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

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

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

critical points

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

x = 0

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 2, x = 2

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = - 2, x = 2

, therefore:

x = 0

x^{\frac{4}{3}} - x^{\frac{1}{3}}

f (x) = x^{3} - 3 x^{2} + 6

f (x, y) = e^{(x^{2} + 0.5 y^{2} - 3 x y - 4 x)}

f (x) = \frac{2 x ^{2}}{x ^{2} - 1}

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

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