What is the extreme f(x)=((x^2))/(x^2-81) ?

The extreme f(x)=((x^2))/(x^2-81) is Maximum(0,0)

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extreme f(x)=((x^2))/(x^2-81)

extreme

Maximum (0, 0)

Solution steps

f^{'} (x) = - \frac{162 x}{( x ^{2} - 81 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 9,

Increasing

: - 9 < x < 0,

Decreasing

: 0 < x < 9,

Decreasing

: 9 < x < \infty

Plug

x = 0

into

\frac{x ^{2}}{x ^{2} - 81} : 0

Maximum (0, 0)

What is the extreme f(x)=((x^2))/(x^2-81) ?
The extreme f(x)=((x^2))/(x^2-81) is Maximum(0,0)