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

The extreme f(x)=(x^2)/((x^2-16)) is Maximum(0,0)

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

extreme points

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

Maximum (0, 0)

Solution steps

f^{'} (x) = - \frac{32 x}{( x ^{2} - 16 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = 0

into

\frac{x ^{2}}{x ^{2} - 16} : 0

Maximum (0, 0)

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