This is Part 3 of our AP Calculus AB unit test on derivatives. These questions cover tangent lines, mean value theorem, Rolle’s theorem, and inverses of functions. Practicing the application of derivatives theorems to a function will help you solve complicated problems on the AP exam asking you to analyze the behavior of a given function.

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Question 1 |

**What is the slope of the tangent line at $x = 2$ for the function below.**$f(x) = \ln(2x) + x^2$

$\ln(4) + 4$ | |

$\dfrac{9}{2}$ | |

$\dfrac{17}{4}$ | |

$\text{There}$ $\text{is}$ $\text{no}$ $\text{tangent}$ $\text{line}$ $\text{at}$ $x = 2$ |

Question 1 Explanation:

The correct answer is (B). It is important to recognize that the slope of the tangent line is simply the value of the derivative of the function. Therefore, we must first identify the derivative of the function.

$f'(x) = \dfrac{1}{2x} \cdot 2 + 2x$ $= \dfrac{1}{x} + 2x$ $= \dfrac{1 + 2x^2} {x}$

Then, to find the slope of the tangent line at $x = 2$, we just plug in $x = 2$ into the derivative function.

$f'(2) = \dfrac{1 +2(2)^2}{2}$ $= \dfrac{9}{2}$

$f'(x) = \dfrac{1}{2x} \cdot 2 + 2x$ $= \dfrac{1}{x} + 2x$ $= \dfrac{1 + 2x^2} {x}$

Then, to find the slope of the tangent line at $x = 2$, we just plug in $x = 2$ into the derivative function.

$f'(2) = \dfrac{1 +2(2)^2}{2}$ $= \dfrac{9}{2}$

Question 2 |

**Which of the following is the equation for the line tangent to the function below at $x = 3$.**$f(x) = \sin\left(\dfrac{π}{2}x\right) +6x^3$

$y = 162x + 158$ | |

$y = $ $\left(162 + \dfrac{π}{2}\right)x − 325 − \dfrac{3π}{2}$ | |

$y = 162x$ | |

$y = 162x − 325$ |

Question 2 Explanation:

The correct answer is (D). To find the equation of the tangent line, we must first find the derivative of the function to find the tangent line’s slope.

$f'(x) = \dfrac{π}{2}\cos\left(\dfrac{π}{2}x\right) + 18x^2$

Then, the slope of the tangent line at $x = 3$ is the value of the derivative function at $x = 3$.

$f'(3) = $ $\dfrac{π}{2}\cos\left(\dfrac{π}{2} \cdot 3\right) + 18(3)^2$ $= \dfrac{π}{2}(0) + 18(9) = 162$

It is important not to forget that the equation for the tangent line in point slope form takes the general form below, where $(x_1, y_1)$ is a given or known point on the function and m is the slope of the line.

$y − y_1 = m(x − x_1)$

Thus, we need a point on the function. However, for the tangent line, we need more than just any point on the function, but actually the specific point on the function that the line is tangent to. We know that the tangent line at $x = 3$ is tangent to only the point on the function where $x = 3$. Therefore, we must identify the point on the function where $x = 3$ to find the known point for the point-slope formula. We can identify the function value at $x = 3$ by plugging in this value of $x$.

$f'(3) = \sin\left(\dfrac{π}{2} \cdot 3\right) + 6(3)^3$ $= −1 + 162 = 161$

Thus, we know that the tangent line is tangent to the point $(3, 28)$ on the function. Therefore, the point-slope formula for the tangent line is as follows.

$y − 161 = 162(x − 3)$

This can then be simplified to the form below, which matches answer choice (D).

$y = 162x − 325$

$f'(x) = \dfrac{π}{2}\cos\left(\dfrac{π}{2}x\right) + 18x^2$

Then, the slope of the tangent line at $x = 3$ is the value of the derivative function at $x = 3$.

$f'(3) = $ $\dfrac{π}{2}\cos\left(\dfrac{π}{2} \cdot 3\right) + 18(3)^2$ $= \dfrac{π}{2}(0) + 18(9) = 162$

It is important not to forget that the equation for the tangent line in point slope form takes the general form below, where $(x_1, y_1)$ is a given or known point on the function and m is the slope of the line.

