# Calculus/Differentiation/Applications of Derivatives/Exercises

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## Relative Extrema

Find the relative maximum(s) and minimum(s), if any, of the following functions.

1. ${\displaystyle f(x)={\frac {x}{x+1}}}$
none
none
2. ${\displaystyle f(x)={\sqrt[{3}]{(x-1)^{2}}}}$
Minimum at the point ${\displaystyle (1,0)}$
Minimum at the point ${\displaystyle (1,0)}$
3. ${\displaystyle f(x)=x^{2}+{\frac {2}{x}}}$
Relative minimum at ${\displaystyle x=1}$
Relative minimum at ${\displaystyle x=1}$
4. ${\displaystyle f(s)={\frac {s}{1+s^{2}}}}$
Relative minimum at ${\displaystyle s=-1}$
Relative maximum at ${\displaystyle s=1}$
Relative minimum at ${\displaystyle s=-1}$
Relative maximum at ${\displaystyle s=1}$
5. ${\displaystyle f(x)=x^{2}-4x+9}$
Relative minimum at ${\displaystyle x=2}$
Relative minimum at ${\displaystyle x=2}$
6. ${\displaystyle f(x)={\frac {x^{2}+x+1}{x^{2}-x+1}}}$
Relative minimum at ${\displaystyle x=-1}$
Relative maximum at ${\displaystyle x=1}$
Relative minimum at ${\displaystyle x=-1}$
Relative maximum at ${\displaystyle x=1}$

## Range of Function

7. Show that the expression ${\displaystyle x+{\frac {1}{x}}}$ cannot take on any value strictly between 2 and -2.

{\displaystyle {\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 ${\displaystyle f''(-1)}$ is negative, ${\displaystyle x=-1}$ corresponds to a relative maximum.
{\displaystyle {\begin{aligned}&f(-1)=-2\\&\lim \limits _{x\to -\infty }f(x)=-\infty \end{aligned}}}

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

The above analysis shows that there is a gap in the function's range between ${\displaystyle -2}$ and ${\displaystyle 2}$ .

{\displaystyle {\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 ${\displaystyle f''(-1)}$ is negative, ${\displaystyle x=-1}$ corresponds to a relative maximum.
{\displaystyle {\begin{aligned}&f(-1)=-2\\&\lim \limits _{x\to -\infty }f(x)=-\infty \end{aligned}}}

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

The above analysis shows that there is a gap in the function's range between ${\displaystyle -2}$ and ${\displaystyle 2}$ .

## Absolute Extrema

Determine the absolute maximum and minimum of the following functions on the given domain

8. ${\displaystyle f(x)={\frac {x^{3}}{3}}-{\frac {x^{2}}{2}}+1}$ on ${\displaystyle [0,3]}$
Maximum at ${\displaystyle (3,{\tfrac {11}{2}})}$ ; minimum at ${\displaystyle (1,{\tfrac {5}{6}})}$
Maximum at ${\displaystyle (3,{\tfrac {11}{2}})}$ ; minimum at ${\displaystyle (1,{\tfrac {5}{6}})}$
9. ${\displaystyle f(x)=\left({\frac {4}{3}}x^{2}-1\right)x}$ on ${\displaystyle [-{\tfrac {1}{2}},2]}$
Maximum at ${\displaystyle (2,{\tfrac {26}{3}})}$ ; minimum at ${\displaystyle ({\tfrac {1}{2}},-{\tfrac {1}{3}})}$
Maximum at ${\displaystyle (2,{\tfrac {26}{3}})}$ ; minimum at ${\displaystyle ({\tfrac {1}{2}},-{\tfrac {1}{3}})}$

