# Find out the the inflexion point of the function f(x) = x/(x^2 + 1)

Expert Answers
hala718 | Certified Educator

The inflection point is the point where the second derivative of the function changes signs. Or where the second derivative equals zero.

First we will determine the second derivative f''(x)

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

u= x ==> u'= 1

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

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

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

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

Now we will calculate f''(x)

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

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

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

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

= (2x^5-2x^3-6x)/(x^2+1)^4

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

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

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

Now f''(x) =0

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

==> x1=0

or x^2-3=0

==> x^2= 3

==> x2= sqrt3

x3= -sqrt3

Now we will calculate f(x1), f(x2), and f(x3)

f(X1)= f(0)= 0/1 = 0

f(x2)= f(sqrt3)= sqrt3/ 4

f(x3)= f(-sqrt3)= -sqrt3/ 4

Then the inflection points are:

(0,0), (sqrt3, sqrt3/4), and (-sqrt3, -sqrt3/4)

giorgiana1976 | Student

The inflexion points could be found by calculating the roots of the second derivative (if there are any).

First of all, we'll calculate the first derivative following the fraction derivation 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

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

After f"(x) calculus, we'll try to determine the roots of f"(x).

For this, f"(x)=0

Because (x^2 + 1)^2 > 0, only 2x(x^2 - 3) = 0.

2x = 0, for x = 0

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

So, the inflexion points are:

(0, f(0)), (-sqrt3, f(-sqrt3)), (sqrt3,f (sqrt3))

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