$y − y_1 = m(x − x_1)$

Thus, we need a point on the function. However, for the tangent line, we need more than just any point on the function, but actually the specific point on the function that the line is tangent to. We know that the tangent line at $x = 3$ is tangent to only the point on the function where $x = 3$. Therefore, we must identify the point on the function where $x = 3$ to find the known point for the point-slope formula. We can identify the function value at $x = 3$ by plugging in this value of $x$.

$f'(3) = \sin\left(\dfrac{π}{2} \cdot 3\right) + 6(3)^3$ $= −1 + 162 = 161$

Thus, we know that the tangent line is tangent to the point $(3, 28)$ on the function. Therefore, the point-slope formula for the tangent line is as follows.

$y − 161 = 162(x − 3)$

This can then be simplified to the form below, which matches answer choice (D).

$y = 162x − 325$

Question 3 |

**Which of the following equations is the equation of the line tangent to the graph below at $x = 0$.**$f(x) = \dfrac{1}{x^3 + 4x + 6}$

$y = −\frac{1}{9}x$ | |

$y = \frac{1}{6}x$ | |

$y = −\frac{1}{9}x + \frac{1}{6}$ | |

$y = −\frac{1}{6}x − \frac{1}{9}$ |

Question 3 Explanation:

The correct answer is (C). We first find the slope of any tangent line by identifying the function’s derivative.

$f'(x) =$ $−(x^3 + 4x + 6)^{−2} \cdot (3x^2 + 4)$ $= \dfrac{−3x^2 – 4}{(x^3 + 4x + 6)^2}$

Then, we can identify the slope of the tangent line at $x = 0$ by identifying the value of the derivative at $x = 0$.

$f'(0) =$ $\dfrac{−3(0) – 4}{(0^3 + 4(0) + 6)^2}$ $= −\dfrac{4}{6^2} = −\dfrac{1}{9}$

It is important not to forget that the equation for the tangent line in point slope form takes the general form below, where $(x_1, y_1 )$ is a given or known point on the function and $m$ is the slope of the line.

$y − y_1 = m(x − x_1)$

Thus, we need a point on the function. However, for the tangent line, we need more than just any point on the function, but actually the specific point on the function that the line is tangent to. We know that the tangent line at $x = 3$ is tangent to only the point on the function where $x = 3$. Therefore, we must identify the point on the function where $x = 3$ to find the known point for the point-slope formula. We can identify the function value at $x = 3$ by plugging in this value of $x$.

$f(0) = \dfrac{1}{0^3 + 4(0) + 6)}$ $= \dfrac{1}{6}$

Thus, we know that the tangent line is tangent to the point $(0, \frac{1}{6})$ on the function. Therefore, the point-slope formula for the tangent line is as follows.

$y − \dfrac{1}{6} = −\dfrac{1}{9}(x – 0)$

This can then be simplified to the form below, which matches answer choice (C).

$y = −\frac{1}{9}x + \frac{1}{6}$

$f'(x) =$ $−(x^3 + 4x + 6)^{−2} \cdot (3x^2 + 4)$ $= \dfrac{−3x^2 – 4}{(x^3 + 4x + 6)^2}$

Then, we can identify the slope of the tangent line at $x = 0$ by identifying the value of the derivative at $x = 0$.

$f'(0) =$ $\dfrac{−3(0) – 4}{(0^3 + 4(0) + 6)^2}$ $= −\dfrac{4}{6^2} = −\dfrac{1}{9}$

It is important not to forget that the equation for the tangent line in point slope form takes the general form below, where $(x_1, y_1 )$ is a given or known point on the function and $m$ is the slope of the line.

$y − y_1 = m(x − x_1)$

Thus, we need a point on the function. However, for the tangent line, we need more than just any point on the function, but actually the specific point on the function that the line is tangent to. We know that the tangent line at $x = 3$ is tangent to only the point on the function where $x = 3$. Therefore, we must identify the point on the function where $x = 3$ to find the known point for the point-slope formula. We can identify the function value at $x = 3$ by plugging in this value of $x$.

$f(0) = \dfrac{1}{0^3 + 4(0) + 6)}$ $= \dfrac{1}{6}$

Thus, we know that the tangent line is tangent to the point $(0, \frac{1}{6})$ on the function. Therefore, the point-slope formula for the tangent line is as follows.

