What is the critical \sqrt[3]{(x^2-4)^2} ?

The critical \sqrt[3]{(x^2-4)^2} is x=-2,x=0,x=2

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critical \sqrt[3]{(x^2-4)^2}

critical points

x = - 2, x = 0, x = 2

Solution steps

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 2, x = 2

Identify critical points not in

f (x)

domain

Domain of

All critical points are in domain

x = - 2, x = 0, x = 2

x^{2} e^{x}

f (x) = x^{8} (x - 1)^{7}

e^{- y} + e^{2}

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

y = 2 x - 6

What is the critical \sqrt[3]{(x^2-4)^2} ?
The critical \sqrt[3]{(x^2-4)^2} is x=-2,x=0,x=2