What is the critical f(x)=x^4-8x^2+4 ?

The critical f(x)=x^4-8x^2+4 is x=-2,x=0,x=2

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critical f(x)=x^4-8x^2+4

critical points

f (x) = x^{4} - 8 x^{2} + 4

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

Solution steps

x^{4} - 8 x^{2} + 4

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^{4} - 8 x^{2} + 4 : - \infty < x < \infty

All critical points are in domain

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

f (x) = \frac{- 3 x ^{2} - 12 x}{2 x + 8}

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

y = 2 x^{(2)} + 8 x + 5

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

What is the critical f(x)=x^4-8x^2+4 ?
The critical f(x)=x^4-8x^2+4 is x=-2,x=0,x=2