Find the inflection point of the function y=x/(x^2+1)

2 Answers | Add Yours

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Tofind the  inflexion point of y = x/(x^2+1).

At the infection point d^2/dx^2 = 0.

So we find d2y/dx^2 and equate it ti zero and  solve for x and y .

dy/dx=y' =  (x/x^2+1)'

y' = { (x)'(x^2+1)-x(x^2+1)}/(x^2+1

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

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

y' = -(x^2-1)/(x^2+1) = (-1) { 1  -2/(x^2+1)

y" =  {-1+2/(x^2+1)}'

y" = -2*2x/(x^2+1)^2.

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

Equating to zero, we get:   -4x = 0 or x = 0 and corresponding y = 0/(0+1) = 0

So the inflection point is at (0,0)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The inflection 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 applying 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

f"(x) = [f'(x)]'

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 the numerator could be cancelled.

 2x(x^2-3)=0.

2x=0, for x=0

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

So, the inflection points are:

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

We’ve answered 318,957 questions. We can answer yours, too.

Ask a question