# What is the domain and range of y=(4+x)/(1-4x)?

May 29, 2017

The domain is $\mathbb{R} - \left\{\frac{1}{4}\right\}$
The range is $\mathbb{R} - \left\{- \frac{1}{4}\right\}$

#### Explanation:

$y = \frac{4 + x}{1 - 4 x}$

As you cannot divide by $0$, $\implies$, $1 - 4 x \ne 0$

So,

$x \ne \frac{1}{4}$

The domain is $\mathbb{R} - \left\{\frac{1}{4}\right\}$

To find the range, we calculate the inverse function ${y}^{-} 1$

We interchange $x$ and $y$

$x = \frac{4 + y}{1 - 4 y}$

We express $y$ in terms of $x$

$x \left(1 - 4 y\right) = 4 + y$

$x - 4 x y = 4 + y$

$y + 4 x y = x - 4$

$y \left(1 + 4 x\right) = x - 4$

$y = \frac{x - 4}{1 + 4 x}$

The inverse is ${y}^{-} 1 = \frac{x - 4}{1 + 4 x}$

The range of $y$ is $=$ to the domain of ${y}^{-} 1$

$1 + 4 x \ne 0$

The range is $\mathbb{R} - \left\{- \frac{1}{4}\right\}$

May 29, 2017

$x \in \mathbb{R} , x \ne \frac{1}{4}$
$y \in \mathbb{R} , y \ne - \frac{1}{4}$

#### Explanation:

$\text{the domain is defined for all real values of x, except}$
$\text{those values which make the denominator zero}$

$\text{to find excluded values equate the denominator to zero}$
$\text{and solve for x}$

$\text{solve " 1-4x=0rArrx=1/4larrcolor(red)"excluded value}$

$\Rightarrow \text{domain is } x \in \mathbb{R} , x \ne \frac{1}{4}$

$\text{to find any excluded values in the range, change the subject}$
$\text{of the function to x}$

$y \left(1 - 4 x\right) = 4 + x$

$\Rightarrow y - 4 x y = 4 + x$

$\Rightarrow - 4 x y - x = 4 - y$

$\Rightarrow x \left(- 4 y - 1\right) = 4 - y$

$\Rightarrow x = \frac{4 - y}{- 4 y - 1}$

$\text{the denominator cannot equal zero}$

$\Rightarrow - 4 y - 1 = 0 \Rightarrow y = - \frac{1}{4} \leftarrow \textcolor{red}{\text{ excluded value}}$

$\Rightarrow \text{range is } y \in \mathbb{R} , y \ne - \frac{1}{4}$