Determine the number of extreme values for the function 2x/(x^2+1).

In order to determine the number of local extremes of the
function, we'll have to find out the extreme values of a function,
first.

We'll have to do the first derivative test of a
function.

We'll apply 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

We'll simplify the ratio by
2:

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

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

f'(x)=0

It's
obvious that the denominator is positive, for any value of
x.

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

The numerator 1-x^2 is a difference of
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 0.

1-x=0,
x=1

1+x=0,x=-1

So, the extreme
values of the function
are:

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

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

The
number of local extremes of f(x) is 2.