Question

How do you find increasing, decreasing intervals, local max mins, concave up and down for `f(x) = (x^2)/(x^2 +3)`?

Answer 1

`f(x)` is concave up on the interval `(-1,1)` and concave down on `(-oo,-1) uu (1, oo)`.

Explanation:

Start by calculating the first derivative of `f(x)` - use the quotient rule

`d/dx(f(x)) = ( [d/dx(x^2)] * (x^2 + 3) - x^2 * d/dx(x^2 + 3))/(x^2 + 3)^2`

`f^' = (2x * (x^2 + 3) - x^2 * 2x)/(x^2 + 3)^2`

`f^' = (color(red)(cancel(color(black)(2x^3))) + 6x - color(red)(cancel(color(black)(2x^3))))/(x^2 + 3)^2 = (6x)/(x^2 + 3)^2`

Before calculating the second derivative, find the critical points of the function by having `f^' = 0`. These points will help you determine the local minimum and local maximum.

`(6x)/(x^2 + 3)^2 = 0 <=> 6x = 0 => x = color(green)(0) ->` critical point

To determine where the function is increasing and where it's decreasing, examine the sign of the first derivative around the critical point `x=0`.

Since the numerator of `f^'` will always be positive, the sign of the first derivative will be determined by the numerator.

This means that you have `6x<0` for `x<0`, so the function is decreasing on `(-oo,0)`.

Likewise, because `6x>0` for `x>0`, the function is increasing on `(0, oo)`.

Now calculate the second derivative - use the quotient and the chain rules

`d/dx(f^') = ([d/dx(6x)] * (x^2 + 3)^2 - 6x * d/dx(x^2 + 3)^2)/[(x^2+3)^2]^2`

`f^('') = (6 * (x^2 + 3)^color(red)(cancel(color(black)(2))) - 6x * 2 * color(red)(cancel(color(black)((x^2+3)))) * 2x)/(x^2+3)^color(red)(cancel(color(black)(4)))`

`f^('') = (6 *(x^2 + 3) - 24x^2)/(x^2 + 3)^3`

`f^('') = (-18x^2 + 18)/(x^2 + 3)^3 = -(18 * (x^2-1))/(x^2 + 3)^3`

You can determine the local minimum and local maximum of the function by using the second derivative test, which tells you that if you have `c` as a critical point of `f(x)`, then the sign of `f^('')(c)` will tell you

• `f^('')(c)<0 ->` local maximum

• `f^('')(c)>0 ->` local minimum

So, your critical point is `c=0`, which means that you have

`f^('')(0) = -18 * (0^2 -1)/(0^2 + 3)^3`

`f^('')(0) = -18 * ((-1))/27 = 18/27 = 2/3`

The point `(0, f(0))` will be a local minimum. Moreover, the point `(0, f(0))` will be an absolute minimum as well, since

`f(x) = x^2/(x^2 + 3) > 0,(AA) x !=0` on `(-oo,oo)`

To determine where the function is concave up and where it's concave down, analyze the behavior of `f^('')` around the Inflection points, where `f^('')=0`.

`f^('') = -(18(x^2-1))/(x^2 + 3)^2=0`

This implies that

`-18(x^2-1) = 0 <=> x^2 = 1 => x = +-1`

Now you're going to see what the sign of the second derivative is for the intervals

• `(-oo,-1)`

It's clear that `f^('')` will be negative on this interval, so `f` will be concave down.

• `(-1,1)`

On this interval, `f^('')` will be positive, since `(x^2-1)` will be negative. As a result, `f` will be concave up.

• `(1,oo)`

Once again, `f^('')` will be negative for this interval, so `f` will be concave down.