Question

What is the domain and range of `(x+3)/(x^2+9)`?

Answer 1

`-oo < x < oo`

`-1 <= y <= 1`

Explanation:

The domain is the set of real values that `x` can take to give a real value.

The range is the set of real values you can get out of the equation.

With fractions you often have to make sure that the denominator is not `0`, because you can't divide by `0`. However, here the denominator cannot equal `0`, because if

`x^2 + 9 = 0`

`x^2 = -9`

`x = sqrt(-9)`, which doesn't exist as a real number.

Therefore, we know we can put pretty much anything into the equation.

The domain is `-oo < x < oo`.

The range is found by recognising that `abs(x^2 + 9) >= abs(x + 3)` for any real value of `x`, which means that `abs((x+3)/(x^2+9)) <= 1`

This means that the range is

`-1 <= y <= 1`

Answer 2

The domain is `x in RR` and the range is `y in [-0.069, 0.402]`

Explanation:

The domain is `x in RR` as the denominator is

`(x^2+9)>0, AA x in RR`

For the range, proceed as follows,

Let `y=(x+3)/(x^2+9)`

Then,

`yx^2+9y=x+3`

`yx^2-x+9y-3=0`

This is a quadratic equation in `x`

In order for this equation to have solutions, the discriminant `Delta>=0`

Therefore,

`Delta=b^2-4ac=(-1)^2-4(y)(9y-3)>=0`

`1-36y^2+12y>=0`

`-36y^2+12y+1>=0`

`y=(-12+-sqrt(12^2-4(-36)(1)))/(2*-36)`

`y=(-12+-sqrt288)/(-72)=-((-1+-sqrt2)/(6))`

`y_1=(1+sqrt2)/6=0.402`

`y_2=(1-sqrt2)/6=-0.069`

Therefore,

The range is `y in [-0.069, 0.402]`

You can cofirm this with a sign chart and a graph

graph{(x+3)/(x^2+9) [-7.9, 7.9, -3.95, 3.95]}