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

The critical f(x)=(x^2(x-4))/(x+6) is x=(-7-sqrt(145))/2 ,x=0,x=(-7+sqrt(145))/2

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

critical points

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

x = \frac{- 7 - 145}{2}, x = 0, x = \frac{- 7 + 145}{2}

Solution steps

\frac{x ^{2} ( x - 4 )}{x + 6}

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{- 7 - 145}{2}, x = 0, x = \frac{- 7 + 145}{2}, x = - 6

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2} ( x - 4 )}{x + 6} : x < - 6 or x > - 6

\frac{x ^{2} ( x - 4 )}{x + 6}

is not defined at

x = - 6

, therefore:

x = \frac{- 7 - 145}{2}, x = 0, x = \frac{- 7 + 145}{2}

What is the critical f(x)=(x^2(x-4))/(x+6) ?
The critical f(x)=(x^2(x-4))/(x+6) is x=(-7-sqrt(145))/2 ,x=0,x=(-7+sqrt(145))/2