Verify if the function f=2x/(x^2+1) has one or two extreme values.

To establish the number of extreme values of the given
function, we have to apply the first derivative test.

We
notice that the function is a ratio, so we'll calculate the first derivative using the
quotient rule:

f'(x)=
[(2x)'*(x^2+1)-(2x)*(x^2+1)']/(x^2+1)^2

f'(x)=
[2(x^2+1)-2x*2x]/(x^2+1)^2

f'(x)= (2x^2 +2
-4x^2)/(x^2+1)^2

f'(x)=
(-2x^2+2)/(x^2+1)^2

f'(x)=
(1-x^2)/(x^2+1)^2

In order to calculate the extreme values
of the function, we have to determine the roots of the first
derivative.

f'(x)=0

The
denominator is a sum of squares, so, it will be positive (it won't be zero) for any
value of x.

So, only the numerator could have roots, if the
delta>0.

1-x^2 it's a difference between
squares:

a^2-b^2=(a-b)(a+b)

1-x^2=(1-x)(1+x)

(1-x)(1+x)=0

We'll
set each factor as zero.

1-x=0,
x=1

1+x=0,x=-1

So, there are 2
extreme values of f(x) and the values
are:

f(1)=2*1/(1^2+1)=2/2=1

f(-1)=2*(-1)/(-1^2+1)=-2/2=-1