \(A'(t) = 1.5t^2\). 3. At \(t = 4\), \(A'(4) = 1.5(4)^2 = 24\,\frac{\text{mm}^2}{\text{h}}\). 4. Set the growth rate equal to \(54\): \(1.5t^2 = 54\), so \(t^2 = 36\). Because \(t \ge 0\), \(t = 6\,\text{h}\).

a) Area: \(42\,\text{mm}^2\); instantaneous growth rate: \(24\,\frac{\text{mm}^2}{\text{h}}\) b) \(t = 6\,\text{h}\)

- Recall the algebraic tests for symmetry using \(f(-x)\). - Consider what happens to the exponent of \(x^n\) when you differentiate it. - Substitute \(-x\) into the derivative and compare the result with \(f'(x)\) and \(-f'(x)\).

1. Evaluate \(f(-x)\): \(f(-x)=-(-x)^4+4(-x)^2-2=-x^4+4x^2-2=f(x)\). Therefore, \(f\) is even, and its graph is symmetric about the \(y\)-axis. 2. Apply the power rule and sum rule: \(f'(x)=-4x^3+8x\). 3. Evaluate the derivative at \(-x\): \(f'(-x)=-4(-x)^3+8(-x)=4x^3-8x=-f'(x)\). Therefore, \(f'\) is odd, and its graph is symmetric about the origin. 4. Differentiating an even polynomial changes each positive even exponent to an odd exponent, while the constant term disappears. Thus, the derivative of an even polynomial is odd.

a) The graph of \(f\) is symmetric about the \(y\)-axis because \(f(-x)=f(x)\). b) \(f'(x)=-4x^3+8x\). Its graph is symmetric about the origin because \(f'(-x)=-f'(x)\). c) The derivative of an even polynomial is an odd polynomial.

- Recall which powers can appear in an odd polynomial. - Choose simple coefficients for your cubic example. - Compare the parity of an odd integer \(k\) with the parity of \(k-1\).

1. A polynomial symmetric about the origin is an odd function, so every nonzero term has an odd exponent. One possible example is \(g(x)=x^3+2x\). 2. For this example, \(g'(x)=3x^2+2\). Since \(g'(-x)=3(-x)^2+2=3x^2+2=g'(x)\), the derivative is even and its graph is symmetric about the \(y\)-axis. 3. In general, every nonzero term of an odd polynomial has the form \(a_kx^k\) with odd \(k\). Its derivative is \(ka_kx^{k-1}\), and \(k-1\) is even. Therefore, the derivative contains only even powers and is symmetric about the \(y\)-axis.

a) Every nonzero term must have an odd exponent. One possible example is \(g(x)=x^3+2x\). b) For this example, \(g'(x)=3x^2+2\), which is symmetric about the \(y\)-axis. c) Differentiating each odd power produces an even power, so the derivative is an even polynomial.

- Apply the derivative rules you know for sums, powers, and constant multiples. - Interpret the derivative as the slope of a graph. - Test a general claim with a simple example such as \(f(x)=x^2\). - Keep the function value and the slope at a point distinct.

1. Statement a) is true. By the power rule, differentiating the leading term \(ax^n\) gives \(nax^{n-1}\), so the degree decreases by \(1\). 2. Statement b) is true on an interval. If \(g'(x)=h'(x)\) for every \(x\), then the derivative of \(g(x)-h(x)\) is \(0\), so \(g(x)-h(x)\) is constant. 3. Statement c) is false. Derivatives of factors are not simply multiplied. Here, first expand: \(p(x)=x^5+x^3\), so \(p'(x)=5x^4+3x^2\), not \(6x^3\). 4. Statement d) is true. At a differentiable local minimum at an interior point of the domain, the tangent line is horizontal, so \(f'(x_0)=0\). Thus, \((x_0, 0)\) is an \(x\)-intercept of the derivative graph.

a) True b) True c) False d) True

- Does a graph that stays above the \(x\)-axis have to be increasing everywhere? - What happens to all tangent slopes when a graph is translated vertically? - Recall that the derivative gives a local slope. - Compare the slopes of an even function at inputs \(x\) and \(-x\).

