What is the critical f(x)=(4x^2)/(x^2-9) ?

The critical f(x)=(4x^2)/(x^2-9) is x=0

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critical f(x)=(4x^2)/(x^2-9)

critical points

f (x) = \frac{4 x ^{2}}{x ^{2} - 9}

x = 0

Solution steps

\frac{4 x ^{2}}{x ^{2} - 9}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 3, x = 3

Identify critical points not in

f (x)

domain

Domain of

\frac{4 x ^{2}}{x ^{2} - 9} : x < - 3 or - 3 < x < 3 or x > 3

\frac{4 x ^{2}}{x ^{2} - 9}

is not defined at

x = - 3, x = 3

, therefore:

x = 0

f (x) = x^{4} - 6 x^{3} - 12 x^{2} - 8 x^{1}

y = x (4 - x)^{3}

f (x) = x^{8} (x - 2)^{7}

x^{2} + y^{2} - 3 x y

f (x, y) = 4 x^{2} + \frac{1}{4} y^{3} - 2 x y

What is the critical f(x)=(4x^2)/(x^2-9) ?
The critical f(x)=(4x^2)/(x^2-9) is x=0