$y − \dfrac{1}{6} = −\dfrac{1}{9}(x – 0)$

This can then be simplified to the form below, which matches answer choice (C).

$y = −\frac{1}{9}x + \frac{1}{6}$

Question 4 |

**Using the mean value theorem, identify a value for $c$ which the slope of the line tangent to $f(x)$ at $x = c$ is equal to the average rate of change of the function on the given domain.**$f(x) = x^3 + 3x + 2$ $−1 ≤ x ≤ 3$

$c = 10$ | |

$c = \sqrt{\frac{7}{3}}$ | |

$c = \sqrt{5}$ | |

$c = ± \sqrt{\frac{7}{3}}$ |

Question 4 Explanation:

The correct answer is (B). It is important to recognize first that the given function satisfies the requirements of mean value theorem, meaning that the function is continuous and differentiable on the given domain. Then, we must identify the average rate of change of the function on the domain. This is equal to the slope of the secant line connecting the two endpoints. Therefore, we must identify the two endpoints.

$f(−1) = (−1)^3 + 3(−1) + 2$ $= −2$

$f(3) = (3)^3 + 3(3) + 2$ $= 38$

Then, the secant line has the slope below,

$f'(c) = \dfrac{38 − (−2)}{3 − (−1)}$ $= \dfrac{40}{4} = 10$

Then, to find the value $c$ that corresponds to the tangent line with this slope, we must identify the derivative of the given function.

$f'(x) = 3x^2 + 3$

Then, we can set this derivative function equal to the slope of the secant line to identify the value of $x$ that corresponds to the tangent line with this same slope.

$f'(c) = 3c^2 + 3 = 10$

$c^2 = \dfrac{7}{3}$

$c = ± \sqrt{\dfrac{7}{3}}$

Although we have identified two values of $c$, it is important to remember that according to the mean value theorem, the value $x = c$ must exist within the given domain, and $−\sqrt{\frac{7}{3}}$ is outside the domain. Thus, the value of $c$ is $\sqrt{\frac{7}{3}}$.

$f(−1) = (−1)^3 + 3(−1) + 2$ $= −2$

$f(3) = (3)^3 + 3(3) + 2$ $= 38$

Then, the secant line has the slope below,

$f'(c) = \dfrac{38 − (−2)}{3 − (−1)}$ $= \dfrac{40}{4} = 10$

Then, to find the value $c$ that corresponds to the tangent line with this slope, we must identify the derivative of the given function.

$f'(x) = 3x^2 + 3$

Then, we can set this derivative function equal to the slope of the secant line to identify the value of $x$ that corresponds to the tangent line with this same slope.

$f'(c) = 3c^2 + 3 = 10$

$c^2 = \dfrac{7}{3}$

$c = ± \sqrt{\dfrac{7}{3}}$

Although we have identified two values of $c$, it is important to remember that according to the mean value theorem, the value $x = c$ must exist within the given domain, and $−\sqrt{\frac{7}{3}}$ is outside the domain. Thus, the value of $c$ is $\sqrt{\frac{7}{3}}$.

Question 5 |

**Using mean value theorem, identify a value $x = c$ such that the slope of the line tangent to $f(x)$ at $c$ is equal to the average rate of change of the function on the given domain.**$f(x) = \sin^{−1}(x)$ $−\frac{1}{2} ≤ x ≤ \frac{1}{2}$

$c = ± \sqrt{\dfrac{π^2 − 9}{π^2}}$ | |

$c = \sqrt{\dfrac{π^2 − 9}{π^2}}$ | |

$c = \dfrac{π}{3}$ | |

$\text{There}$ $\text{does}$ $\text{not}$ $\text{exist}$ $\text{a}$ $c$ $\text{that}$ $\text{satisfies}$ $\text{mean}$ $\text{value}$ $\text{theorem.}$ |

Question 5 Explanation:

The correct answer is (A). It is important to recognize first that the given function satisfies the requirements of mean value theorem, meaning that the function is continuous and differentiable on the given domain. Then, we must identify the average rate of change of the function on the domain. This is equal to the slope of the secant line connecting the two endpoints. Therefore, we must identify the two endpoints.

