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

The extreme f(x)=(x^2-1)/(x^2+1) is Minimum(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}

Minimum (0, - 1)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

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

Minimum (0, - 1)

f (x) = - x^{4} - 8 x^{3} - 18 x^{2} + 6

f (x) = x^{3} - 6 x^{2} + 9 x - 2

f (x) = \frac{x}{x ^{2} - x + 9}

f (x, y) = ∣ x ∣ + y^{2} - 1

E, E = I R

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