Calculus/Differentiation/Applications of Derivatives/Exercises

A reviewed version of this page, approved on 8 February 2017, was based off this revision.

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	Differentiation/Applications of Derivatives/Exercises

Relative Extrema

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

1.

f(x)={\frac {x}{x+1}}

none

2.

f(x)={\sqrt[{3}]{(x-1)^{2}}}

Minimum at the point

(1,0)

Minimum at the point

(1,0)

3.

f(x)=x^{2}+{\frac {2}{x}}

Relative minimum at

x=1

Relative minimum at

x=1

4.

f(s)={\frac {s}{1+s^{2}}}

Relative minimum at

s=-1

Relative maximum at

s=1

Relative minimum at

s=-1

Relative maximum at

s=1

5.

f(x)=x^{2}-4x+9

Relative minimum at

x=2

Relative minimum at

x=2

6.

f(x)={\frac {x^{2}+x+1}{x^{2}-x+1}}

Relative minimum at

x=-1

Relative maximum at

x=1

Relative minimum at

x=-1

Relative maximum at

x=1

Solutions

Range of Function

7. Show that the expression

x+{\frac {1}{x}}

cannot take on any value strictly between 2 and -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 $2$ .

The above analysis shows that there is a gap in the function's range between

-2

and

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 $2$ .

The above analysis shows that there is a gap in the function's range between

-2

and

2

.

Absolute Extrema

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

8.

f(x)={\frac {x^{3}}{3}}-{\frac {x^{2}}{2}}+1

on

[0,3]

Maximum at

(3,{\tfrac {11}{2}})

; minimum at

(1,{\tfrac {5}{6}})

Maximum at

(3,{\tfrac {11}{2}})

; minimum at

(1,{\tfrac {5}{6}})

9.

f(x)=\left({\frac {4}{3}}x^{2}-1\right)x

on

[-{\tfrac {1}{2}},2]

Maximum at

(2,{\tfrac {26}{3}})

; minimum at

({\tfrac {1}{2}},-{\tfrac {1}{3}})

Maximum at

(2,{\tfrac {26}{3}})

; minimum at

({\tfrac {1}{2}},-{\tfrac {1}{3}})

Solutions

Determine Intervals of Change

Find the intervals where the following functions are increasing or decreasing

10.

f(x)=10-6x-2x^{2}

Increasing on

(-\infty ,-{\tfrac {3}{2}})

; decreasing on

(-{\tfrac {3}{2}},\infty )

Increasing on

(-\infty ,-{\tfrac {3}{2}})

; decreasing on

(-{\tfrac {3}{2}},\infty )

11.

f(x)=2x^{3}-12x^{2}+18x+15

Decreasing on

(1,3)

; increasing elsewhere

Decreasing on

(1,3)

; increasing elsewhere

12.

f(x)=5+36x+3x^{2}-2x^{3}

Increasing on

(-2,3)

; decreasing elsewhere

Increasing on

(-2,3)

; decreasing elsewhere

13.

f(x)=8+36x+3x^{2}-2x^{3}

Increasing on

(-2,3)

; decreasing elsewhere

Increasing on

(-2,3)

; decreasing elsewhere

14.

f(x)=5x^{3}-15x^{2}-120x+3

Decreasing on

(-2,4)

; increasing elsewhere

Decreasing on

(-2,4)

; increasing elsewhere

15.

f(x)=x^{3}-6x^{2}-36x+2

Decreasing on

(-2,6)

; increasing elsewhere

Decreasing on

(-2,6)

; increasing elsewhere

Solutions

Determine Intervals of Concavity

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

16.

f(x)=10-6x-2x^{2}

Concave down everywhere

17.

f(x)=2x^{3}-12x^{2}+18x+15

Concave down on

(-\infty ,2)

; concave up on

(2,\infty )

Concave down on

(-\infty ,2)

; concave up on

(2,\infty )

18.

f(x)=5+36x+3x^{2}-2x^{3}

Concave up on

(-\infty ,{\tfrac {1}{2}})

; concave down on

({\tfrac {1}{2}},\infty )

Concave up on

(-\infty ,{\tfrac {1}{2}})

; concave down on

({\tfrac {1}{2}},\infty )

19.

f(x)=8+36x+3x^{2}-2x^{3}

Concave up on

(-\infty ,{\tfrac {1}{2}})

; concave down on

({\tfrac {1}{2}},\infty )

Concave up on

(-\infty ,{\tfrac {1}{2}})

; concave down on

({\tfrac {1}{2}},\infty )

20.

f(x)=5x^{3}-15x^{2}-120x+3

Concave down on

(-\infty ,1)

; concave up on

(1,\infty )

Concave down on

(-\infty ,1)

; concave up on

(1,\infty )

21.

f(x)=x^{3}-6x^{2}-36x+2

Concave down on

(-\infty ,2)

; concave up on

(2,\infty )

Concave down on

(-\infty ,2)

; concave up on

(2,\infty )

Solutions

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

4t

meters per second (time

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?