How do you find the domain and range of #y = -sqrt( x^2- 3x -10)#?

1 Answer
Dec 22, 2017

Answer:

Domain: #{x|x<=-2,x>=5}#
Range: #y in RR_(<=0)#

Explanation:

Finding the domain
To find the domain, we want to look at when the function is defined. We can see that the function will not be defined in terms of real numbers if the value inside the square root is negative, so let's look at that inequality:

#x^2-3x-10<0#

We can factor the left hand side:
#x^2-5x+2x-10<0#

#x(x-5)+2(x-5)<0#

#(x+2)(x-5)<0#

We need to determine the intervals where this could be negative. They will be where the factors equal #0#, since that is where they go from positive to negative. These points are at #x=-2# and #x=5#.

This means that our intervals will be #(-oo,-2)#, #(-2,5)# and #(5,oo)#.

Let's start with #(-oo,-2)#. For the original expression to be negative, only one of the factors may be negative. However, in this interval both factors will be negative, making the product positive.

In #(-2,5)#, only one of the factors will be negative #(x-5)#, so here the value will be negative.

And lastly, in #(5,oo)#, both factors will always be positive, so we get a positive product.

This means that the only time the expression in the square root is negative is when #x# is between #-2# and #5#. This means that our domain will be:
#{x|x<=-2,x>=5}#

Finding the Range
A square root can never put out a negative value, and since we concluded that our bits under the square root are defined up to infinity, we can conclude that the values of the square root will range from #0# to #oo#.

However, we have a minus sign in front of the square root, so the range will instead be #-oo# to #0#, or all the real numbers less than or equal to #0#, #RR_(<=0)#