# How do you find the domain and range of f(x) = (x-2)/(x+1)?

May 15, 2017

The domain is restricted as the denominator cannot be zero.

i.e. if x + 1 = 0, then x = -1

Hence, the domain can take all real values except x = -1

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

i.e. $f \left(x\right) = \frac{x + 1}{x + 1} - \frac{3}{x + 1}$

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

as x gets very small or very big, 3/(x + 1) tends towards zero

i.e. f(x) approaches a limit of 1

Hence, the range of f(x) takes all real values except 3

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May 15, 2017

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

#### Explanation:

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

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

To find the range, we need ${f}^{-} 1 \left(x\right)$

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

Interchange $y$ and $x$

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

We rewrite $y$ as function of $x$

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

$x y + x = y - 2$

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

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

So,

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

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

This is the range of $f \left(x\right)$, ${R}_{f} \left(x\right) = \mathbb{R} - \left\{1\right\}$