What is the critical 8x^4-8x^2+1 ?

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

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critical 8x^4-8x^2+1

critical points

8 x^{4} - 8 x^{2} + 1

x = - \frac{1}{2}, x = 0, x = \frac{1}{2}

Solution steps

8 x^{4} - 8 x^{2} + 1

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{1}{2}, x = 0, x = \frac{1}{2}

Identify critical points not in

f (x)

domain

Domain of

8 x^{4} - 8 x^{2} + 1 : - \infty < x < \infty

All critical points are in domain

x = - \frac{1}{2}, x = 0, x = \frac{1}{2}

y = - 2 (5)^{x - 4} + 3

f (x) = \frac{x ^{2} + 6 x - 8}{x - 4}

f (x) = \frac{x + 3}{x - 7}

f (x) = 2 x^{2} - 8 x - 3

f (x) = \frac{x - 9}{x ^{2} + 81}

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