$f\left(−\dfrac{1}{2}\right) = \sin^{–1}\left(–\dfrac{1}{2}\right)$ $= −\dfrac{π}{6}$

$f\left(\dfrac{1}{2}\right) = \sin^{−1}\left(\dfrac{1}{2}\right)$ $= \dfrac{π}{6}$

Be sure when using inverse trigonometric functions to have your answers in radians. Also, be sure to identify the correct radian values by recalling the domain and range of inverse trigonometric functions, as there will always be multiple radian values that correspond to a specific radius length.

Then, the secant line has the slope below:

$f'(c) = \dfrac{(\frac{π}{6} − (− \frac{π}{6}))}{\frac{1}{2} − (− \frac{1}{2})} $ $ = \dfrac{\frac{π}{3}}{1} = \dfrac{π}{3}$

Then, to find the value $c$ that corresponds to the tangent line with this slope, we must identify the derivative of the given function.

$f'(x) = \dfrac{1}{\sqrt{1 − x^2}}$

Then, we can set this derivative function equal to the slope of the secant line to identify the value of $x$ that corresponds to the tangent line with this same slope.

$f'(c) = \dfrac{1}{\sqrt{1 − c^2}} = \dfrac{π}{3}$

Then, after some basic algebra, we find that c has the value below.

$c = ± \sqrt{\dfrac{π^2 − 9}{π^2}}$

We have identified two values of $c$, but it is important to remember that according to the mean value theorem, the value $x = c$ must exist within the given domain. Nevertheless, both values exist within the given domain, therefore, there are two values of $c$.

$f\left(−\dfrac{1}{2}\right) = \sin^{–1}\left(–\dfrac{1}{2}\right)$ $= −\dfrac{π}{6}$

$f\left(\dfrac{1}{2}\right) = \sin^{−1}\left(\dfrac{1}{2}\right)$ $= \dfrac{π}{6}$

Be sure when using inverse trigonometric functions to have your answers in radians. Also, be sure to identify the correct radian values by recalling the domain and range of inverse trigonometric functions, as there will always be multiple radian values that correspond to a specific radius length.

Then, the secant line has the slope below:

$f'(c) = \dfrac{(\frac{π}{6} − (− \frac{π}{6}))}{\frac{1}{2} − (− \frac{1}{2})} $ $ = \dfrac{\frac{π}{3}}{1} = \dfrac{π}{3}$

Then, to find the value $c$ that corresponds to the tangent line with this slope, we must identify the derivative of the given function.

$f'(x) = \dfrac{1}{\sqrt{1 − x^2}}$

Then, we can set this derivative function equal to the slope of the secant line to identify the value of $x$ that corresponds to the tangent line with this same slope.

$f'(c) = \dfrac{1}{\sqrt{1 − c^2}} = \dfrac{π}{3}$

Then, after some basic algebra, we find that c has the value below.

$c = ± \sqrt{\dfrac{π^2 − 9}{π^2}}$

We have identified two values of $c$, but it is important to remember that according to the mean value theorem, the value $x = c$ must exist within the given domain. Nevertheless, both values exist within the given domain, therefore, there are two values of $c$.

Question 6 |

**Identify if the function below can be applied to mean value theorem, and, if so, identify the value $c$ such that the slope of the line tangent to $f(x)$ at $x = c$ is equal to the average rate of change of the function on the given domain.**$f(x) = \frac{1}{x}$ $−1 ≤ x ≤ 3$

$\text{Mean}$ $\text{Value}$ $\text{Theorem}$ $\text{can}$ $\text{be}$ $\text{applied.}$ $c = \sqrt{3}$ | |

$\text{Mean}$ $\text{Value}$ $\text{Theorem}$ $\text{can}$ $\text{be}$ $\text{applied.}$ $c = ±\sqrt{3}$ | |

$\text{Mean}$ $\text{Value}$ $\text{Theorem}$ $\text{cannot}$ $\text{be}$ $\text{applied}$ $\text{to}$ $\text{this}$ $\text{function.}$ | |

$\text{Mean}$ $\text{Value}$ $\text{Theorem}$ $\text{can}$ $\text{be}$ $\text{applied.}$ $\text{There}$ $\text{does}$ $\text{not}$ $\text{exist}$ $\text{a}$ $\text{value}$ $c$ $\text{that}$ $\text{satisfies}$ $\text{the}$ $\text{theorem.}$ |

Question 6 Explanation:

The correct answer is (C). Do not forget that mean value theorem applies only to functions that are continuous and differentiable on the given domain. The given function is undefined at $x = 0$, and therefore is not continuous or differentiable on the domain $−1 ≤ x ≤ 3$.

