What is the critical f(x)=x^{-1/3}(x-12) ?

The critical f(x)=x^{-1/3}(x-12) is x=-6

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critical f(x)=x^{-1/3}(x-12)

critical points

f (x) = x^{- \frac{1}{3}} (x - 12)

x = - 6

Solution steps

x^{- \frac{1}{3}} (x - 12)

Find where

f^{'} (x)

is equal to zero or undefined

x = - 6, x = 0

Identify critical points not in

f (x)

domain

Domain of

x^{- \frac{1}{3}} (x - 12) : x < 0 or x > 0

x^{- \frac{1}{3}} (x - 12)

is not defined at

x = 0

, therefore:

x = - 6

x^{3} + 6 x^{2} - 13 x + 42

5 x^{2} ln (x)

f (x) = I n^{2} 2 x

f (x) = - 20 - 40 x - 25 x^{2}

x^{3} - 6 x^{2} - 15 x + 40

What is the critical f(x)=x^{-1/3}(x-12) ?
The critical f(x)=x^{-1/3}(x-12) is x=-6