1. Statement a) is true. Symmetry about the \(y\)-axis means \(f(-x)=f(x)\). The slopes at opposite inputs have opposite signs, so \(f'(-x)=-f'(x)\), which means the derivative is symmetric about the origin. 2. Statement b) is false. For example, \(f(x)=(x-1)^2+1\) is positive for every real \(x\), but \(f'(x)=2x-2\), and \(f'(0)=-2\). 3. Statement c) is true. An upward translation gives \(g(x)=f(x)+5\). Since the derivative of a constant is \(0\), \(g'(x)=f'(x)\). 4. Statement d) is false. For example, if \(f(x)=x^2\) and \(x_0=3\), then \(f(3)=9\), but \(f'(3)=6\). A function value and a slope are generally different quantities.

a) True b) False; for example, \(f(x)=(x-1)^2+1\). c) True d) False; for example, \(f(x)=x^2\) at \(x_0=3\).

- Track how the exponents of power terms change after each differentiation. - Does knowing every slope determine the vertical position of a graph? - Test the statements with simple lines or parabolas. - Distinguish between a zero of a function and a zero of its derivative.

1. Statement a) is true. A polynomial symmetric about the origin contains only odd powers. Differentiating once changes those powers to even powers, and differentiating a second time changes the remaining positive powers back to odd powers. Therefore, \(f''\) is odd and its graph is symmetric about the origin. 2. Statement b) is false. Equal derivatives imply that the functions differ by a constant, not necessarily that they are identical. For example, \(f(x)=x^2\) and \(g(x)=x^2+1\) both have derivative \(2x\), but they are different functions. 3. Statement c) is false. A zero of \(f\) gives information about the function value, not necessarily the slope. For example, \(f(x)=x\) has a zero at \(x_0=0\), but \(f'(0)=1\ne 0\).

a) True b) False c) False

- Apply the power rule to the leading term \(a_nx^n\). - Can a function be strictly increasing while having a horizontal tangent at one point? - Which powers occur in an even polynomial, and what happens to them after differentiation? - What derivative value corresponds to a horizontal tangent?

1. Statement a) is true. If the leading term is \(a_nx^n\), where \(a_n\ne 0\), then its derivative is \(na_nx^{n-1}\). Since \(n\ge 1\), the new leading coefficient is nonzero, so the derivative has degree \(n-1\). 2. Statement b) is false. The function \(f(x)=x^3\) is strictly increasing on all real numbers, but \(f'(x)=3x^2\), and \(f'(0)=0\). 3. Statement c) is true. Such a polynomial is even and contains only even powers. Its derivative contains only odd powers, so \(f'(0)=0\). Therefore, the tangent line at \(x=0\) is horizontal.

a) True b) False c) True

- Simplify the function rule by expanding the square first. - Identify the differentiation variable and treat the parameter as a constant. - Apply the sum rule and power rule. - After finding the derivative, substitute the specified input.

1. Expand the square: \(f(x)=x^2+4kx+4k^2-k^2x\). 2. Combine the linear terms: \(f(x)=x^2+(4k-k^2)x+4k^2\). 3. Differentiate with respect to \(x\), treating \(k\) as a constant: \(f'(x)=2x+4k-k^2\). 4. Substitute \(x=0\): \(f'(0)=4k-k^2\).

\(f'(x)=2x+4k-k^2\) \(f'(0)=4k-k^2\)

- Expand the product before differentiating. - Combine only terms that have the same power of \(t\). - Remember that \(a\) is constant when differentiating with respect to \(t\). - Substitute \(t=0\) after finding the derivative.

1. Distribute \(t\): \(g(t)=at^3-2t^2+a^2t-\frac{1}{2}at^3\). 2. Combine the cubic terms: \(g(t)=\frac{1}{2}at^3-2t^2+a^2t\). 3. Differentiate with respect to \(t\), treating \(a\) as a constant: \(g'(t)=\frac{3}{2}at^2-4t+a^2\). 4. Substitute \(t=0\): \(g'(0)=a^2\).

\(g'(t)=\frac{3}{2}at^2-4t+a^2\) \(g'(0)=a^2\)

- Use a secant slope for the average rate over the first five days. - Use the derivative for the instantaneous rate. - Apply the power rule term by term. - For part c, set the derivative equal to the target rate and solve the quadratic equation.

