# Calculus/Differentiation/Applications of Derivatives/Exercises
## Relative Extrema

Find the relative maximum(s) and minimum(s), if any, of the following functions.
$f(x)={\frac {x}{x+1}}$
$f(x)={\sqrt[{3}]{(x-1)^{2}}}$
$(1,0)$
$(1,0)$
$f(x)=x^{2}+{\frac {2}{x}}$
$x=1$
$x=1$
$f(s)={\frac {s}{1+s^{2}}}$
$s=-1$
$s=1$
$s=-1$
$s=1$
$f(x)=x^{2}-4x+9$
$x=2$
$x=2$
$f(x)={\frac {x^{2}+x+1}{x^{2}-x+1}}$
$x=-1$
$x=1$
$x=-1$
$x=1$
Solutions

## Range of Function

$x+{\frac {1}{x}}$
${\begin{aligned}&f(x)=x+{\frac {1}{x}}\\&f'(x)=1-{\frac {1}{x^{2}}}\\&1-{\frac {1}{x^{2}}}=0\implies x=\pm 1\\&f''(x)={\frac {2}{x^{3}}}\\&f''(-1)=-2\end{aligned}}$
Since $f''(-1)$ is negative, $x=-1$ corresponds to a relative maximum. ${\begin{aligned}&f(-1)=-2\\&\lim \limits _{x\to -\infty }f(x)=-\infty \end{aligned}}$
For $x<-1$ , $f'(x)$ is positive, which means that the function is increasing. Coming from very negative $x$-values, $f$ increases from a very negative value to reach a relative maximum of $-2$ at $x=-1$ . For $-1<x<1$ , $f'(x)$ is negative, which means that the function is decreasing. $\lim _{x\to 0^{-}}f(x)=-\infty$ $\lim _{x\to 0^{+}}f(x)=\infty$ $f''(1)=2$ Since $f''(1)$ is positive, $x=1$ corresponds to a relative minimum. $f(1)=2$ Between $[-1,0)$ the function decreases from $-2$ to $-\infty$ , then jumps to $+\infty$ and decreases until it reaches a relative minimum of $2$ at $x=1$ . For $x>1$ , $f'(x)$ is positive, so the function increases from a minimum of $2$ .
$-2$
$2$
${\begin{aligned}&f(x)=x+{\frac {1}{x}}\\&f'(x)=1-{\frac {1}{x^{2}}}\\&1-{\frac {1}{x^{2}}}=0\implies x=\pm 1\\&f''(x)={\frac {2}{x^{3}}}\\&f''(-1)=-2\end{aligned}}$
Since $f''(-1)$ is negative, $x=-1$ corresponds to a relative maximum. ${\begin{aligned}&f(-1)=-2\\&\lim \limits _{x\to -\infty }f(x)=-\infty \end{aligned}}$
For $x<-1$ , $f'(x)$ is positive, which means that the function is increasing. Coming from very negative $x$-values, $f$ increases from a very negative value to reach a relative maximum of $-2$ at $x=-1$ . For $-1<x<1$ , $f'(x)$ is negative, which means that the function is decreasing. $\lim _{x\to 0^{-}}f(x)=-\infty$ $\lim _{x\to 0^{+}}f(x)=\infty$ $f''(1)=2$ Since $f''(1)$ is positive, $x=1$ corresponds to a relative minimum. $f(1)=2$ Between $[-1,0)$ the function decreases from $-2$ to $-\infty$ , then jumps to $+\infty$ and decreases until it reaches a relative minimum of $2$ at $x=1$ . For $x>1$ , $f'(x)$ is positive, so the function increases from a minimum of $2$ .
$-2$
$2$

## Absolute Extrema

Determine the absolute maximum and minimum of the following functions on the given domain
$f(x)={\frac {x^{3}}{3}}-{\frac {x^{2}}{2}}+1$
$[0,3]$
$(3,{\tfrac {11}{2}})$
$(1,{\tfrac {5}{6}})$
$(3,{\tfrac {11}{2}})$
$(1,{\tfrac {5}{6}})$
$f(x)=\left({\frac {4}{3}}x^{2}-1\right)x$
$[-{\tfrac {1}{2}},2]$
$(2,{\tfrac {26}{3}})$
$({\tfrac {1}{2}},-{\tfrac {1}{3}})$
$(2,{\tfrac {26}{3}})$
$({\tfrac {1}{2}},-{\tfrac {1}{3}})$
Solutions

## Determine Intervals of Change

Find the intervals where the following functions are increasing or decreasing
$f(x)=10-6x-2x^{2}$
$(-\infty ,-{\tfrac {3}{2}})$
$(-{\tfrac {3}{2}},\infty )$
$(-\infty ,-{\tfrac {3}{2}})$
$(-{\tfrac {3}{2}},\infty )$
$f(x)=2x^{3}-12x^{2}+18x+15$
$(1,3)$
$(1,3)$
$f(x)=5+36x+3x^{2}-2x^{3}$
$(-2,3)$
$(-2,3)$
$f(x)=8+36x+3x^{2}-2x^{3}$
$(-2,3)$
$(-2,3)$
$f(x)=5x^{3}-15x^{2}-120x+3$
$(-2,4)$
$(-2,4)$
$f(x)=x^{3}-6x^{2}-36x+2$
$(-2,6)$
$(-2,6)$
Solutions

