What are the critical points of $f (x) = x^{\frac{2}{3}} x^{2} - 16$ $f(x)=x^(2/3)x^2 - 16$ ?

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Answer 1 · 2015-12-23T04:21:22

$c = 0$ $c=0$

Critical points ( $c$ $c$ ) occur when $f'(c)=0$ or when $f'(c)$ doesn't exist.

In order to find $f'(x)$ , first simplify $f(x)$ .

$f(x)=x^(2/3)x^2-16$

$f(x)=x^(2/3)x^(6/3)-16$

$f(x)=x^(8/3)-16$

Use the power rule: $d/dx[x^n]=nx^(n-1)$

$f'(x)=8/3x^(5/3)$

$f'(x)=0$ when $x=0$ and is never undefined.

Thus, $c=0$ .

graph{x^(2/3)x^2-16 [-32.06, 32.9, -20.27, 12.2]}

What are the critical points of f(x)=x^(2/3)x^2 - 16f(x)=x23x2−16 f(x)=x^(2/3)x^2 - 16?