Question 7 |

**Using Rolle’s Theorem, identify a value $x = c$ such that the slope of the line tangent to $f(x)$ at $c$ is equal to the average rate of change of the function on the given domain.**$f(x) = x^2 + 3$ $−2 ≤ x ≤ 2$

$c = 0$ | |

$c = 1$ | |

$c = 3$ | |

$c = −\dfrac{3}{2}$ |

Question 7 Explanation:

The correct answer is (A). We can first verify that this function satisfies Rolle’s Theorem, as it is continuous and differentiable on the given domain, and the secant line between the two endpoints of the domain has a slope of $0$.

$\dfrac{f(b) − f(a)}{b − a} = \dfrac{f(2) − f(−2)}{2 − (−2)}$

$= \dfrac{((−2)^2 + 3 − (2^2 + 3)}{4}$ $= \dfrac{0}{4} = 0$

Thus, we can find the value of $c$ by first finding the derivative of the function and setting it equal to zero.

$f'(x) = 2x$

$f'(c) = 2c = 0$

$c = 0$

$\dfrac{f(b) − f(a)}{b − a} = \dfrac{f(2) − f(−2)}{2 − (−2)}$

$= \dfrac{((−2)^2 + 3 − (2^2 + 3)}{4}$ $= \dfrac{0}{4} = 0$

Thus, we can find the value of $c$ by first finding the derivative of the function and setting it equal to zero.

$f'(x) = 2x$

$f'(c) = 2c = 0$

$c = 0$

Question 8 |

**Using Rolle’s Theorem, identify a value $x = c$ such that the slope of the line tangent to $f(x)$ at $c$ is equal to the average rate of change of the function on the given domain.**$f(x) = \ln(x^2)$ $−1 ≤ x ≤ 1$

$c = 0$ | |

$c = 1$ | |

$c = −1$ | |

$\text{Rolle's}$ $\text{Theorem}$ $\text{is}$ $\text{not}$ $\text{applicable.}$ |

Question 8 Explanation:

The correct answer is (D). It is important to recognize that this function does not satisfy Rolle’s Theorem even though the secant line connecting the two end-points does have a slope of $0$. As Rolle’s Theorem is an adaptation of the mean value theorem, the function must be continuous and differentiable on the given domain. $f(0)$ is undefined, and the function is not continuous or differentiable on the domain.

Question 9 |

**For the**$y = x^2 + 2x + 3$, for $x \ge −1$

*inverse*of the function below, what is the slope of the line tangent to the curve when $x = 6?$$\text{The}$ $\text{slope}$ $\text{is}$ $\text{undefined.}$ | |

$\dfrac{1}{4}$ | |

$\dfrac{1}{14}$ | |

$\dfrac{1}{6}$ |

Question 9 Explanation:

The correct answer is (B). Begin by finding the inverse of the function:

$x = y^2 + 2y + 3$, for $x \ge 2$ and $y \ge −1$

(The range of the original function is the domain of its inverse and the domain of the original function is the range of its inverse.)

Complete the square and solve for $y$:

$x − 3 = y^2 +2y$

$x − 3 + 1 = y^2 + 2y + 1$

$x − 2 = (y + 1) ^2$

$\pm (x − 2)^{\frac{1}{2}} = ((y + 1)^2)^{\frac{1}{2}}$

$\pm (x − 2)^{\frac{1}{2}} = y + 1$

$−1 \pm (x − 2)^{\frac{1}{2}} = y$

$y = −1 + (x − 2)^{\frac{1}{2}}$

(Note: $y = −1 − (x − 2)^{\frac{1}{2}}$ is not valid since it produces values for y outside the range, $y \ge −1$, of the inverse function.)