## Determine Intervals of Concavity

Find the intervals where the following functions are concave up or concave down
$f(x)=10-6x-2x^{2}$
$f(x)=2x^{3}-12x^{2}+18x+15$
$(-\infty ,2)$
$(2,\infty )$
$(-\infty ,2)$
$(2,\infty )$
$f(x)=5+36x+3x^{2}-2x^{3}$
$(-\infty ,{\tfrac {1}{2}})$
$({\tfrac {1}{2}},\infty )$
$(-\infty ,{\tfrac {1}{2}})$
$({\tfrac {1}{2}},\infty )$
$f(x)=8+36x+3x^{2}-2x^{3}$
$(-\infty ,{\tfrac {1}{2}})$
$({\tfrac {1}{2}},\infty )$
$(-\infty ,{\tfrac {1}{2}})$
$({\tfrac {1}{2}},\infty )$
$f(x)=5x^{3}-15x^{2}-120x+3$
$(-\infty ,1)$
$(1,\infty )$
$(-\infty ,1)$
$(1,\infty )$
$f(x)=x^{3}-6x^{2}-36x+2$
$(-\infty ,2)$
$(2,\infty )$
$(-\infty ,2)$
$(2,\infty )$
Solutions

## Word Problems

$4t$
$t$
$10{\tfrac {\rm {m}}{\rm {s}}}$
$10{\tfrac {\rm {m}}{\rm {s}}}$
$12{\rm {\,mph}}$
$5{\rm {\,mph}}$
$13{\rm {\,mph}}$
$13{\rm {\,mph}}$
$200{\rm {\,m^{3}}}$
${\rm {m^{2}}}$
${\rm {m^{2}}}$
$216{\rm {\,cm}}$
$972{\rm {\,cm^{2}}}$
$972{\rm {\,cm^{2}}}$
$r$
$2r^{2}$
$2r^{2}$
$R=3{\rm {\,in}}$
$12\pi {\sqrt {3}}{\rm {\,in^{3}}}$
$12\pi {\sqrt {3}}{\rm {\,in^{3}}}$
$6\,{\rm {ft}}$
$15$
$6$
$4{\rm {\,ft/s}}$
$4{\rm {\,ft/s}}$
29. A canoe is being pulled toward a dock (normal to the water) using a taut rope. The canoe is normal to the water while it is being pulled. The rope is hauled in at a constant $5\,{\rm {ft/s}}$. The dock is $5\,{\rm {ft}}$ above the water. Answer items (a) through (b).
$13\,{\rm {ft}}$
$13{\rm {\,ft/s}}$
$13{\rm {\,ft/s}}$
$0.2{\rm {\,rad/s}}$
$0.2{\rm {\,rad/s}}$
$100\,{\rm {m}}$
$2\,{\rm {m}}$
$4\,{\rm {m/s}}$
$1\,{\rm {rad/s}}$
$1\,{\rm {rad/s}}$
Solutions

## Graphing Functions

For each of the following, graph a function that abides by the provided characteristics
$f(1)=f(-2)=0,\ \lim _{x\to \infty }f(x)=\lim _{x\to -\infty }f(x)=0,\ {\mbox{ vertical asymptote at }}x=-3,\ f'(x)>0{\mbox{ on }}(0,2),$
$f'(x)<0{\mbox{ on }}(-\infty ,-3)\cup (-3,0)\cup (2,\infty ),\ f''(x)>0{\mbox{ on }}(-3,1)\cup (3,\infty ),\ f''(x)<0{\mbox{ on }}(-\infty ,-3)\cup (1,3).$
$f{\mbox{ has domain }}[-1,1],\ f(-1)=-1,\ f(-{\tfrac {1}{2}})=-2,\ f'(-{\tfrac {1}{2}})=0,\ f''(x)>0{\mbox{ on }}(-1,1)$
Solutions

## Approximation Problems

By assumption, for these problems, assume $\pi \approx 3.14$ and $e\approx 2.718$ unless stated otherwise. One may use a calculator or design a computer program, but one must indicate the method and reasoning behind every step where necessary.
$\sin(3)$
$\sin(3)\approx 0.14$
$\Delta x=0.14$
$f(x,y)=1-y^{2}$
$\sin(3)\approx 0.14$
$\Delta x=0.14$
$f(x,y)=1-y^{2}$
${\sqrt {2}}$
${\sqrt {2}}\approx 1.4167$
$2$
${\sqrt {2}}\approx 1.4167$
$2$
$\ln(3)$
$\ln(3)\approx 1.10{\text{ OR }}1.09$
$\ln(3)\approx 1.10{\text{ OR }}1.09$
Solutions

## Advanced Understanding

45. Consider the differentiable function $f(x)$ for all $x>-2$ and continuous function $g(x)$ below, where $g(x)$ is linear for all $x\geq 1$ and differentiable for all $x\in (-2,1)\cup (1,\infty )$, and $f(x)$ and $g(x)$ are continuous for all $x\geq -2$.
$f^{\prime }(1.6)$
$f^{\prime }(1.6)\approx -5$
$f^{\prime }(1.6)\approx -5$
$\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}$
$\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}={\frac {3}{7}}$
$\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}={\frac {3}{7}}$
$f^{\prime }(1.6)=f^{\prime }(2)$
$f(x)$
$c\in \mathbb {R}$
$f(c)=0$
$c=2.2$
$c\in \mathbb {R}$
$f(c)=0$
$c=2.2$
$f(x)$
$g(x)$
Solutions
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