## Determine Intervals of Change

Find the intervals where the following functions are increasing or decreasing

10. ${\displaystyle f(x)=10-6x-2x^{2}}$
Increasing on ${\displaystyle (-\infty ,-{\tfrac {3}{2}})}$ ; decreasing on ${\displaystyle (-{\tfrac {3}{2}},\infty )}$
Increasing on ${\displaystyle (-\infty ,-{\tfrac {3}{2}})}$ ; decreasing on ${\displaystyle (-{\tfrac {3}{2}},\infty )}$
11. ${\displaystyle f(x)=2x^{3}-12x^{2}+18x+15}$
Decreasing on ${\displaystyle (1,3)}$ ; increasing elsewhere
Decreasing on ${\displaystyle (1,3)}$ ; increasing elsewhere
12. ${\displaystyle f(x)=5+36x+3x^{2}-2x^{3}}$
Increasing on ${\displaystyle (-2,3)}$ ; decreasing elsewhere
Increasing on ${\displaystyle (-2,3)}$ ; decreasing elsewhere
13. ${\displaystyle f(x)=8+36x+3x^{2}-2x^{3}}$
Increasing on ${\displaystyle (-2,3)}$ ; decreasing elsewhere
Increasing on ${\displaystyle (-2,3)}$ ; decreasing elsewhere
14. ${\displaystyle f(x)=5x^{3}-15x^{2}-120x+3}$
Decreasing on ${\displaystyle (-2,4)}$ ; increasing elsewhere
Decreasing on ${\displaystyle (-2,4)}$ ; increasing elsewhere
15. ${\displaystyle f(x)=x^{3}-6x^{2}-36x+2}$
Decreasing on ${\displaystyle (-2,6)}$ ; increasing elsewhere
Decreasing on ${\displaystyle (-2,6)}$ ; increasing elsewhere

## Determine Intervals of Concavity

Find the intervals where the following functions are concave up or concave down

16. ${\displaystyle f(x)=10-6x-2x^{2}}$
Concave down everywhere
Concave down everywhere
17. ${\displaystyle f(x)=2x^{3}-12x^{2}+18x+15}$
Concave down on ${\displaystyle (-\infty ,2)}$ ; concave up on ${\displaystyle (2,\infty )}$
Concave down on ${\displaystyle (-\infty ,2)}$ ; concave up on ${\displaystyle (2,\infty )}$
18. ${\displaystyle f(x)=5+36x+3x^{2}-2x^{3}}$
Concave up on ${\displaystyle (-\infty ,{\tfrac {1}{2}})}$ ; concave down on ${\displaystyle ({\tfrac {1}{2}},\infty )}$
Concave up on ${\displaystyle (-\infty ,{\tfrac {1}{2}})}$ ; concave down on ${\displaystyle ({\tfrac {1}{2}},\infty )}$
19. ${\displaystyle f(x)=8+36x+3x^{2}-2x^{3}}$
Concave up on ${\displaystyle (-\infty ,{\tfrac {1}{2}})}$ ; concave down on ${\displaystyle ({\tfrac {1}{2}},\infty )}$
Concave up on ${\displaystyle (-\infty ,{\tfrac {1}{2}})}$ ; concave down on ${\displaystyle ({\tfrac {1}{2}},\infty )}$
20. ${\displaystyle f(x)=5x^{3}-15x^{2}-120x+3}$
Concave down on ${\displaystyle (-\infty ,1)}$ ; concave up on ${\displaystyle (1,\infty )}$
Concave down on ${\displaystyle (-\infty ,1)}$ ; concave up on ${\displaystyle (1,\infty )}$
21. ${\displaystyle f(x)=x^{3}-6x^{2}-36x+2}$
Concave down on ${\displaystyle (-\infty ,2)}$ ; concave up on ${\displaystyle (2,\infty )}$
Concave down on ${\displaystyle (-\infty ,2)}$ ; concave up on ${\displaystyle (2,\infty )}$

