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

The critical f(x)=x^4-2x^2+8 is x=-1,x=0,x=1

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

critical points

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

x = - 1, x = 0, x = 1

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = 0, x = 1

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = - 1, x = 0, x = 1

s (t) = \frac{5 t}{sin ( t )}

ln (x - 5)

f (x) = \frac{3 x}{x ^{2} + 4}

3 x^{2} + 6 x

- 3 cos (3 x - 5)

What is the critical f(x)=x^4-2x^2+8 ?
The critical f(x)=x^4-2x^2+8 is x=-1,x=0,x=1