What is the critical f(x)=x^{2/3}-x/6 ?

The critical f(x)=x^{2/3}-x/6 is x=0,x=64

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critical f(x)=x^{2/3}-x/6

critical points

f (x) = x^{\frac{2}{3}} - \frac{x}{6}

x = 0, x = 64

Solution steps

x^{\frac{2}{3}} - \frac{x}{6}

Find where

f^{'} (x)

is equal to zero or undefined

x = 64, x = 0

Identify critical points not in

f (x)

domain

Domain of

x^{\frac{2}{3}} - \frac{x}{6} : - \infty < x < \infty

All critical points are in domain

x = 0, x = 64

f (x) = x^{- 8} ln (x)

\frac{x ^{2} - 6 x + 12}{x - 4}

f (x) = x^{3} + 4 x

(x^{2} + 12) (4 - x^{2})

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

What is the critical f(x)=x^{2/3}-x/6 ?
The critical f(x)=x^{2/3}-x/6 is x=0,x=64