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

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

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critical f(x)=(x^2-4)^2

critical points

f (x) = (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 = - 2, x = 0, 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

\frac{1}{2} 4^{x}

f (x) = \frac{1}{1 + x}

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

2 x - y - 7 = 0

Y (x) = - \frac{2}{3} x + \frac{1}{3}

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