What is the critical f(x)=(5x^2)/(x^2-16) ?

The critical f(x)=(5x^2)/(x^2-16) is x=0

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critical f(x)=(5x^2)/(x^2-16)

critical points

f (x) = \frac{5 x ^{2}}{x ^{2} - 16}

x = 0

Solution steps

\frac{5 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{5 x ^{2}}{x ^{2} - 16} : x < - 4 or - 4 < x < 4 or x > 4

\frac{5 x ^{2}}{x ^{2} - 16}

is not defined at

x = - 4, x = 4

, therefore:

x = 0

\frac{1}{64} x^{3}

x^{3} - 2 x^{2} - 3 x

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

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

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

What is the critical f(x)=(5x^2)/(x^2-16) ?
The critical f(x)=(5x^2)/(x^2-16) is x=0