What is the critical f(x)=x^4-32x^2 ?

The critical f(x)=x^4-32x^2 is x=-4,x=0,x=4

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critical f(x)=x^4-32x^2

critical points

f (x) = x^{4} - 32 x^{2}

x = - 4, x = 0, x = 4

Solution steps

x^{4} - 32 x^{2}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 4, x = 0, x = 4

Identify critical points not in

f (x)

domain

Domain of

x^{4} - 32 x^{2} : - \infty < x < \infty

All critical points are in domain

x = - 4, x = 0, x = 4

f (x) = \frac{4}{3} x - \frac{5}{3}

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

f (x) = \frac{3 x ^{2} - 6 x}{x ^{2} - 25}

3 = 3 (2) - 12

f (x) = - (x - 2)^{2} + 5

What is the critical f(x)=x^4-32x^2 ?
The critical f(x)=x^4-32x^2 is x=-4,x=0,x=4