# Question #44ffb

Sep 12, 2017

The domain is $x < 0 \cup 0 < x < 3 \cup x > 3$, or $\left(- \infty , 0\right) \cup \left(0 , 3\right) \cup \left(3 , \infty\right)$
The range is $y \in \mathbb{R} \mathmr{and} \left(- \infty , \infty\right)$

#### Explanation:

First, the domain:
Since the function $f \left(x\right) = \frac{1}{x} + \frac{5}{x - 3}$ has two terms with different denominators, I want to get like denominators to find the domain.
Multiply $\frac{1}{x}$ by $\frac{x - 3}{x - 3}$ and $\left(\frac{5}{x - 3}\right)$ by $\frac{x}{x}$:

$f \left(x\right) = \frac{1}{x} \cdot \frac{x - 3}{x - 3} + \frac{5}{x - 3} \cdot \frac{x}{x}$

Multiply the fractions:
$\frac{x - 3}{x \left(x - 3\right)} + \frac{5 x}{x \left(x - 3\right)}$

$\frac{6 x - 3}{x \left(x - 3\right)} = \frac{6 x - 3}{{x}^{2} - 3 x}$
Since fractions don't exist where the denominators $= 0$, set $x \left(x - 3\right) = 0$. 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 $y = 0$.