Answer to Question #161101 in Calculus for dsfsdds

Question

Locate the absolute maximum and minimum for each of the following functions and justify your responses. (Show all work and justify)

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

Answer 1

Given: $f(x)=\sqrt[3]{x^2}-x$ , $\left [ -1,1 \right ]$

Find the first derivative of the given function.

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

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

For critical values, set $f'(x)=0\Rightarrow \frac{2}{3}x^{\frac{-1}{3}}-1=0$

$\Rightarrow x^{\frac{-1}{3}}=\frac{3}{2}\Rightarrow x=\left ( \frac{2}{3} \right )^3=\frac{8}{27}$

Also observe that $x=\frac{8}{27}\in \left [ -1,1 \right ]$

Now find the function values at the end values $x=-1,1$

and at the critical value $x=\frac{8}{27}$

$x=-1\Rightarrow f(-1)=(-1)^{\frac{2}{3}}-(-1)=1+1=2$ (Absolute maximum)

$x=1\Rightarrow f(1)=(1)^{\frac{2}{3}}-(1)=1-1=0$ (Absolute minimum)

$x=\frac{8}{27}\Rightarrow f(\frac{8}{27})=(\frac{8}{27})^{\frac{2}{3}}-(1)=\frac{4}{9}$

Therefore $f(x)$ has absolute maximum at $x=-1$ and

absolute maximum value $=2$

$f(x)$ has absolute minimum value at $x=1$ and

absolute minimum value $=0$