Question #44ffb

1 Answer
Sep 12, 2017

The domain is #x<0 uu 0< x<3 uu x>3#, or #(-oo, 0) uu (0, 3) uu (3,oo)#
The range is #yinRR or (-oo, oo)#

Explanation:

First, the domain:
Since the function #f(x)=1/x+5/(x-3)# has two terms with different denominators, I want to get like denominators to find the domain.
Multiply #1/x# by #(x-3)/(x-3)# and #(5/(x-3))# by #x/x#:

#f(x)=1/x*(x-3)/(x-3) + 5/(x-3)*x/x#

Multiply the fractions:
#(x-3)/(x(x-3))+(5x)/(x(x-3))#

Add like terms:
#(6x-3)/(x(x-3))=(6x-3)/(x^2-3x)#

Since fractions don't exist where the denominators #=0#, set #x(x-3)=0#. Here, #x=0,3#, so we know that our function doesn't have points where #x=0,3#. These are vertical asymptotes.
Also, because the degree of the denominator is greater than the degree of the numerator, there is a horizontal asymptote at #y=0#.

If you don't know about the rules for asymptotes, reference this page: http://www.purplemath.com/modules/asymtote4.htm

Finally,
Have fun! You can do this!