Question

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

Answer 1

You can use the second derivative test.

Explanation:

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`.

• `(-oo, -sqrt(3)/3)`

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.

• `(sqrt(3)/3, +oo)`

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]}