The second derivative test allows you to determine the concavity of a function by analyzing the behavior of the function's second derivative around inflexion points, which are points at which #f^('') = 0#.
If #f^('')# is positive on a given interval, then #f(x)# will be concave up. LIkewise, if #f^('')# 8s negative on a given interval, then #f(x)# will be concave down.
The inflexion points will determine on which intervals the sign of the second derivative should be examined.
So, start by calculating the function's first derivative - use the quotient rule
#d/dx(f(x)) = ([d/dx(x^2)] * (x^2 + 1) - x^2 * d/dx(x^2+1))/(x^2+1)^2#
#f^' = (2x * (x^2 + 1) - x^2 * 2x)/(x^2 + 1)^2#
#f^' = (color(red)(cancel(color(black)(2x^3))) + 2x - color(red)(cancel(color(black)(2x^3))))/(x^2 + 1)^2 = (2x)/(x^2 + 1)^2#
Next, calculate the second derivative - use the quotient and chain rules
#d/dx(f^') = ([d/dx(2x)] * (x^2 + 1)^2 - 2x * d/dx(x^2 + 1)^2)/([(x^2 + 1)^2]^2#
#f^('') = (2 * (x^2 + 1)^color(red)(cancel(color(black)(2))) - 2x * 2 * color(red)(cancel(color(black)((x^2 + 1)))) * 2x)/(x^2+1)^color(red)(cancel(color(black)(4)))#
#f^('') = (2x^2 + 2 - 8x^2)/(x^2 + 1)^3 = -(2(3x^2 - 1))/(x^2 + 1)^3#
Next, find the inflexion points by calculating #f^('')=0#.
#-(2(3x^2-1))/(x^2+1)^3 = 0#
This is equivalent to
#3x^2 - 1 = 0 => x^2 = 1/3#
Take the square root of both sides to get
#sqrt(x^2) = sqrt(1/3) => x = +- sqrt(3)/3#
So, the function's graph has two inflexion points at #x=-sqrt(3)/3# and #x=sqrt(3)/3#, so you're going to have to look at three intervals to determine its concavity.
Since #(x^2 + 1)^3>0, (AA) x# on #(-oo, +oo)#, the sign of the second derivative will depend exclusively on the sign of #3x^2-1#.
More specifically, if #3x^2-1>0#, then #f^('')<0#, and if #3x^2-1<0#, then #f^('')>0#.
On this interval, #3x^2-1>0#, which means that #f^('')<0# and the function is concave down.
- #(-sqrt(3)/3, sqrt(3)/3)#
On this interval, #3x^2-1<0#, which means tha t#f^('')>0# and the function is concave up.
Once again, #3x^2-1>0#, #f^('')<0# and the function is concave down.
So, your function is concave down on #(-oo, -sqrt(3)/3) uu (sqrt(3)/3, +oo)# and concave up on #(-sqrt(3)/3, sqrt(3)/3)#.
The function's graph will have two inflexion points at #(-sqrt(3)/3, f(-sqrt(3)/3))# and #(sqrt(3)/3, f(sqrt(3)/3))#.
graph{x^2/(x^2+1) [-4.932, 4.934, -2.465, 2.467]}
