Determine the roots of f"(x) if f(x)= x/(x^2+1) .

First, we'll determine the first derivative 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

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

Now,
we'll calculate the second derivative, using the quotient rule,
again:

f''(x)={(1-x^2)'(x^2+1)^2-(1-x^2)[(x^2+1)^2]'}/(x^2+1)^4

f''(x)=2x(x^2-3)/(x^2+1)^2

Now,
we'll solve the equation f"(x) = 0:

Because the denominator
is always positive, for any value of x, namely (x^2+1)^2>0, only the numerator is
cancelling.

2x*(x^2-3)=0

We'll
set each factor as zero:

2x=0, for
x=0

x^2-3=0, for x=-sqrt3 or
x=sqrt3

The roots of f"(x) = 0 are: {-sqrt3 ,
0 , sqrt3}.