What is the extreme f(x)=(x^2+9)/(x^2-64) ?

The extreme f(x)=(x^2+9)/(x^2-64) is Maximum(0,-9/64)

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extreme f(x)=(x^2+9)/(x^2-64)

extreme points

f (x) = \frac{x ^{2} + 9}{x ^{2} - 64}

Maximum (0, - \frac{9}{64})

Solution steps

f^{'} (x) = - \frac{146 x}{( x ^{2} - 64 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 8,

Increasing

: - 8 < x < 0,

Decreasing

: 0 < x < 8,

Decreasing

: 8 < x < \infty

Plug

x = 0

into

\frac{x ^{2} + 9}{x ^{2} - 64} : - \frac{9}{64}

Maximum (0, - \frac{9}{64})

f (x) = \frac{2 x + 6}{x - 4}

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

\frac{x ^{2} - 2 x}{x ^{4} - 16}

f (x) = \frac{x + 4}{9 x + 4}

- x^{2} - 4

What is the extreme f(x)=(x^2+9)/(x^2-64) ?
The extreme f(x)=(x^2+9)/(x^2-64) is Maximum(0,-9/64)