Answer to Question #160500 in Calculus for yjyt

Question

Locate the absolute maximum and minimum for each of the following functions and justify your responses. Be careful on your differentiation and show all work with calculus justification!

 a) 𝑓(𝑥) = (Cube root of x^2) − 𝑥 on the interval [−1,1]

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

For each of the following functions, respond to the given prompts. Show all work that leads to your responses. With calculus justification!

a. On what interval(s), is 𝑓(𝑥) increasing? Show all work with calculus justification

b. At what value(s) of 𝑥 does 𝑓(𝑥) have a relative minimum? Show all work with calculus justification

c. On what interval(s), is 𝑓(𝑥) decreasing and concave up? Show all work with calculus justification

Answer 1

a)

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

$$f'(x)=\dfrac{2}{3\sqrt[3]{x}}-1$$

$$f'(x)=0=>\dfrac{2}{3\sqrt[3]{x}}-1=0=>x=\dfrac{8}{27}$$

$f'(x)$ is undefined at $x=0$

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

$$f(1)=\sqrt[3]{(1)^2 }-(1)=0$$

$$f(0)=\sqrt[3]{(0)^2 }-0=0$$

$$f(\dfrac{8}{27})=\sqrt[3]{(\dfrac{8}{27})^2 }-\dfrac{8}{27}=\dfrac{4}{27}$$

The function $f$ the absolute maximum with value of $2$ at $x=-1$ on $[-1,1].$

The function $f$ the absolute minimum with value of at $x=0$ and $x=1$ on $[-1,1].$

b)

$$g(x)=xe^{2x}$$

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

$$g'(x)=0=>e^{2x}+2xe^{2x}=0=>x=-\dfrac{1}{2}$$

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

$$g(0)=0(e^{2(0)})=0$$

$$g(-\dfrac{1}{2})=-\dfrac{1}{2}(e^{2(-{1 \over 2})})=-\dfrac{1}{2e}$$

The function $g$ the absolute maximum with value of at $x=0$ on $[-2,0].$

The function $g$ the absolute minimum with value of $-\dfrac{1}{2e}$ at $x=-\dfrac{1}{2}$ on $[-2,0].$

a. $f(x)=\sqrt[3]{x^2 }-x$

$Domain: (-\infin, \infin)$

$$f'(x)=\dfrac{2}{3\sqrt[3]{x}}-1$$

$$f'(x)=0=>\dfrac{2}{3\sqrt[3]{x}}-1=0=>x=\dfrac{8}{27}$$

$f'(x)$ is undefined at $x=0.$

If $0<x<\dfrac{8}{27}, f'(x)>0$

The function $f(x)$ is increasing on $(0, \dfrac{8}{27}).$

b. If $x<0, f'(x)<0$

The function $f(x)$ is decreasing on $(-\infin, 0).$

If $x>\dfrac{8}{27}, f'(x)<0$

The function $f(x)$ is decreasing on $(\dfrac{8}{27},\infin).$

The function $f(x)$ has a relative minimum with value of at $x=0.$

c. The function $f(x)$ is decreasing on $(-\infin,0)\cup(\dfrac{8}{27},\infin).$

$f''(x)=-\dfrac{2}{9\sqrt[3]{x^4}}$

$f''(x)<0, x\not=0$

The function $f(x)$ is concave down on $(-\infin,0)\cup(0,\infin).$

The function $f(x)$ is never concave up.