What is the number of extremes of the function f(x)=x/(x^2+1)?

To determine the number of extremes of the function, we'll
have to find out how many critical values the function does
have.

The critical values of the function are the zeroes of
the first derivative of the function.

We'll determine the
1st derivative of the function, using the quotient
rule:

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

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

We'll combine like term inside
brackets:

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

We'll cancel f'(x):

(-x^2
+ 1)/(x^2 + 1)^2 = 0

Since the denominator is always
positive for any x, we'll cancel only the numerator:

-x^2 +
1 = 0

-x^2 = -1

x^2 = 1
=> x1 = 1 and x2 = -1

Since the 1st derivative has
two zeroes, therefore the function has two critical
values.

The number of extremes of the
function is of 2.