## Word Problems

22. You peer around a corner. A velociraptor 64 meters away spots you. You run away at a speed of 6 meters per second. The raptor chases, running towards the corner you just left at a speed of ${\displaystyle 4t}$ meters per second (time ${\displaystyle t}$ measured in seconds after spotting). After you have run 4 seconds the raptor is 32 meters from the corner. At this time, how fast is death approaching your soon to be mangled flesh? That is, what is the rate of change in the distance between you and the raptor?
${\displaystyle 10{\tfrac {\rm {m}}{\rm {s}}}}$
${\displaystyle 10{\tfrac {\rm {m}}{\rm {s}}}}$
23. Two bicycles leave an intersection at the same time. One heads north going ${\displaystyle 12{\rm {\,mph}}}$ and the other heads east going ${\displaystyle 5{\rm {\,mph}}}$ . How fast are the bikes getting away from each other after one hour?
${\displaystyle 13{\rm {\,mph}}}$
${\displaystyle 13{\rm {\,mph}}}$
24. You're making a can of volume ${\displaystyle 200{\rm {\,m^{3}}}}$ with a gold side and silver top/bottom. Say gold costs 10 dollars per ${\displaystyle {\rm {m^{2}}}}$ and silver costs 1 dollar per ${\displaystyle {\rm {m^{2}}}}$ . What's the minimum cost of such a can?
$878.76$878.76
25. A farmer is investing in ${\displaystyle 216{\rm {\,cm}}}$ of fencing so that he can create an outdoor pen to display three different animals to sell. To make it cost effective, he used one of the walls of the outdoor barn as one of the sides of the fenced in area, which is able to enclose the entire area. He wants the internal areas for the animals to roam in to be congruent (i.e. he wants to segment the total area into three equal areas). What is the maximum internal area that the animals can roam in, given these conditions?
${\displaystyle 972{\rm {\,cm^{2}}}}$
${\displaystyle 972{\rm {\,cm^{2}}}}$
26. What is the maximum area of a rectangle inscribed (fitted so that the corners of the rectangle are on the circumference) inside a circle of radius ${\displaystyle r}$?
${\displaystyle 2r^{2}}$.
${\displaystyle 2r^{2}}$.
27. A cylinder is to be fitted inside a glass spherical display case with a radius of ${\displaystyle R=3{\rm {\,in}}}$. (The sphere will form around the cylinder.) What is the largest volume that a cylinder will have inside such a display case?
${\displaystyle 12\pi {\sqrt {3}}{\rm {\,in^{3}}}}$.
${\displaystyle 12\pi {\sqrt {3}}{\rm {\,in^{3}}}}$.
28. A ${\displaystyle 6\,{\rm {ft}}}$ tall man is walking away from a light that is ${\displaystyle 15}$-feet above the ground. The man is walking away from the light at ${\displaystyle 6}$ feet per second. How fast (speed not velocity) is the shadow, cast by the man, changing its length with respect to time?
${\displaystyle 4{\rm {\,ft/s}}}$.
${\displaystyle 4{\rm {\,ft/s}}}$.

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 ${\displaystyle 5\,{\rm {ft/s}}}$. The dock is ${\displaystyle 5\,{\rm {ft}}}$ above the water. Answer items (a) through (b).

(a) How fast is the boat approaching the dock when ${\displaystyle 13\,{\rm {ft}}}$ of rope are out?
${\displaystyle 13{\rm {\,ft/s}}}$.
${\displaystyle 13{\rm {\,ft/s}}}$.
(b) Hence, what is the rate of change of the angle between the rope and the dock?
${\displaystyle 0.2{\rm {\,rad/s}}}$.
${\displaystyle 0.2{\rm {\,rad/s}}}$.
30. A very enthusiastic parent is video taping a runner in your class during a ${\displaystyle 100\,{\rm {m}}}$ race. The parent has the runner center frame and is recording ${\displaystyle 2\,{\rm {m}}}$ from the straight-line track. The runner in your class is running at a constant ${\displaystyle 4\,{\rm {m/s}}}$. What is the rate of change of the shooting angle if the runner passes the parent half a second after the parent's direct shot (after the point in which the runner's motion and the parent's line of sight are perpendicular)?
${\displaystyle 1\,{\rm {rad/s}}}$
${\displaystyle 1\,{\rm {rad/s}}}$

## Graphing Functions

For each of the following, graph a function that abides by the provided characteristics

30. ${\displaystyle f(1)=f(-2)=0,\ \lim _{x\to \infty }f(x)=\lim _{x\to -\infty }f(x)=0,\ {\mbox{ vertical asymptote at }}x=-3,\ f'(x)>0{\mbox{ on }}(0,2),}$ ${\displaystyle f'(x)<0{\mbox{ on }}(-\infty ,-3)\cup (-3,0)\cup (2,\infty ),\ f''(x)>0{\mbox{ on }}(-3,1)\cup (3,\infty ),\ f''(x)<0{\mbox{ on }}(-\infty ,-3)\cup (1,3).}$
There are many functions that satisfy all the conditions. Here is one example:
There are many functions that satisfy all the conditions. Here is one example:
31. ${\displaystyle f{\mbox{ has domain }}[-1,1],\ f(-1)=-1,\ f(-{\tfrac {1}{2}})=-2,\ f'(-{\tfrac {1}{2}})=0,\ f''(x)>0{\mbox{ on }}(-1,1)}$
There are many functions that satisfy all the conditions. Here is one example:
There are many functions that satisfy all the conditions. Here is one example:

