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

The critical f(x)=4x^2-6x^4 is x=-1/(sqrt(3)),x=0,x= 1/(sqrt(3))

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

critical points

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

x = - \frac{1}{3}, x = 0, x = \frac{1}{3}

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{1}{3}, x = 0, x = \frac{1}{3}

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = - \frac{1}{3}, x = 0, x = \frac{1}{3}

f (x) = - 40 - 8 x - 4

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

y = x - 1

f (x) = 4 x + 9

f (x) = - \frac{1}{2} tan (2 π x)

What is the critical f(x)=4x^2-6x^4 ?
The critical f(x)=4x^2-6x^4 is x=-1/(sqrt(3)),x=0,x= 1/(sqrt(3))