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

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

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

extreme points

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

Maximum (0, - 2)

Solution steps

f^{'} (x) = - \frac{6 x}{( x ^{2} - 1 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 0

into

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

Maximum (0, - 2)

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