Answer to Question #159101 in Calculus for dfsd

Question

Locate the absolute maximum and minimum for each of the following functions and justify your responses.

Question 1: 𝑓(𝑥) = cube root of x^2 on the interval [−1,1] 

Question2: 𝑓(𝑥)= 𝑔(𝑥) = 𝑥𝑒^2𝑥 on the interval [−2,0] 

Question 3: Given 𝑓(𝑥) = 3𝑥^3 − 18𝑥^2 − 45𝑥 + 10

a) On what interval(s), is 𝑓(𝑥) increasing? Justify.

b) what value(s) of 𝑥 does 𝑓(𝑥) have a relative minimum?

c) On what interval(s), is 𝑓(𝑥) decreasing and concave up? Justify.

Answer 1

Question 1:

$f(x)=(x^2)^{\frac{1}{3}}$ on the interval [−1,1]

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

$f'(x)<0$ when $x<0$ and $f'(x)>0$ when $x>0$

$f'(x)$ does not exists when $x=0$

$f(-1)=f(1)=1$ - absolute maximum on the interval [−1,1]

$f(0)=0$ - absolute minimum on the interval [−1,1]

Question2: $g(x)=xe^{2x}$ on the interval [−2,0]

$g'(x)=e^{2x}+2xe^{2x}=(1+2x)e^{2x}$

$g'(x)=0$ at the point $x=-\frac{1}{2}$ . This is point of minimum because $g'(x)<0$ when $x<-\frac{1}{2}$ ,

and $g'(x)>0$ when $x>-\frac{1}{2}$ .

$g(-\frac{1}{2})=-\frac{1}{2}e^{-1}$ - absolute minimum on the interval [−2,0].

$g(-2)=-2e^{-4}$ and $g(0)=0$

$g(0)>g(-2)$ , so g(0)=0 - absolute maximum on the interval [−2,0].

Question 3: Given $f(x)=3x^3-18x^2-45x+10$

a) On what interval(s), is 𝑓(𝑥) increasing? Justify.

$f'(x)=9x^2-36x-45=9(x^2-4x-5)=9(x+1)(x-5)$

$f'(x)>0$ on the intervals $(-\infty, -1)$ and $(5,+\infty)$ , so $f(x)$ is increasing on these intervals

b) what value(s) of 𝑥 does 𝑓(𝑥) have a relative minimum?

at the point $x=5$ function $f(x)$ has a relative minimum because $f'(x)$ changes its sign from

minus to plus

c) On what interval(s), is 𝑓(𝑥) decreasing and concave up? Justify.

$f'(x)<0$ on the interval $(-1,5)$ , so $f(x)$ is decreasing on this interval.

$f''(x)=18x-36=18(x-2)$

$f''(x)>0$ on the interval $(2, +\infty)$ , so the function is concave up on this interval