What are the points of inflection of #f(x)=x/(1+x^2)#?

Question

William Abdalla · Accepted Answer

There are 3 points of inflexion #(0.0)# and #(sqrt3,sqrt3/4)# and #(-sqrt3,-sqrt3/4)#

The domain of #f(x)# is #RR#

To determine the points of inflexion, we calculte #f''(x)# and find out when it is #=0#

We start by calculating the first derivative #u(x)=x# #=># #u'(x)=1# #v(x)=1+x^2# #=># #v'(x)=2x#

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

#f(x)=0# when #1-x^2=0# #=># #x=+-1# from #x=oo# to #x=-1# #f'(x)# is negative

from #x=-1# to #x=1# #f'(x)# is positive

from #x=1# to #x=oo# #f'(x)# is negative

So we have a minimum at #(-1,-1/2)# and a maximum at # (1,1/2)#

Let's calculate #f''(x)# #u=1-x^2# , #u'=-2x# #v=(1+x^2)^2# so #v'=4x(1+x^2)#

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

#f''(x)=0# when #x=0# and tis correspond to the point #(0,0)#

#f''(x)=(-2x(1+x^2)(1+x^2+2-2x^2))/(1+x^2)^4# This is #=0# for #x=0#

and #3-x^2=0# # => # #x=+-sqrt3#

diagram{x/(1+x^2) [-5, 5, -2.5, 2.5]}

Emilia Aguilar · Answer

undefined To find the points of inflection of $ f(x) = \frac{x}{1 + x^2} $, we need to locate where the concavity changes.

1. Find the second derivative of $ f(x) $.
2. Set the second derivative equal to zero and solve for $ x $ to find potential points of inflection.
3. Test the concavity of $ f(x) $ around each potential point of inflection to confirm whether it changes.

First, find the first derivative:

$$ f(x) = \frac{x}{1 + x^2} $$
$$ f'(x) = \frac{(1 + x^2)(1) - (x)(2x)}{(1 + x^2)^2} $$
$$ f'(x) = \frac{1 + x^2 - 2x^2}{(1 + x^2)^2} $$
$$ f'(x) = \frac{1 - x^2}{(1 + x^2)^2} $$

Now, find the second derivative:

$$ f''(x) = \frac{(1 + x^2)^2(0) - (1 - x^2)(2(1 + x^2)(2x))}{(1 + x^2)^4} $$
$$ f''(x) = \frac{-4x(1 - x^2)}{(1 + x^2)^3} $$

Set $ f''(x) = 0 $ to find potential points of inflection:

$$ -4x(1 - x^2) = 0 $$

Solve for $ x $:

$$ -4x(1 - x^2) = 0 $$
$$ -4x + 4x^3 = 0 $$
$$ 4x^3 - 4x = 0 $$
$$ 4x(x^2 - 1) = 0 $$
$$ x(x - 1)(x + 1) = 0 $$

So, potential points of inflection are $ x = 0, x = 1, x = -1 $.

Test the concavity around each potential point of inflection to confirm if they are points of inflection. If the concavity changes signs around a point, it is a point of inflection.