1. Since \(h(0) = 2\) and \(h(5) = -0.008(5)^3+0.24(5)^2+2 = 7\), the average growth rate on \([0, 5]\) is \(\frac{7-2}{5} = 1\,\frac{\text{in.}}{\text{day}}\). 2. Differentiate: \(h'(t) = -0.024t^2+0.48t\). Then \(h'(10) = -0.024(10)^2+0.48(10) = 2.4\,\frac{\text{in.}}{\text{day}}\). 3. Set the derivative equal to \(1.8\): \(-0.024t^2+0.48t = 1.8\). This is equivalent to \(t^2-20t+75 = 0\), which factors as \((t-5)(t-15)=0\). 4. Therefore, the growth rate is \(1.8\,\frac{\text{in.}}{\text{day}}\) on days \(5\) and \(15\).

a) \(1\,\frac{\text{in.}}{\text{day}}\) b) \(2.4\,\frac{\text{in.}}{\text{day}}\) c) \(t = 5\) days and \(t = 15\) days

- Average velocity uses a difference quotient over an interval. - Instantaneous velocity is the derivative of height with respect to time. - For part c, set the derivative equal to the target velocity and solve the quadratic equation. - Check that each solution lies in the stated time interval.

1. Evaluate the endpoint heights: \(H(2) = \frac{124}{3}\) and \(H(5) = \frac{475}{3}\). 2. The average upward velocity is \(\frac{H(5)-H(2)}{5-2} = \frac{117}{3} = 39\,\frac{\text{m}}{\text{s}}\). 3. Differentiate: \(H'(t) = -t^2+12t+10\). Then \(H'(4) = -16+48+10 = 42\,\frac{\text{m}}{\text{s}}\). 4. Set the derivative equal to \(43\): \(-t^2+12t+10 = 43\), or \(t^2-12t+33 = 0\). 5. The solutions are \(t = 6\pm\sqrt{3}\), approximately \(4.27\,\text{s}\) and \(7.73\,\text{s}\). Both lie in the model’s domain.

a) \(39\,\frac{\text{m}}{\text{s}}\) b) \(42\,\frac{\text{m}}{\text{s}}\) c) \(t = 6-\sqrt{3} \approx 4.27\,\text{s}\) and \(t = 6+\sqrt{3} \approx 7.73\,\text{s}\)

- Treat the parameter as a constant when differentiating with respect to \(x\). - Write the derivative in terms of \(k\). - Use the given rate at \(x = 3\) to form an equation.

1. Differentiate with respect to \(x\), treating \(k\) as a constant: \(h_k'(x) = 3x^2-2kx\). 2. Apply the condition \(h_k'(3) = 15\): \(3(3)^2-2k(3) = 15\). 3. Solve \(27-6k = 15\), which gives \(k = 2\).

\(k = 2\)

- Average change over an interval is found with a secant slope. - Instantaneous change at one time is found with the derivative. - Differentiate the quadratic function. - Set the derivative equal to the target rate and solve for \(t\).

1. Evaluate the endpoint temperatures: \(T(0) = 80\,{}^\circ\text{F}\) and \(T(2) = 4-24+80 = 60\,{}^\circ\text{F}\). 2. The average rate of change on \([0, 2]\) is \(\frac{60-80}{2-0} = -10\,\frac{{}^\circ\text{F}}{\text{h}}\). 3. Differentiate: \(T'(t) = 2t-12\). 4. Set the derivative equal to \(-9\): \(2t-12 = -9\), so \(2t = 3\) and \(t = 1.5\,\text{h}\).

a) \(-10\,\frac{{}^\circ\text{F}}{\text{h}}\) b) \(t = 1.5\,\text{h}\)

- Average velocity is the slope through two position-time points. - Instantaneous velocity is the derivative of position. - Use the power rule to find the velocity function. - Set the velocity function equal to the given value and solve.

1. Evaluate the endpoint positions: \(s(1) = 6\) and \(s(4) = 36\). 2. The average velocity on \([1, 4]\) is \(\frac{s(4)-s(1)}{4-1} = \frac{36-6}{3} = 10\,\frac{\text{m}}{\text{s}}\). 3. Differentiate to obtain instantaneous velocity: \(s'(t) = 2t+5\). 4. Solve \(2t+5 = 12\), giving \(t = 3.5\,\text{s}\).

a) \(10\,\frac{\text{m}}{\text{s}}\) b) \(t = 3.5\,\text{s}\)

- The derivative gives the slope at a specific time. - What do positive and negative slopes mean for temperature? - If a quantity is momentarily neither increasing nor decreasing, what must its derivative equal?

