Question

How do you find the domain and range of `y =sqrt (x^2 - 9)`?

Answer 1

The domain is `(-oo,-3]uu[3,oo)`
The range is `[0,oo)`

Explanation:

One way to find it is by graphing it, and then looking for the `y` axis (which shows the range) and the `x` axis (which shows the domain)

graph{sqrt(x^2-9) [-10, 10, -5, 5]}

It looks like the range is `[0,oo)` (the graph isn't defined for negative numbers)

The domain is `(-oo,-3]uu[3,oo)` (For `x=-3` to `x=3`, it is not defined).

But let's solve this algebraically.

The square root function is only defined when the number or expression under the radical sign is greater than or equal to `0`.

Therefore, `y=sqrt(x^2-9) rarr x^2-9>=0`
Now, we set it equal to `0`
`x^2-9=0`
`x^2=9`
`x=-3, x=3`

Now we have these points. Let us call them boundary points of the real number line. So we have 3 intervals - `(-oo,-3), (-3,3), and (3,oo)`
Just choose a point in each interval and substitute it into the original equation `(x^2-9>=0)`to check if it is valid or not, and each boundary point to see if it is in the domain or not. We find that `-3` and `3` is defined, `(-oo,-3)` and `(3,oo)` is also defined but `(-3,3)` is not. So since `[ ]` means including, we get the domain is `(-oo,-3]uu[3,oo)`. We put `( )` around the infinity symbols because infinity is never reached, so it is not an included value.

For the range, just think about this - the lowest value a square root function can give is `0`, because the lowest point where the square root function is defined is when it is `0`. `sqrt(0)` IS defined, whereas `sqrt(-0.000001)` is not. So it is simply `[0,oo)` - and including `0`.