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 points

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

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)

f (x) = 3 x^{3} - 12 x^{2} + 12 x - 6, 0 \leq x \leq 4

y = 3 x^{3} (x - 5)^{2}

y = \frac{1}{x - 8} - 9

y = x + \frac{25}{x}

f (x) = \frac{3 x ^{2}}{x - 2}, - 2 \leq x \leq 1

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