What is the critical f(x)=-x^4+6x^2 ?

The critical f(x)=-x^4+6x^2 is x=-sqrt(3),x=0,x=sqrt(3)

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critical f(x)=-x^4+6x^2

critical points

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

x = - 3, x = 0, x = 3

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 3, x = 0, x = 3

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = - 3, x = 0, x = 3

f (x) = \frac{4 ^{y}}{4}

f (x) = - 4 (x - 2)^{2} (x^{2} - 9)

6 x^{4} + 16 x^{3}

f (x) = \frac{2 x}{2 x - 4}

f (x) = 2 + 4 + 6 x

What is the critical f(x)=-x^4+6x^2 ?
The critical f(x)=-x^4+6x^2 is x=-sqrt(3),x=0,x=sqrt(3)