What is the critical f(x)=(8x^2)/(x-2) ?

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

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critical f(x)=(8x^2)/(x-2)

critical points

f (x) = \frac{8 x ^{2}}{x - 2}

x = 0, x = 4

Solution steps

\frac{8 x ^{2}}{x - 2}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 4, x = 2

Identify critical points not in

f (x)

domain

Domain of

\frac{8 x ^{2}}{x - 2} : x < 2 or x > 2

\frac{8 x ^{2}}{x - 2}

is not defined at

x = 2

, therefore:

x = 0, x = 4

f (x) = 4 x - 3 y = 8

\frac{e ^{x} - e ^{- x}}{5}

f (x) = \frac{( x + 1 )}{2 x + 1}

f (x) = x - 11

f (x) = x^{6} (x - 2)^{5}

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