What is the domain and range of #f(x) = abs((9-x^2)/(x+3))#?

1 Answer
Mar 16, 2016

Answer:

In this case the range is pretty clear. Because of the absolute bars #f(x)# can never be negative

Explanation:

We see from the fraction that #x!=-3# or we divide by zero.
Otherwise:
#9-x^2# can be factored into #(3-x)(3+x)=(3-x)(x+3)# and we get:
#abs(((3-x)cancel(x+3))/cancel(x+3))=abs(3-x)#

This gives no restriction on the domain, except the earlier one:
So:
Domain: #x!=-3#
Range: #f(x)>=0#