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AP Calculus AB : AP Calculus AB

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3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #103 : Derivative At A Point

$\begin{align*}&t&p(t)\\&5&24\\&10&-45\\&15&-43\\&20&-42\\&25&30\\&\text{The above table lists a particle's postion at different times.}\\&\text{Approximate its velocity with a forward difference at }t=20\end{align*}$

Possible Answers:

$-\frac{42}{5}$

$\frac{72}{5}$

$\frac{36}{5}$

$\frac{1}{5}$

Correct answer:

$\frac{72}{5}$

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(20)\text{ and }f(25)\\&\text{Applying the formula above, with }h=5\text{, we find:}\\&v(20)\approx\frac{30-(-42)}{5}\\&v(20)\approx\frac{72}{5}\end{align*}$

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Example Question #111 : Derivative At A Point

$\begin{align*}&t&p(t)\\&0&14\\&1&-17\\&2&16\\&3&25\\&4&8\\&5&24\\&6&-27\\&\text{Using the above table of positions,}\\&\text{make a forward difference velocity approximation at }t=4\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(4)\text{ and }f(5)\\&\text{Applying the formula above, with }h=1\text{, we find:}\\&v(4)\approx\frac{24-(8)}{1}\\&v(4)\approx16\end{align*}$

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Example Question #112 : Derivative At A Point

$\begin{align*}&t&p(t)\\&2&19\\&4&10\\&6&29\\&\text{The above table lists a particle's postion at different times.}\\&\text{Approximate its velocity with a forward difference at }t=2\end{align*}$

Possible Answers:

$-\frac{9}{4}$

$-\frac{9}{2}$

$\frac{9}{4}$

$\frac{9}{2}$

Correct answer:

$-\frac{9}{2}$

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(2)\text{ and }f(4)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&v(2)\approx\frac{10-(19)}{2}\\&v(2)\approx-\frac{9}{2}\end{align*}$

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Example Question #113 : Derivative At A Point

$\begin{align*}&\text{Approximate a particle's velocity with a forward difference at }t=4\\&\text{Using the following table of positions:}\\&t&p(t)\\&2&5\\&4&-27\\&6&14\\&8&-2\\&10&-35\\&12&28\\&14&-40\end{align*}$

Possible Answers:

$\frac{41}{4}$

$-\frac{27}{2}$

$\frac{41}{2}$

Correct answer:

$\frac{41}{2}$

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(4)\text{ and }f(6)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&v(4)\approx\frac{14-(-27)}{2}\\&v(4)\approx\frac{41}{2}\end{align*}$

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Example Question #114 : Derivative At A Point

$\begin{align*}&\text{Approximate a particle's velocity with a forward difference at }t=3\\&\text{Using the following table of positions:}\\&t&p(t)\\&0&-39\\&3&29\\&6&-21\end{align*}$

Possible Answers:

$-\frac{25}{3}$

$\frac{68}{3}$

$-\frac{50}{3}$

$\frac{29}{3}$

Correct answer:

$-\frac{50}{3}$

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(3)\text{ and }f(6)\\&\text{Applying the formula above, with }h=3\text{, we find:}\\&v(3)\approx\frac{-21-(29)}{3}\\&v(3)\approx-\frac{50}{3}\end{align*}$

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Example Question #115 : Derivative At A Point

$\begin{align*}&t&p(t)\\&3&16\\&7&-39\\&11&44\\&15&-32\\&19&-24\\&23&30\\&27&-1\\&\text{The above table lists a particle's postion at different times.}\\&\text{Approximate its velocity with a forward difference at }t=19\end{align*}$

Possible Answers:

$\frac{27}{4}$

$\frac{27}{2}$

Correct answer:

$\frac{27}{2}$

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(19)\text{ and }f(23)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&v(19)\approx\frac{30-(-24)}{4}\\&v(19)\approx\frac{27}{2}\end{align*}$

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Example Question #116 : Derivative At A Point

$\begin{align*}&t&p(t)\\&3&-5\\&7&11\\&11&-45\\&\text{Using the above table of positions,}\\&\text{make a forward difference velocity approximation at }t=3\end{align*}$

