What is the critical f(x)=(x^2-2x-8)^{1/3} ?

The critical f(x)=(x^2-2x-8)^{1/3} is x=-2,x=1,x=4

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critical f(x)=(x^2-2x-8)^{1/3}

critical points

f (x) = (x^{2} - 2 x - 8)^{\frac{1}{3}}

x = - 2, x = 1, x = 4

Solution steps

(x^{2} - 2 x - 8)^{\frac{1}{3}}

Find where

f^{'} (x)

is equal to zero or undefined

x = 1, x = - 2, x = 4

Identify critical points not in

f (x)

domain

Domain of

(x^{2} - 2 x - 8)^{\frac{1}{3}} : - \infty < x < \infty

All critical points are in domain

x = - 2, x = 1, x = 4

f (x) = x + 3

y = - x + 1 - 3

f (x) = \frac{5 x ( x - 6 )}{5 x ^{2} - 29 x - 6}

f (x) = x^{2} + 2 x - 1

f (x) = 8 - 6 e^{- x}

What is the critical f(x)=(x^2-2x-8)^{1/3} ?
The critical f(x)=(x^2-2x-8)^{1/3} is x=-2,x=1,x=4