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AP Calculus AB : Approximate rate of change from graphs and tables of values

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Example Questions

Example Question #841 : Ap Calculus Ab

$\begin{align*}&\text{Approximate a forward derivative at }x=0\\&\text{Using the following table:}\\&x&f(x)\\&0&-31\\&4&-1\\&8&-36\end{align*}$

Possible Answers:

$-\frac{31}{4}$

$-\frac{35}{4}$

$\frac{15}{4}$

$\frac{15}{2}$

Correct answer:

$\frac{15}{2}$

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(0)\text{ and }f(4)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&f'(0)\approx\frac{-1-(-1)}{4}\\&f'(0)\approx\frac{15}{2}\end{align*}$

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Example Question #11 : Approximate Rate Of Change From Graphs And Tables Of Values

$\begin{align*}&x&f(x)\\&1&32\\&5&38\\&9&49\\&13&-50\\&17&37\\&21&11\\&25&49\\&29&3\\&\text{Approximate a forward derivative at }x=13\end{align*}$

Possible Answers:

$-\frac{25}{2}$

$\frac{87}{8}$

$-\frac{99}{4}$

$\frac{87}{4}$

Correct answer:

$\frac{87}{4}$

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(13)\text{ and }f(17)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&f'(13)\approx\frac{37-(37)}{4}\\&f'(13)\approx\frac{87}{4}\end{align*}$

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Example Question #843 : Ap Calculus Ab

$\begin{align*}&x&f(x)\\&3&27\\&8&8\\&13&43\\&18&8\\&23&-49\\&28&-38\\&33&37\\&\text{Approximate a backward derivative at }x=33\end{align*}$

Possible Answers:

$-\frac{37}{5}$

$\frac{37}{5}$

Correct answer:

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(33)\text{ and }f(28)\\&\text{Applying the formula above, with }h=5\text{, we find:}\\&f'(33)\approx\frac{37-(-38)}{5}\\&f'(33)\approx15\end{align*}$

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Example Question #844 : Ap Calculus Ab

$\begin{align*}&x&f(x)\\&-3&12\\&1&-14\\&5&-45\\&9&-1\\&13&-31\\&17&-38\\&\text{Approximate a backward derivative at }x=1\end{align*}$

Possible Answers:

$\frac{13}{2}$

$-\frac{1}{2}$

$\frac{1}{2}$

$-\frac{13}{2}$

Correct answer:

$-\frac{13}{2}$

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(1)\text{ and }f(-3)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&f'(1)\approx\frac{-14-(12)}{4}\\&f'(1)\approx-\frac{13}{2}\end{align*}$

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Example Question #845 : Ap Calculus Ab

$\begin{align*}&x&f(x)\\&-2&4\\&2&20\\&6&0\\&\text{Approximate a backward derivative at }x=2\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(2)\text{ and }f(-2)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&f'(2)\approx\frac{20-(4)}{4}\\&f'(2)\approx4\end{align*}$

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Example Question #846 : Ap Calculus Ab

$\begin{align*}&x&f(x)\\&3&-23\\&8&-29\\&13&7\\&18&14\\&23&-8\\&\text{Approximate a backward derivative at }x=23\end{align*}$

Possible Answers:

$\frac{6}{5}$

$-\frac{22}{5}$

$\frac{22}{5}$

$-\frac{6}{5}$

Correct answer:

$-\frac{22}{5}$

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(23)\text{ and }f(18)\\&\text{Applying the formula above, with }h=5\text{, we find:}\\&f'(23)\approx\frac{-8-(14)}{5}\\&f'(23)\approx-\frac{22}{5}\end{align*}$

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Example Question #847 : Ap Calculus Ab

$\begin{align*}&x&f(x)\\&-2&12\\&-1&7\\&0&-45\\&\text{Using the above table, make a backward derivative approximation at }x=0\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(0)\text{ and }f(-1)\\&\text{Applying the formula above, with }h=1\text{, we find:}\\&f'(0)\approx\frac{-45-(7)}{1}\\&f'(0)\approx-52\end{align*}$

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Example Question #848 : Ap Calculus Ab

$\begin{align*}&\text{Approximate a backward derivative at }x=8\\&\text{Using the following table:}\\&x&f(x)\\&3&49\\&8&36\\&13&29\end{align*}$

Possible Answers:

$-\frac{13}{5}$

$-\frac{7}{5}$

Correct answer:

$-\frac{13}{5}$

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(8)\text{ and }f(3)\\&\text{Applying the formula above, with }h=5\text{, we find:}\\&f'(8)\approx\frac{36-(49)}{5}\\&f'(8)\approx-\frac{13}{5}\end{align*}$

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Example Question #849 : Ap Calculus Ab

$\begin{align*}&x&f(x)\\&3&-20\\&4&-21\\&5&-17\\&\text{Approximate a backward derivative at }x=5\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(5)\text{ and }f(4)\\&\text{Applying the formula above, with }h=1\text{, we find:}\\&f'(5)\approx\frac{-17-(-21)}{1}\\&f'(5)\approx4\end{align*}$

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Example Question #850 : Ap Calculus Ab

$\begin{align*}&x&f(x)\\&2&-15\\&5&-5\\&8&-45\\&11&-33\\&14&16\\&17&-17\\&20&40\\&23&-39\\&\text{Using the above table, make a backward derivative approximation at }x=14\end{align*}$

Possible Answers:

$\frac{49}{3}$

$-\frac{17}{3}$

$-\frac{1}{3}$

Correct answer:

$\frac{49}{3}$

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(14)\text{ and }f(11)\\&\text{Applying the formula above, with }h=3\text{, we find:}\\&f'(14)\approx\frac{16-(-33)}{3}\\&f'(14)\approx\frac{49}{3}\end{align*}$

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AP Calculus AB : Approximate rate of change from graphs and tables of values

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #841 : Ap Calculus Ab

Example Question #11 : Approximate Rate Of Change From Graphs And Tables Of Values

Example Question #843 : Ap Calculus Ab

Example Question #844 : Ap Calculus Ab

Example Question #845 : Ap Calculus Ab

Example Question #846 : Ap Calculus Ab

Example Question #847 : Ap Calculus Ab

Example Question #848 : Ap Calculus Ab

Example Question #849 : Ap Calculus Ab

Example Question #850 : Ap Calculus Ab

All AP Calculus AB Resources

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