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

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

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

critical points

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

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

Solution steps

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

All critical points are in domain

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

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

y = \frac{x + 5}{3 x}

(x - 1)^{2}

y = - x + 7

f (x) = x^{2} e^{x} - 6

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