Find the derivative of the inverse function:

$y' = \dfrac{1}{2} (x − 2)^{−\frac{1}{2}}$

$y' = \dfrac{1}{2(x − 2)^{\frac{1}{2}}}$

Evaluate the derivative of the inverse function at $x = 6$:

$y' = \dfrac{1}{2(6− 2)^{\frac{1}{2}}}$

$y' = \dfrac{1}{4}$

Note: the derivative of the inverse function can also be found using implicit differentiation:

$x = y^2 + 2y + 3$

$\dfrac{dx}{dx} = 2y\dfrac{dy}{dx} + 2\dfrac{dy}{dx}$

$1 = \dfrac{dy}{dx} (2y + 2)$

$\dfrac{dy}{dx} = \dfrac{1}{(2y + 2)}$

$y' = \dfrac{1}{2y + 2}$

$x = y^2 + 2y + 3$, for $x \ge 2$ and $y \ge −1$

(The range of the original function is the domain of its inverse and the domain of the original function is the range of its inverse.)

Complete the square and solve for $y$:

$x − 3 = y^2 +2y$

$x − 3 + 1 = y^2 + 2y + 1$

$x − 2 = (y + 1) ^2$

$\pm (x − 2)^{\frac{1}{2}} = ((y + 1)^2)^{\frac{1}{2}}$

$\pm (x − 2)^{\frac{1}{2}} = y + 1$

$−1 \pm (x − 2)^{\frac{1}{2}} = y$

$y = −1 + (x − 2)^{\frac{1}{2}}$

(Note: $y = −1 − (x − 2)^{\frac{1}{2}}$ is not valid since it produces values for y outside the range, $y \ge −1$, of the inverse function.)

Find the derivative of the inverse function:

$y' = \dfrac{1}{2} (x − 2)^{−\frac{1}{2}}$

$y' = \dfrac{1}{2(x − 2)^{\frac{1}{2}}}$

Evaluate the derivative of the inverse function at $x = 6$:

$y' = \dfrac{1}{2(6− 2)^{\frac{1}{2}}}$

$y' = \dfrac{1}{4}$

Note: the derivative of the inverse function can also be found using implicit differentiation:

$x = y^2 + 2y + 3$

$\dfrac{dx}{dx} = 2y\dfrac{dy}{dx} + 2\dfrac{dy}{dx}$

$1 = \dfrac{dy}{dx} (2y + 2)$

$\dfrac{dy}{dx} = \dfrac{1}{(2y + 2)}$

$y' = \dfrac{1}{2y + 2}$

Question 10 |

**For the**$y = \ln(x^6)$

*inverse*of the function below, what is the slope of the line tangent to the curve when $x = 6?$$e$ | |

$1$ | |

$\dfrac{e}{6}$ | |

$6$ |

Question 10 Explanation:

The correct answer is (C). Begin by finding the inverse of $y = \ln{(x^6)}$:

Switch the $x$ and $y$ variables to get: $x=\ln{(y^6)}$

Solve for $y$:

$e^x=e^{ln(y^6)}\rightarrow e^x=y^6\rightarrow $ $ {{(e}^x)}^\frac{1}{6}={(y^6)}^\frac{1}{6}\rightarrow $ $ e^{\frac{1}{6}x}=y$

Our inverse function is: $y=e^{\frac{1}{6}x}$

Find the derivative of our inverse function (remember to use the chain rule):

$y' =(e^{\frac{1}{6}x})(\frac{1}{6})\rightarrow $ $ y' =\frac{1}{6}e^{\frac{1}{6}x}$

Find the slope of the tangent line at $x=6$:

$y' = \frac{1}{6}e^{\frac{1}{6}(6)}\rightarrow $ $ y' =\frac{e}{6}$

Switch the $x$ and $y$ variables to get: $x=\ln{(y^6)}$

Solve for $y$:

$e^x=e^{ln(y^6)}\rightarrow e^x=y^6\rightarrow $ $ {{(e}^x)}^\frac{1}{6}={(y^6)}^\frac{1}{6}\rightarrow $ $ e^{\frac{1}{6}x}=y$

Our inverse function is: $y=e^{\frac{1}{6}x}$

Find the derivative of our inverse function (remember to use the chain rule):

$y' =(e^{\frac{1}{6}x})(\frac{1}{6})\rightarrow $ $ y' =\frac{1}{6}e^{\frac{1}{6}x}$

Find the slope of the tangent line at $x=6$:

$y' = \frac{1}{6}e^{\frac{1}{6}(6)}\rightarrow $ $ y' =\frac{e}{6}$

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