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

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

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

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

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

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

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

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

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

$\displaystyle \begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(3)\text{ and }f(1)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&f'(3)\approx\frac{32-(27)}{2}\\&f'(3)\approx\frac{5}{2}\end{align*}$