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

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

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

critical points

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

x = 0

Solution steps

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

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

is not defined at

x = - 4, x = 4

, therefore:

x = 0

F (a, b) = a^{6} - 64 b^{6}

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

f (x) = \frac{( x ^{2} - 1 )}{( x ^{2} + 1 )}

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

(x^{2} + 10) (1 - x^{2})

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