What is the critical f(x)=(x^2)/(x^2-16) ?

The critical f(x)=(x^2)/(x^2-16) is x=0

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critical f(x)=(x^2)/(x^2-16)

critical points

f (x) = \frac{x ^{2}}{x ^{2} - 16}

x = 0

Solution steps

\frac{x ^{2}}{x ^{2} - 16}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 4, x = 4

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2}}{x ^{2} - 16} : x < - 4 or - 4 < x < 4 or x > 4

\frac{x ^{2}}{x ^{2} - 16}

is not defined at

x = - 4, x = 4

, therefore:

x = 0

f (x) = \frac{x + 2}{x}

y = x^{2} - 2 x

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

f (x) = (x - 2) (x - 5)^{3} + 9

4 x + 6

What is the critical f(x)=(x^2)/(x^2-16) ?
The critical f(x)=(x^2)/(x^2-16) is x=0