Possible Answers:

$-\frac{5}{4}$

Correct answer:

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(3)\text{ and }f(7)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&v(3)\approx\frac{11-(-5)}{4}\\&v(3)\approx4\end{align*}$

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Example Question #117 : Derivative At A Point

$\begin{align*}&t&p(t)\\&0&-50\\&1&-8\\&2&16\\&\text{The above table lists a particle's postion at different times.}\\&\text{Approximate its velocity with a forward difference at }t=1\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(1)\text{ and }f(2)\\&\text{Applying the formula above, with }h=1\text{, we find:}\\&v(1)\approx\frac{16-(-8)}{1}\\&v(1)\approx24\end{align*}$

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Example Question #871 : Ap Calculus Ab

Which of the following functions has exactly one point of inflection?

Possible Answers:

$f(x) = -12e^{-x} + 7$

f(x) = -15 x +3

$f(x) = 3x^{2} - 10x + 20$

$f(x) = 10x^{4} - 4x^{3} + 3$

$f(x) = x^{3} - x -1$

Correct answer:

$f(x) = x^{3} - x -1$

Explanation:

For any function, a point of inflection occurs when f''(x) = 0 . For a function to have exactly one point of inflection, it has to be of third order (i.e. have an x^3 term as its highest order term). The function $f(x) = x^{3} - x- 1$ only has one point of inflection when x = 0 .

A linear equation does not have any points of inflection, nor does a quadratic equation or exponential function.

A function with an order higher than three has multiple points of inflection.

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Example Question #872 : Ap Calculus Ab

Over what interval is the graph of $f(x)=e^{1-x^2}$ concave upward?

Possible Answers:

$\left ( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right )$

$(-\infty ,-\sqrt{2})$

$\left ( \frac{\sqrt{2}}{2}, \infty \right )$

$\left ( -\infty ,\frac{\sqrt{2}}{2} \right )\bigcup \left ( \frac{\sqrt{2}}{2},\infty \right )$

$(0,\sqrt{2})$

Correct answer:

$\left ( -\infty ,\frac{\sqrt{2}}{2} \right )\bigcup \left ( \frac{\sqrt{2}}{2},\infty \right )$

Explanation:

The concavity of a function is determined by its second derivative. Whenever the second derivative of a function is positive, the function is said to be concave upward.

We will find the first derivative of f(x) using the Chain Rule.

$f'(x)=\frac{d}{dx}[e^{1-x^2}]=e^{1-x^2}\cdot \frac{d}{dx}[1-x^2]$

$=-2xe^{1-x^2}$

Next, we will differentiate the first derivative of f(x) by employing the Product Rule.

$f''(x)=\frac{d}{dx}[-2xe^{1-x^2}]=e^{1-x^2}\cdot \frac{d}{dx}[-2x] + -2x\cdot \frac{d}{dx}[e^{1-x^2}]$

$=-2e^{1-x^2}-2xe^{1-x^2}(-2x)=-2e^{1-x^2}+4x^2e^{1-x^2}$

$=e^{1-x^2}(4x^2-2)$

We must now determine where $e^{1-x^{2}}\left ( 4x^{2}-2 \right )>0$ .

We can divide both sides of the equation by $e^{1-x^2}$ . Since is positive no matter what power it is raised to, we are left with the much simpler inequality:

4x^2-2>0

$x^2>\frac{1}{2}$

$|x|>\frac{\sqrt{2}}{2}$

This is only true when is less than or greater than $\frac{\sqrt{2}}{2}$ . Thus, the interval over which f(x) is concave upward is $(-\infty ,\frac{\sqrt{2}}{2})\cup (\frac{\sqrt{2}}{2},\infty )$ .

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #103 : Derivative At A Point

Example Question #111 : Derivative At A Point

Example Question #112 : Derivative At A Point

Example Question #113 : Derivative At A Point

Example Question #114 : Derivative At A Point

Example Question #115 : Derivative At A Point

Example Question #116 : Derivative At A Point

Example Question #117 : Derivative At A Point

Example Question #871 : Ap Calculus Ab

Example Question #872 : Ap Calculus Ab

All AP Calculus AB Resources

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