What is the critical x^4-8x^3 ?

The critical x^4-8x^3 is x=0,x=6

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critical x^4-8x^3

critical points

x^{4} - 8 x^{3}

x = 0, x = 6

Solution steps

x^{4} - 8 x^{3}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 6

Identify critical points not in

f (x)

domain

Domain of

x^{4} - 8 x^{3} : - \infty < x < \infty

All critical points are in domain

x = 0, x = 6

f (x) = - x - 1 - 2

f (x) = x^{4} - 32 x^{2} + 1

(x + 4)^{3} - 2

\frac{1}{x ^{2}}

(0, - 2), (- 4, 4)