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AP Calculus AB : Derivative at a point

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Example Questions

Example Question #71 : Derivative At A Point

$\begin{align*}&x&f(x)\\&1&-27\\&3&-38\\&5&11\\&7&-5\\&9&-4\\&11&16\\&13&27\\&15&-15\\&\text{Using the above table, make a forward derivative approximation at }x=9\end{align*}$

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(9)\text{ and }f(11)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&f'(9)\approx\frac{16-(16)}{2}\\&f'(9)\approx10\end{align*}$

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Example Question #72 : Derivative At A Point

$\begin{align*}&\text{Approximate a forward derivative at }x=7\\&\text{Using the following table:}\\&x&f(x)\\&1&8\\&3&4\\&5&37\\&7&-24\\&9&-18\\&11&-38\\&13&44\end{align*}$

Possible Answers:

$-\frac{61}{2}$

$\frac{3}{2}$

Correct answer:

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(7)\text{ and }f(9)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&f'(7)\approx\frac{-18-(-18)}{2}\\&f'(7)\approx3\end{align*}$

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Example Question #73 : Derivative At A Point

$\begin{align*}&\text{Approximate a forward derivative at }x=0\\&\text{Using the following table:}\\&x&f(x)\\&0&22\\&4&2\\&8&50\\&12&-28\\&16&-40\\&20&-39\end{align*}$

Possible Answers:

$-\frac{5}{2}$

$\frac{11}{2}$

$-\frac{39}{2}$

Correct answer:

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(0)\text{ and }f(4)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&f'(0)\approx\frac{2-(2)}{4}\\&f'(0)\approx-5\end{align*}$

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Example Question #74 : Derivative At A Point

$\begin{align*}&\text{Approximate a forward derivative at }x=9\\&\text{Using the following table:}\\&x&f(x)\\&1&27\\&5&44\\&9&48\\&13&-31\\&17&-36\end{align*}$

Possible Answers:

$-\frac{79}{4}$

$-\frac{79}{8}$

Correct answer:

$-\frac{79}{4}$

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(9)\text{ and }f(13)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&f'(9)\approx\frac{-31-(-31)}{4}\\&f'(9)\approx-\frac{79}{4}\end{align*}$

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Example Question #75 : Derivative At A Point

$\begin{align*}&x&f(x)\\&0&-11\\&5&17\\&10&24\\&15&2\\&20&-15\\&25&-35\\&\text{Approximate a forward derivative at }x=10\end{align*}$

Possible Answers:

$-\frac{11}{5}$

$-\frac{22}{5}$

$\frac{24}{5}$

$\frac{7}{5}$

Correct answer:

$-\frac{22}{5}$

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(10)\text{ and }f(15)\\&\text{Applying the formula above, with }h=5\text{, we find:}\\&f'(10)\approx\frac{2-(2)}{5}\\&f'(10)\approx-\frac{22}{5}\end{align*}$

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Example Question #76 : Derivative At A Point

$\begin{align*}&\text{Approximate a forward derivative at }x=0\\&\text{Using the following table:}\\&x&f(x)\\&0&19\\&2&-14\\&4&24\\&6&-11\\&8&19\\&10&21\\&12&-6\end{align*}$

Possible Answers:

$\frac{19}{2}$

$-\frac{33}{2}$

$-\frac{33}{4}$

Correct answer:

$-\frac{33}{2}$

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(0)\text{ and }f(2)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&f'(0)\approx\frac{-14-(-14)}{2}\\&f'(0)\approx-\frac{33}{2}\end{align*}$

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Example Question #481 : Derivatives

$\begin{align*}&x&f(x)\\&2&-7\\&3&39\\&4&-11\\&5&27\\&\text{Using the above table, make a forward derivative approximation at }x=3\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(3)\text{ and }f(4)\\&\text{Applying the formula above, with }h=1\text{, we find:}\\&f'(3)\approx\frac{-11-(-11)}{1}\\&f'(3)\approx-50\end{align*}$

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Example Question #482 : Derivatives

$\begin{align*}&x&f(x)\\&2&45\\&4&-17\\&6&17\\&8&-6\\&10&34\\&\text{Using the above table, make a forward derivative approximation at }x=8\end{align*}$

Possible Answers:

$-\frac{23}{2}$

Correct answer:

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(8)\text{ and }f(10)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&f'(8)\approx\frac{34-(34)}{2}\\&f'(8)\approx20\end{align*}$

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Example Question #77 : Derivative At A Point

$\begin{align*}&x&f(x)\\&3&9\\&6&-35\\&9&-30\\&12&-9\\&15&25\\&18&33\\&21&29\\&24&-18\\&\text{Approximate a forward derivative at }x=12\end{align*}$

Possible Answers:

$\frac{17}{3}$

$\frac{34}{3}$

Correct answer:

$\frac{34}{3}$

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(12)\text{ and }f(15)\\&\text{Applying the formula above, with }h=3\text{, we find:}\\&f'(12)\approx\frac{25-(25)}{3}\\&f'(12)\approx\frac{34}{3}\end{align*}$

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Example Question #71 : Derivative At A Point

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

Possible Answers:

$\frac{15}{4}$

$-\frac{31}{4}$

$\frac{15}{2}$

$-\frac{35}{4}$

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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AP Calculus AB : Derivative at a point

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #71 : Derivative At A Point

Example Question #72 : Derivative At A Point

Example Question #73 : Derivative At A Point

Example Question #74 : Derivative At A Point

Example Question #75 : Derivative At A Point

Example Question #76 : Derivative At A Point

Example Question #481 : Derivatives

Example Question #482 : Derivatives

Example Question #77 : Derivative At A Point

Example Question #71 : Derivative At A Point

All AP Calculus AB Resources

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