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

1 Answer
Jul 30, 2017

Answer:

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#.