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)

f (x, y) = - x^{2} - y^{2} + 8 x + 6 y

f (x, y) = x^{3} + y^{3} + 3 y^{2} - 3 x - 9 y + 2

f (x, y) = 2 x + 4 y - x^{2} y^{4}

f (x) = \frac{5}{2} x - \frac{x ^{2}}{2}

f (x, y) = ln (4 - x^{2} - y^{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)