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

First let us assume that:

f(x) = u/v such that:

u= x ==> u' = 1

v= x^2 + 1 ==> v'= 2x

f'(x) = (u'v-uv')/v^2

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

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

Now to calculate f''(x)

f'(x) = u/v such that:

u= 1-x^2 ==> u'= 2x

v= (x^2+1)^2 ==> v'= 4x(x^2+1)= 4x^3 + 4x

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

= 2x(x^2+1)(x^2+1 - 2(1-x^2)]/(1+x^2)^4

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

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

f''(x) = 0

==> 2x(3x^2 -1) = 0

==> x1= 0

==> x2= sqrt(1/3)

===? x3= - sqrt(1/3)

=

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}.**

f(x) = x/(1+x^2). To find the roots of f"(x).

Solution:

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

We find f'(x) using u(x)/v(x)}' = {u'(x)*v(x) - u(x)*v'(x)}/(1+x^2)^2.

So f'(x) {x/(1+x^2)}' = {(x)'*(1+x^2)- x(1+x^2)}/(1+x^2)^2

= {1(1+x^2) - x*2x}/(1+x^2)^2

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

Differentiate again.

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

f"(x) = {-2x(1+x^2)^2 - (1-x^2) (1+x^2)*2x}/(1+x^2)^2

= (1+x^2) {-2x(1+x^2) - (1-x^2)*2x)}/(1+x^2)^4

= {-2x-2x^3 - 2x+2x^3}/(1+x^2)^3

f"(x) = (-4x)/(1+x^2)^3...........(1)

The roots of f"(x) is obtained by solving f"(x) = 0

-4x/(1+x^2)^3 = 0 for numerator -4x= 0. Or x = 0

So x=0 is the root of f"(x) = 0.