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

The critical f(x)=(2x^2)/(x-4) is x=0,x=8

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

critical points

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

x = 0, x = 8

Solution steps

\frac{2 x ^{2}}{x - 4}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 8, x = 4

Identify critical points not in

f (x)

domain

Domain of

\frac{2 x ^{2}}{x - 4} : x < 4 or x > 4

\frac{2 x ^{2}}{x - 4}

is not defined at

x = 4

, therefore:

x = 0, x = 8

f (x) = x^{2} e^{9 x}

y = x^{\frac{4}{5}} (x - 4)^{2}

f (x) = 4 x^{3} - 36 x^{2} + 96 x - 64

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

f (x) = I n (x - 5)

What is the critical f(x)=(2x^2)/(x-4) ?
The critical f(x)=(2x^2)/(x-4) is x=0,x=8