## Approximation Problems

By assumption, for these problems, assume ${\displaystyle \pi \approx 3.14}$ and ${\displaystyle e\approx 2.718}$ 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.

35. Approximate ${\displaystyle \sin(3)}$ using whatever method. If you use Newton's or Euler's method, do it in a maximum of THREE (3) iterations.
Example: ${\displaystyle \sin(3)\approx 0.14}$ using Euler's method with step size ${\displaystyle \Delta x=0.14}$ and ${\displaystyle f(x,y)=1-y^{2}}$. See solution page for details
Example: ${\displaystyle \sin(3)\approx 0.14}$ using Euler's method with step size ${\displaystyle \Delta x=0.14}$ and ${\displaystyle f(x,y)=1-y^{2}}$. See solution page for details
36. Approximate ${\displaystyle {\sqrt {2}}}$ using whatever method. If you use Newton's or Euler's method, do it in a maximum of THREE (3) iterations.
Example: ${\displaystyle {\sqrt {2}}\approx 1.4167}$ using Newton-Rhapson method through ${\displaystyle 2}$ iterations. See solution page for details
Example: ${\displaystyle {\sqrt {2}}\approx 1.4167}$ using Newton-Rhapson method through ${\displaystyle 2}$ iterations. See solution page for details
37. Approximate ${\displaystyle \ln(3)}$ using whatever method. If you use Newton's or Euler's method, do it in a maximum of THREE (3) iterations.
Example: ${\displaystyle \ln(3)\approx 1.10{\text{ OR }}1.09}$ using local-point linearization. See solution page for details
Example: ${\displaystyle \ln(3)\approx 1.10{\text{ OR }}1.09}$ using local-point linearization. See solution page for details

## Advanced Understanding

45. Consider the differentiable function ${\displaystyle f(x)}$ for all ${\displaystyle x>-2}$ and continuous function ${\displaystyle g(x)}$ below, where ${\displaystyle g(x)}$ is linear for all ${\displaystyle x\geq 1}$ and differentiable for all ${\displaystyle x\in (-2,1)\cup (1,\infty )}$, and ${\displaystyle f(x)}$ and ${\displaystyle g(x)}$ are continuous for all ${\displaystyle x\geq -2}$.

a. Approximate ${\displaystyle f^{\prime }(1.6)}$.
${\displaystyle f^{\prime }(1.6)\approx -5}$
${\displaystyle f^{\prime }(1.6)\approx -5}$
b. Using your answer from (a), find ${\displaystyle \lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}}$.
${\displaystyle \lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}={\frac {3}{7}}}$
${\displaystyle \lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}={\frac {3}{7}}}$
c. Assume ${\displaystyle f^{\prime }(1.6)=f^{\prime }(2)}$. Find an approximation of the first positive root of ${\displaystyle f(x)}$ shown on the graph. Use only ONE (1) iteration.
Let ${\displaystyle c\in \mathbb {R} }$ allow ${\displaystyle f(c)=0}$. Using only one iteration of the Newton-Rhapson method, ${\displaystyle c=2.2}$.
Let ${\displaystyle c\in \mathbb {R} }$ allow ${\displaystyle f(c)=0}$. Using only one iteration of the Newton-Rhapson method, ${\displaystyle c=2.2}$.
d. A computer program found that there exists only one local maximum and minimum on the function ${\displaystyle f(x)}$ and found no local maximum or minimum for ${\displaystyle g(x)}$. Based on this finding, what flaw exists in the program and how can it be fixed?
Flaw: the program fails to consider the case where the derivative does not exist. Fix: add additional code considering this case. More details in the solutions page.
Flaw: the program fails to consider the case where the derivative does not exist. Fix: add additional code considering this case. More details in the solutions page.
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