# How do you find the domain and range of f(x)=(x+7)/(x^2-49)?

Oct 7, 2017

The domain is $x \in \mathbb{R} - \left\{7\right\}$
The range is $f \left(x\right) \in \left(- \infty , {0}^{-}\right) \cup \left({0}^{+} , + \infty\right)$

#### Explanation:

Let's simplify the function

$f \left(x\right) = \frac{x + 7}{{x}^{2} - 49} = \frac{\cancel{x + 7}}{\cancel{x + 7} \left(x - 7\right)}$

$f \left(x\right) = \frac{1}{x - 7}$

The denominator is $\ne 0$

Therefore,

$x - 7 \ne 0$, $\implies$,$x \ne 7$

The domain is $I = \mathbb{R} - \left\{7\right\}$

The function is bijective over the domain $I$

$f \left(- \infty\right) = \frac{1}{- \infty - 7} = {0}^{-}$

$f \left({7}^{-}\right) = \frac{1}{{7}^{-} - 7} = - \infty$

$f \left({7}^{+}\right) = \frac{1}{{7}^{+} - 7} = + \infty$

$f \left(+ \infty\right) = \frac{1}{+ \infty - 7} = {0}^{+}$

Therefore,

The range is $f \left(x\right) \in \left(- \infty , {0}^{-}\right) \cup \left({0}^{+} , + \infty\right)$

graph{(1/(x-7)) [-13.34, 27.2, -10.06, 10.22]}