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)

(x - 2)^{2} - 3

f (x) = 2 x

e^{x} \cdot (x^{2} + 4 x + 1)

y = \frac{5}{6} x - 6, (- 2, 4)

f (x) = \frac{1}{3 x + 6}

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)