1. Differentiate: \(T'(t) = -2t+12\). 2. Evaluate the derivative: \(T'(2) = -2(2)+12 = 8\,\frac{{}^\circ\text{C}}{\text{min}}\), and \(T'(8) = -2(8)+12 = -4\,\frac{{}^\circ\text{C}}{\text{min}}\). 3. The positive rate at \(t = 2\) means the solution is warming. The negative rate at \(t = 8\) means it is cooling. 4. Set the rate equal to zero: \(-2t+12 = 0\). Solving gives \(t = 6\,\text{min}\).

a) \(T'(2) = 8\,\frac{{}^\circ\text{C}}{\text{min}}\); \(T'(8) = -4\,\frac{{}^\circ\text{C}}{\text{min}}\) b) Positive means the temperature is rising; negative means it is falling. c) \(t = 6\,\text{min}\)

- Velocity is the derivative of position. - The sign of velocity indicates direction. - An object is momentarily at rest when its velocity is zero.

1. Differentiate the position function: \(v(t) = s'(t) = t^2-8t+12\). 2. Evaluate at \(t = 3\): \(v(3) = 9-24+12 = -3\,\frac{\text{m}}{\text{s}}\). The negative sign means the vehicle is moving in the negative coordinate direction. 3. The vehicle is at rest when \(v(t) = 0\). Solve \(t^2-8t+12 = 0\), which factors as \((t-2)(t-6)=0\). 4. Therefore, the vehicle is momentarily at rest at \(t = 2\,\text{s}\) and \(t = 6\,\text{s}\).

a) \(-3\,\frac{\text{m}}{\text{s}}\); the vehicle is moving in the negative coordinate direction. b) \(t = 2\,\text{s}\) and \(t = 6\,\text{s}\)

- Write the secant slope in terms of the unknown parameter \(a\). - Instantaneous rate of change means derivative. - After finding \(a\), set the derivative equal to the required rate and solve.

1. The secant slope is \(\frac{f_a(3)-f_a(1)}{3-1} = \frac{9a-a}{2} = 4a\). 2. Set the secant slope equal to \(4\): \(4a = 4\), so \(a = 1\). 3. For \(f_1(x) = x^2\), the derivative is \(f_1'(x) = 2x\). 4. Set the instantaneous rate equal to \(4\): \(2x = 4\), so \(x = 2\).

1. \(a = 1\) 2. \(x = 2\)

- A difference quotient gives average velocity over a short interval. - Use the sign of velocity to determine the direction of motion. - The exact velocity is the derivative of the height function. - At the maximum height, the exact velocity is zero.

1. With \(\Delta t = 0.001\), the forward difference quotient simplifies to \(-32t+64-16(0.001)\). 2. At \(t = 1\), the approximate velocity is \(31.984\,\frac{\text{ft}}{\text{s}}\). 3. At \(t = 2\), the approximate velocity is \(-0.016\,\frac{\text{ft}}{\text{s}}\). 4. At \(t = 3\), the approximate velocity is \(-32.016\,\frac{\text{ft}}{\text{s}}\). 5. A positive velocity means the ball is rising, and a negative velocity means the ball is falling. The small negative approximation at \(t = 2\) occurs because the forward interval lies just after the highest point. 6. Differentiate using the power rule: \(H'(t) = -32t+64\). 7. Set the exact velocity equal to zero: \(-32t+64 = 0\), so \(t = 2\). 8. Evaluate the height: \(H(2) = -16(2)^2+64(2)+3 = 67\). The maximum height is \(67\,\text{ft}\).

a) \(v(1)\approx 31.984\,\frac{\text{ft}}{\text{s}}\) \(v(2)\approx -0.016\,\frac{\text{ft}}{\text{s}}\) \(v(3)\approx -32.016\,\frac{\text{ft}}{\text{s}}\) b) A positive value means the ball is rising; a negative value means it is falling. The small negative approximation at \(t = 2\) results from using a forward difference quotient at the highest point. c) The exact velocity is \(0\,\frac{\text{ft}}{\text{s}}\) at \(t = 2\,\text{s}\), and the maximum height is \(67\,\text{ft}\).

