What is the critical 2x^4-4x^2+6 ?

The critical 2x^4-4x^2+6 is x=-1,x=0,x=1

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critical 2x^4-4x^2+6

critical points

2 x^{4} - 4 x^{2} + 6

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

Solution steps

2 x^{4} - 4 x^{2} + 6

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

2 x^{4} - 4 x^{2} + 6 : - \infty < x < \infty

All critical points are in domain

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

2 x - 5 y = 10

f (x) = - \frac{5}{2 t ^{\frac{3}{2}}}

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

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

f (x) = \frac{x + 5}{x + 3}