What is the critical f(x)=(3x^2)/(x^2-4) ?

The critical f(x)=(3x^2)/(x^2-4) is x=0

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critical f(x)=(3x^2)/(x^2-4)

critical points

f (x) = \frac{3 x ^{2}}{x ^{2} - 4}

x = 0

Solution steps

\frac{3 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{3 x ^{2}}{x ^{2} - 4} : x < - 2 or - 2 < x < 2 or x > 2

\frac{3 x ^{2}}{x ^{2} - 4}

is not defined at

x = - 2, x = 2

, therefore:

x = 0

f (x, y) = 25 - x^{2} - y^{2}

f (x) = x^{4} - 4 x^{3}, - 1 \leq x \leq 4

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

f (x) = x^{3} - 12 x y^{2} + 72 y^{2}

f (x, y) = 6 x + 12 y - x^{2} - y^{2} - 41

What is the critical f(x)=(3x^2)/(x^2-4) ?
The critical f(x)=(3x^2)/(x^2-4) is x=0