What is the extreme f(x)=(2x^2+3)/(x^2-1) ?

The extreme f(x)=(2x^2+3)/(x^2-1) is Maximum(0,-3)

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extreme f(x)=(2x^2+3)/(x^2-1)

extreme points

f (x) = \frac{2 x ^{2} + 3}{x ^{2} - 1}

Maximum (0, - 3)

Solution steps

f^{'} (x) = - \frac{10 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} + 3}{x ^{2} - 1} : - 3

Maximum (0, - 3)

f (x) = 4 x + 8 cos (x)

f (y) = x^{2} + y^{2} - 20 x + 16 y - 9

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

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

f (x) = 2 x - 0.3 y

What is the extreme f(x)=(2x^2+3)/(x^2-1) ?
The extreme f(x)=(2x^2+3)/(x^2-1) is Maximum(0,-3)