What is the domain and range of #y=sqrt((x+5) (x-5))#?

The domain of the function will include all the values that #x# can take for which #y# is defined.

In this case, the fact that you're dealing with a square root tells you that the expression that's under the square root sign must be positive. That is the case because when working with real numbers, you can only take the square root of a positive number.

This means that you must have

#(x + 5)(x - 5) >=0#

Now, you know that for # x = {-5, 5}#, you have

#(x+5)(x - 5) = 0#

In order to determine the values of #x# that will make

#(x+5)(x-5) > 0#

you need to look at two possible scenarios.

#color(white)(a)#

• #x+5 > 0 " " ul(and) " " x-5 > 0#

In this case, you must have

#x + 5 > 0 implies x > - 5#

and

# x - 5 > 0 implies x > 5#

The solution interval will be

#(-5, + oo) nn (5, + oo) = (5, + oo)#

#color(white)(a)#

• #x + 5 < 0 " " ul(and) " " x- 5 < 0#

This time, you must have

#x + 5 < 0 implies x < -5#

and

# x - 5 < 0 implies x < 5#

The solution interval will be

#(-oo, - 5) nn (-oo, 5) = (-oo, - 5)#

#color(white)(a)#

You can thus say that the domain of the function will be--do not forget that #-5# and #5# are part of the domain#

#"domain: " color(darkgreen)(ul(color(black)(x in (-oo, - 5] uu [5, + oo)#

For the range of the function, you need to find the values that #y# can take for all the values of #x# that a part of its domain.

You know that for real numbers, taking the square root of a positive number will produce a positive number, so you can say that

#y >= 0 " "(AA)color(white)(.) x in (-oo, -5] uu [5, + oo)#

Now, you know that when # x= {-5, 5}#, you have

#y = sqrt((-5 + 5)(-5 - 5)) = 0" " and " " y = sqrt((5 + 5)(5 - 5)) = 0#

Moreover, for every value of #x in (-oo, -5] uu [5, + oo)#, you have

#y >= 0#

This means that the range of the function will be

#"range: " color(darkgreen)(ul(color(black)(y in (-oo"," + oo)))#

graph{sqrt((x+5)(x-5)) [-20, 20, -10, 10]}