# What is the domain and range of f(x) = (x-1)/(x^2+1)?

Mar 15, 2018

$\text{Domain} : x \in \mathbb{R}$
$\text{Range} : f \left(x\right) \in \left[- \frac{\sqrt{2} + 1}{2} , \frac{\sqrt{2} - 1}{2}\right]$

#### Explanation:

Considering that all real values of $x$ will give a non-zero value for ${x}^{2} + 1$, we can say that for $f \left(x\right)$, domain = $x \in \mathbb{R}$

For range, we need the maximum and minimum.
$f \left(x\right) = \frac{x - 1}{{x}^{2} + 1}$

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

The maximum and minimum values occur when $f ' \left(x\right) = 0$

${x}^{2} - 2 x - 1 = 0$

$x = \frac{2 \pm \sqrt{{\left(- 2\right)}^{2} - 4 \left(- 1\right)}}{2}$

$x = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2 \sqrt{2}}{2} = 1 \pm \sqrt{2}$

Now, we input our $x$ values into $f \left(x\right)$:
$\frac{1 + \sqrt{2} - 1}{{\left(1 + \sqrt{2}\right)}^{2} + 1} = \frac{\sqrt{2} - 1}{2}$
$\frac{1 - \sqrt{2} - 1}{{\left(1 - \sqrt{2}\right)}^{2} + 1} = - \frac{\sqrt{2} + 1}{2}$

$f \left(x\right) \in \left[- \frac{\sqrt{2} + 1}{2} , \frac{\sqrt{2} - 1}{2}\right]$