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

The critical f(x)=(6x^2)/(x^2-4) is x=0

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

critical points

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

x = 0

Solution steps

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

\frac{6 x ^{2}}{x ^{2} - 4}

is not defined at

x = - 2, x = 2

, therefore:

x = 0

x^{2} + 2 x - 1

y = sin^{2} (x) + cos (x)

f (x, y) = 2 x^{3} - 21 x^{2} + y^{3} + 12 y^{2}

f (x) = e^{x^{2} - 8 x - 1}

f (x) = x^{3} - 2 x^{2} + x + 1

What is the critical f(x)=(6x^2)/(x^2-4) ?
The critical f(x)=(6x^2)/(x^2-4) is x=0