What is the extreme f(x)=(x^2+1)/(x^2-9) ?

The extreme f(x)=(x^2+1)/(x^2-9) is Maximum(0,-1/9)

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extreme f(x)=(x^2+1)/(x^2-9)

extreme points

f (x) = \frac{x ^{2} + 1}{x ^{2} - 9}

Maximum (0, - \frac{1}{9})

Solution steps

f^{'} (x) = - \frac{20 x}{( x ^{2} - 9 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = 0

into

\frac{x ^{2} + 1}{x ^{2} - 9} : - \frac{1}{9}

Maximum (0, - \frac{1}{9})

\frac{2 x}{x ^{2} - 1}

5 sin (4 x - π)

\frac{1}{( x - 4 ) ^{2}}

- 3 x^{4} + 20 x^{3} - 24 x^{2}

- \frac{9}{( x - 7 ) ^{2}}

What is the extreme f(x)=(x^2+1)/(x^2-9) ?
The extreme f(x)=(x^2+1)/(x^2-9) is Maximum(0,-1/9)