What is the critical f(x)=(2x-8)^{2/3} ?

The critical f(x)=(2x-8)^{2/3} is x=4

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

critical points

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

x = 4

Solution steps

(2 x - 8)^{\frac{2}{3}}

Find where

f^{'} (x)

is equal to zero or undefined

x = 4

Identify critical points not in

f (x)

domain

Domain of

(2 x - 8)^{\frac{2}{3}} : - \infty < x < \infty

All critical points are in domain

x = 4

x^{3} - \frac{3}{2} x^{2}, - 5 \leq x \leq 4

f (x) = \frac{x ^{2} + 1}{2 x ^{2} + 7}

4 (\frac{1}{2})^{x - 3} - 2

y = 5 x^{2}

- e^{x - 1} - 3