# What is the range of the function f(x) = (3x-4)/(1+2x)?

Jan 5, 2017

The range is $= \mathbb{R} - \left\{\frac{3}{2}\right\}$

#### Explanation:

As you cannot divide by $0$, $1 + 2 x \ne 0$, $\implies$, $x \ne - \frac{1}{2}$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- \frac{1}{2}\right\}$

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} \frac{3 x}{2 x} = {\lim}_{x \to \pm \infty} \frac{3}{2} = \frac{3}{2}$

There is a horizontal asymptote $y = \frac{3}{2}$

Therefore the range is ${R}_{f} \left(x\right) = \mathbb{R} - \left\{\frac{3}{2}\right\}$

graph{(y-(3x-4)/(1+2x))(y-3/2)=0 [-18.02, 18.01, -9.01, 9.01]}