- At a local maximum or minimum, what is the slope of the graph? What does that imply about the first derivative? - Use the shape of the original graph to estimate where its slopes are positive, negative, or zero. - Consider where the graph changes concavity and what that implies about the second derivative. - Apply the power rule to each term of the polynomial.

1. From the graph, \(f\) has local extrema at about \(x=-2\) and \(x=2\), so \(f'\) should be \(0\) there. The graph increases outside those inputs and decreases between them, so \(f'\) should look like an upward-opening parabola. The graph changes concavity at \(x=0\), so \(f''\) should change sign there and look like an increasing line through the origin. 2. Apply the power rule: \(f'(x)=0.2\cdot 3x^2-2.4=0.6x^2-2.4\). 3. Differentiate again: \(f''(x)=0.6\cdot 2x=1.2x\). 4. The formulas confirm the graph descriptions. Solving \(0.6x^2-2.4=0\) gives \(x=-2\) and \(x=2\), and \(f''(0)=0\), so the second derivative has \(y\)-intercept \((0, 0)\).

a) The graph of \(f'\) is an upward-opening parabola with vertex \((0, -2.4)\) and zeros at \(x=-2\) and \(x=2\). The graph of \(f''\) is a line through the origin with slope \(1.2\). b) \(f'(x)=0.6x^2-2.4\) and \(f''(x)=1.2x\). c) The formulas confirm the descriptions. The zeros of \(f'\) are \(x=-2\) and \(x=2\). The \(y\)-intercept of the graph of \(f''\) is \((0, 0)\).

- Differentiate the tax function to find the marginal tax rate. - Divide total tax by gross income to find the average tax rate. - Compare the fraction of an additional dollar that remains after tax with the current average fraction of income that remains after tax. - Use \(N(x) = x-T(x)\) to relate the two rates.

1. Differentiate: \(T'(x) = 0.15+0.000008x\). At \(x = 35{,}000\), \(T'(35{,}000) = 0.15+0.000008(35{,}000) = 0.43\), or \(43\%\). 2. Simplify the average tax rate: \(r(x) = \frac{0.15x+0.000004x^2}{x} = 0.15+0.000004x\). Thus, \(r(35{,}000) = 0.29\), or \(29\%\). The marginal rate is higher than the average rate. 3. The marginal net-income rate is \(N'(x) = 1-T'(x)\), so \(N'(35{,}000) = 0.57\). The current average net-income share is \(\frac{N(35{,}000)}{35{,}000} = 1-r(35{,}000) = 0.71\). 4. Because \(0.57<0.71\), each additional dollar contributes a smaller fraction to net income than the current average. Therefore, a small percentage increase in gross income produces a smaller percentage increase in net income.

1. \(T'(x) = 0.15+0.000008x\); \(T'(35{,}000) = 0.43 = 43\%\) 2. \(r(35{,}000) = 0.29 = 29\%\); the marginal tax rate is higher. 3. At \(\$35{,}000\), the marginal net-income rate is \(0.57\), while the average net-income share is \(0.71\). Therefore, net income grows proportionally more slowly than gross income.

- Substitute \(E(t)\) into the tax function first. - Factor \(1+0.02t\) from the numerator before simplifying the fraction. - Rewrite \(R(t)\) as a linear function before differentiating. - A negative derivative means purchasing power is decreasing over time.

1. Substitute the income function into the tax function: \(T(E(t)) = 0.000002[50{,}000(1+0.02t)]^2 = 5000(1+0.02t)^2\). 2. Substitute into real income: \(R(t) = \frac{50{,}000(1+0.02t)-5000(1+0.02t)^2}{1+0.02t}\). 3. Factor and cancel \(1+0.02t\): \(R(t) = 50{,}000-5000(1+0.02t)\). 4. Simplify: \(R(t) = 45{,}000-100t\). Therefore, \(R'(t) = -100\). 5. The negative derivative means real after-tax income decreases by \(\$100\) per year. Also, \(\frac{T(E(t))}{E(t)}=0.1(1+0.02t)\), so tax takes an increasing share of gross income even though gross income and the modeled price level use the same growth factor. This illustrates bracket creep within the model.

1. \(R(t) = 50{,}000-5000(1+0.02t)\) 2. \(R'(t) = -100\,\frac{\$}{\text{year}}\) 3. Real after-tax income decreases by \(\$100\) per year because tax takes an increasing share of gross income even though gross income and the modeled price level use the same linear growth factor.