Derivative at a point - AP Calculus AB

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Question

Find the equation of the line that passes through (2,0) and is perpendicular to the tangent line to the following function at x=0:

$f=\ln(2-x)$

Answer

To determine the equation of the line, we must first find its slope, which is perpendicular to that of the tangent line to the function. The tangent line to the function at any point is given by the first derivative of the function:

$f'=\frac{-1}{2-x}$

which was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Evaluating the derivative at the given point, we find that the slope of the tangent line to the function is

$-\frac{1}{2-0}=-\frac{1}{2}$

However, the line we want has a slope perpendicular to this, so we take the negative reciprocal:

We now can write the equation of the line using point-slope form:

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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 $x\neq -1$ .

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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