What is the critical sqrt((x^2-4)^2) ?

The critical sqrt((x^2-4)^2) is x=-2,x=0,x=2

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critical sqrt((x^2-4)^2)

critical points

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

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

Solution steps

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

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

(x^{2} - 4)^{2} : - \infty < x < \infty

All critical points are in domain

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

f (x) = (- \frac{8 π}{11})

f (x) = e^{3 x} + e^{- 3 x}

x^{3} \cdot 3

f (x) = - 3 x - 6 + 5

f (x) = \frac{- 2 x - 8}{x ^{2} + 6 x + 8}

What is the critical sqrt((x^2-4)^2) ?
The critical sqrt((x^2-4)^2) is x=-2,x=0,x=2