What is the critical (6x^2)/(x^2-9) ?

The critical (6x^2)/(x^2-9) is x=0

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critical (6x^2)/(x^2-9)

critical points

\frac{6 x ^{2}}{x ^{2} - 9}

x = 0

Solution steps

\frac{6 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{6 x ^{2}}{x ^{2} - 9} : x < - 3 or - 3 < x < 3 or x > 3

\frac{6 x ^{2}}{x ^{2} - 9}

is not defined at

x = - 3, x = 3

, therefore:

x = 0

f (x, y) = 2 x^{2} + y^{2} + 8 x - 6 y + 20

f (x) = \frac{2}{3} x^{3} - x^{2} - 84 - 8

f (x) = 8 x^{3} - 30 x^{2} - 504 x + 1

f (x) = I n^{3} x

f (x, y) = x^{4} + y^{4} + 4 a x y + 8 a^{2}