What is the range of the function f(x)=(5x-3)/(2x+1)?

Jul 26, 2017

The range is $y \in \mathbb{R} - \left\{\frac{5}{2}\right\}$

Explanation:

$f \left(x\right) = \frac{5 x - 3}{2 x + 1}$
Let

$y = \frac{5 x - 3}{2 x + 1}$

$y \left(2 x + 1\right) = 5 x - 3$

$2 y x + y = 5 x - 3$

$5 x - 2 y x = y + 3$

$x \left(5 - 2 y\right) = \left(y + 3\right)$

$x = \frac{y + 3}{5 - 2 y}$

The domain of $x = f \left(y\right)$ is $y \in \mathbb{R} - \left\{\frac{5}{2}\right\}$

This is also ${f}^{-} 1 \left(x\right) = \frac{x + 3}{5 - 2 x}$

graph{(5x-3)/(2x+1) [-22.8, 22.83, -11.4, 11.4]}

Jul 26, 2017

$y \in \mathbb{R} , y \ne \frac{5}{2}$

Explanation:

$\text{given } y = \frac{5 x - 3}{2 x + 1}$

$\text{rearrange making x the subject}$

$\Rightarrow y \left(2 x + 1\right) = 5 x - 3 \leftarrow \textcolor{b l u e}{\text{cross-multiplying}}$

$\Rightarrow 2 x y + y = 5 x - 3 \leftarrow \textcolor{b l u e}{\text{ distributing}}$

$\Rightarrow 2 x y - 5 x = - 3 - y \leftarrow \textcolor{b l u e}{\text{ collect terms in x}}$

$\Rightarrow x \left(2 y - 5\right) = - \left(3 + y\right) \leftarrow \textcolor{b l u e}{\text{ factor out x}}$

$\Rightarrow x = - \frac{3 + y}{2 y - 5}$

$\text{the denominator cannot equal zero as this would}$
$\text{be undefined}$

$2 y - 5 = 0 \Rightarrow y = \frac{5}{2} \leftarrow \textcolor{red}{\text{ excluded value}}$

$\text{range is } y \in \mathbb{R} , y \ne \frac{5}{2}$