# How do you find the domain and range of f(x) = x / (3x(x-1))?

Oct 14, 2015

Domain: $\mathbb{R}$\{0,1}

Range: $\mathbb{R}$\{-1/3,0}

#### Explanation:

Domain:
To find the domain of this rational function, simply look for the values of $x$ that will make the the denominator $0$.

$\textcolor{w h i t e}{X X X} 3 x \left(x - 1\right) = 0$

$\Leftrightarrow \textcolor{w h i t e}{X} \textcolor{b l u e}{x = 0} \mathmr{and} \textcolor{red}{x = 1}$

Exclude 0 and 1 from the set of real numbers. The domain is $\mathbb{R}$\{0,1}

Range:
To find the range, start by isolating $x$.

$\textcolor{w h i t e}{X X} y = \frac{x}{3 x \left(x - 1\right)}$

$\textcolor{w h i t e}{X X} y = \frac{\cancel{x}}{3 \cancel{x} \left(x - 1\right)}$

Note: When we cancel $x$ here, we are adding a restriction that $x \ne 0$. The graph of this new equation has a horizontal asymptote $y = 0$. Therefore, $0$ is removed from the range.

$\textcolor{w h i t e}{X X} y = \frac{1}{3 \left(x - 1\right)} , x \ne 0$

$\textcolor{w h i t e}{X X} y = \frac{1}{3 x - 3} , x \ne 0$

$\textcolor{w h i t e}{X X} y \left(3 x - 3\right) = 1 , x \ne 0$

$\textcolor{w h i t e}{X X} 3 x y - 3 y = 1 , x \ne 0$

$\textcolor{w h i t e}{X X} 3 x y = 1 + 3 y , x \ne 0$

$\textcolor{w h i t e}{X X} x = \frac{1 + 3 y}{3 y} , x \ne 0$

Now take note that $x$ must not be equal to $0$ or $1$. We will substitute this into the equation to find out the values that $y$ cannot be.

$\left[1\right] \textcolor{w h i t e}{X} x = \frac{1 + 3 y}{3 y}$

$\textcolor{w h i t e}{\left[1\right] X} \left(0\right) = \frac{1 + 3 y}{3 y}$

$\textcolor{w h i t e}{\left[1\right] X} 0 = 1 + 3 y$

$\textcolor{w h i t e}{\left[1\right] X} 3 y = - 1$

$\textcolor{w h i t e}{\left[1\right] X} y = - \frac{1}{3}$

$\left[2\right] \textcolor{w h i t e}{X} x = \frac{1 + 3 y}{3 y}$

$\textcolor{w h i t e}{\left[2\right] X} \left(1\right) = \frac{1 + 3 y}{3 y}$

$\textcolor{w h i t e}{\left[2\right] X} \cancel{3 y} = 1 + \cancel{3 y}$

$\textcolor{w h i t e}{\left[2\right] X} 0 \ne 1$

From [1], we know that $- \frac{1}{3}$ is excluded from the range. [2] doesn't matter. And looking back at our note, $0$ is also excluded from the range. Therefore, the range is $\mathbb{R}$\{-1/3,0}