How do you determine whether the function #f(x) = (x^2)/(x^2+1)# is concave up or concave down and its intervals?

Question

Scarlett Allison · Accepted Answer

You can use the second derivative test.

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. 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]}

Christopher Agresta · Answer

undefined To determine the concavity of the function $ f(x) = \frac{x^2}{x^2 + 1} $ and its intervals, we need to find the second derivative of the function and examine its sign.

The second derivative $ f''(x) $ will tell us whether the function is concave up or concave down. If $ f''(x) > 0 $, the function is concave up, and if $ f''(x) < 0 $, the function is concave down.

First, find the first derivative of $ f(x) $ with respect to $ x $:
$$ f'(x) = \frac{d}{dx} \left( \frac{x^2}{x^2 + 1} \right) $$

Then, find the second derivative $ f''(x) $:
$$ f''(x) = \frac{d^2}{dx^2} \left( \frac{x^2}{x^2 + 1} \right) $$

After finding $ f''(x) $, determine its sign to identify the intervals where the function is concave up or concave down.