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

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

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

critical points

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

x = 0

Solution steps

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

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

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

is not defined at

x = - 2, x = 2

, therefore:

x = 0

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

\frac{x}{1 + x ^{2}}

f (x) = \frac{x ^{4}}{4} + \frac{x ^{3}}{2} - \frac{x ^{2}}{2} + 8

f (x) = \frac{7 ( - x ^{2} + 49 )}{( x ^{2} + 49 ) ^{2}}

f (x) = - 14 + 5 x^{2} + x y + y^{2}

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