How do you find the domain and range of #sqrt(x-3) - sqrt(x+3)#?

1 Answer
Jun 6, 2017

Answer:

#D_f= x>=3#

#R_f=-sqrt6 <= f(x) < 0#

Explanation:

In order to find the domain and range, we need to find the values of #x# for which the function is defined.

Since we have square roots, we know that they both need to be greater than or equal to #0#. Since the first square root is a subtraction, it is more restrictive so it will control the domain.

#sqrt(x-3) >= 0#

#x-3 >= 0#

#x >= 3#

#therefore D_f= x >= 3#

Now the range is a bit more tricky to find, but if we look at out function, we see that there's a subtraction happening, and this is key to finding the range of #x#.

If we look at what is being subtracted, it's #sqrt(x-3)-sqrt(x+3)#. By looking, we know that this subtraction will always be negative because #sqrt(x+3) > sqrt(x-3)#. Taking the smallest valid value of #x# gives us #sqrt0-sqrt6=-sqrt6#. Thus, our function will always be greater than or equal to #-sqrt6#.

Now, for small values of #x#, the constants will be significant in deciding the value of #f#, so #f# will be more negative as there will be a larger difference between the two square roots. However, as #x# becomes larger and larger, the constants become less significant and the difference between the square roots becomes smaller. Thus #f# becomes less negative and approaches #0#.

However, we know that #sqrt(x-3) != sqrt(x+3)#, so the function will never touch #0#. Thus #0# is the limiting value of #f#.

#R_f=-sqrt6 <= f(x) < 0#