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!