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

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Question

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

Answer

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

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Question

Which of the following functions contains a removeable discontinuity?

Answer

A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.

For example, the function $f(x)=\frac{1+x^3}{1+x}$ contains a removeable discontinuity at . Notice that we could simplify as follows:

$f(x)=\frac{1+x^3}{1+x}=\frac{(1+x)(x^2-x+1)}{1+x}=x^{2}-x+1$ , where $x eq -1$ .

Thus, we could say that $\lim_{x\rightarrow -1}\frac{1+x^3}{1+x}=\lim_{x\rightarrow -1}x^2-x+1=(-1)^2-(-1)+1=3$ .

As we can see, the limit of exists at , even though is undefined.

What this means is that will look just like the parabola with the equation $x^{2}-x+1$ EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .

The functions

$f(x)=\frac{x}{1-x^{2}}$ , and

$f(x)=1-(\ln x)^{2}$

have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.

The functions

$f(x)=x^{2}\sin x^{2}$ and $f(x)=\frac{x+1}{1+x^{2}}$ are continuous over all the real values of ; they have no discontinuities of any kind.

The answer is

$f(x)=\frac{1+x^{3}}{1+x}$ .

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Question

$\begin{align}&\text{Approximate a forward derivative at }x=4\&\text{Using the following table:}\&x&&f(x)\&1&&5\&4&&-8\&7&&15\end{align}$

Answer

$\begin{align}&\text{To find a forward derivative approximation, we consider function}\&\text{two values: one at the point of interest and another succeeding}\&\text{it, and finally that distance between the two. Mathematically,}\&\text{this follows the form:}\&f'(x)=\frac{f(x+h)-f(x)}{h}\&\text{Where h is the size of the increment between points on the x-axis.}\&\text{In approximating a derivative, it is generally prudent to minimize}\&\text{the distance between function values. In other words, it’s a}\&\text{good idea to look at two adjacent points. For our problem, that’d}\&\text{be:}\&f(4)\text{ and }f(7)\&\text{Applying the formula above, with }h=3\text{, we find:}\&f'(4)\approx\frac{15-(15)}{3}\&f'(4)\approx\frac{23}{3}\end{align}$

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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