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)

f (x, y) = x^{2} + x y + y^{2} + 4 x - 4 y + 6

f (x, y) = y ln (x + 2 y)

f (x) = 2 x^{3} - 24 x^{2} + 72 x + 7, (3, 7)

x^{4} + y^{4} - 4 x y + x

y = x^{3} + 3 x^{2} + 1

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)