Question

Question #44ffb

Answer 1

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!