How do you find the domain and range of #g(x)=sqrt((x+3)/(x-2))#?

1 Answer
Aug 19, 2017

Answer:

Range: #" " x in (-oo, -3] uu (2, + oo)#

Domain: #" " g(x) in [0,1 ) uu (1, + oo)#

Explanation:

In order to find the domain of this function, you need to find all the values that #x# can take for which #g(x)# is defined.

Right from the start, you know that the denominator of the fraction cannot be equal to #2# because that would make the function undefined.

#x - 2 != 0 implies x != 2#

Now, you know that when working with real numbers, you can only take the square root of a positive number.

This implies that you must have

#(x+3)/(x-2) >= 0#

You know that when #x = -3#, you have

#(-3 + 3)/(-3 - 2) = 0/(-5) >= 0 #

so you can say that #x = {-3}# will be included in the domain of the function.

Now, in order to have

#(x+3)/(x-2) > 0#

you must look at two possible situations

#color(white)(a)#

  • #x + 3 >0" " ul(and) " " x -2 > 0#

In this case, you must have

#x + 3 > 0 implies x >= -3#

and

#x - 2 > 0 implies x > 2#

This implies that the solution interval will be

#(-3, + oo) nn (2, + oo) = (2, + oo)#

This tells you that any value of #x# that is greater than #2# will satisfy the inequality #(x+3)/(x-2) > 0#.

#color(white)(a)#

  • #x + 3 <0" "ul(and)" " x - 2 < 0#

In this case, you must have

#x + 3 < 0 implies x < -3#

and

#x - 2 < 0 implies x < 2#

This implies that the solution interval will be

#(- oo, -3) nn (-oo, 2) = (-oo, -3)#

This tells you that any value of #x# that is less than #-3# will also satisfy the inequality #(x + 3)/(x - 2) > 0#.

Therefore, the domain of the function will be--remember that #x = -3# is also included in the domain!

#"domain: " color(darkgreen)(ul(color(black)(x in (-oo, -3] uu (2, oo))))#

This tells you that any value of #x# that is less than or equal to #-3# or greater than #2# will get you

#(x+3)/(x-2) >= 0#

Now, to find the range of the function, you must determine the values that #g(x)# can take for any value of #x# that is part of its domain.

Since you're working with real numbers, you can say that taking the square root of a positive number will always return a positive number.

#g(x) >= 0 #

You know that when #x = -3#, you have

#g(-3) = sqrt( (-3 + 3)/(-3 - 2)) = 0#

Now, it's important to realize that the range of the function will not include #1# because you can never have

#color(red)(cancel(color(black)(x))) + 3 = color(red)(cancel(color(black)(x))) -2#

#3 != - 2#

This means that you don't have a value of #x# for which #(x+3)/(x-2) = 1#.

Therefore, the range of the function will be

#"range: " color(darkgreen)(ul(color(black)(g(x) in [0, 1) uu (1, + oo))))#

graph{sqrt( (x+3)/(x-2)) [-16.02, 16.01, -8.01, 8]}