What is the critical f(x)=15x^4-15x^2 ?

The critical f(x)=15x^4-15x^2 is x=-1/(sqrt(2)),x=0,x= 1/(sqrt(2))

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critical f(x)=15x^4-15x^2

critical points

f (x) = 15 x^{4} - 15 x^{2}

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

Solution steps

15 x^{4} - 15 x^{2}

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

15 x^{4} - 15 x^{2} : - \infty < x < \infty

All critical points are in domain

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

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

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

f (x) = cos (4 x)

5 - 2 x

3 x + 5^{2} + 4

What is the critical f(x)=15x^4-15x^2 ?
The critical f(x)=15x^4-15x^2 is x=-1/(sqrt(2)),x=0,x= 1/(sqrt(2))