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?
${\displaystyle 10{\tfrac {m}{s}}}$
${\displaystyle 10{\tfrac {m}{s}}}$
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 ${\displaystyle 200m^{3}}$  with a gold side and silver top/bottom. Say gold costs 10 dollars per m${\displaystyle ^{2}}$  and silver costs 1 dollar per ${\displaystyle m^{2}}$  . What's the minimum cost of such a can?
$878.76$878.76

Graphing Functions

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

25. ${\displaystyle f(1)=f(-2)=0,\ \lim _{x\to \infty }f(x)=\lim _{x\to -\infty }f(x)=0,\ {\mbox{ vertical asymptote at }}x=-3,\ f'(x)>0{\mbox{ on }}(0,2),}$  ${\displaystyle f'(x)<0{\mbox{ on }}(-\infty ,-3)\cup (-3,0)\cup (2,\infty ),\ f''(x)>0{\mbox{ on }}(-3,1)\cup (3,\infty ),\ f''(x)<0{\mbox{ on }}(-\infty ,-3)\cup (1,3).}$
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:
26. ${\displaystyle f{\mbox{ has domain }}[-1,1],\ f(-1)=-1,\ f(-{\tfrac {1}{2}})=-2,\ f'(-{\tfrac {1}{2}})=0,\ f''(x)>0{\mbox{ on }}(-1,1)}$
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.

34. 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
35. 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
36. 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

37. Consider the functions ${\displaystyle f(x)}$  and ${\displaystyle g(x)}$  below, where ${\displaystyle g(x)}$  is linear for all ${\displaystyle x\geq 1}$ .

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, ${\displaystyle c=2.2}$ .
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