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

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

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

extreme points

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

Maximum (0, - 1)

Solution steps

f^{'} (x) = - \frac{4 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{x ^{2} + 1}{x ^{2} - 1} : - 1

Maximum (0, - 1)

\frac{x ^{2}}{x - 5}

f (x, y) = x^{2} y + y^{3} - 12 y

f (x, y) = \frac{x y}{1 + e ^{x}}

f (x, y) = 5 + 9 - x^{2} - y^{2}

f (x, y) = 2 x + y + \frac{2}{x} + \frac{4}{y}

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