# What is the domain and range of f(x) = (10x)/(x(x^2-7))?

Aug 3, 2015

Domain: $\left(- \infty , - \sqrt{7}\right) \cup \left(- \sqrt{7} , \sqrt{7}\right) \cup \left(\sqrt{7} , + \infty\right)$
Range: $\left(- \infty , - \frac{10}{7}\right) \cup \left(0 , + \infty\right)$

#### Explanation:

First, simplify your function to get

$f \left(x\right) = \frac{10 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} \cdot \left({x}^{2} - 7\right)} = \frac{10}{{x}^{2} - 7}$

The domain of the function will be affected by the fact that the denominator cannot be zero.

The two values that will cause the denominator of the function to be
zero are

${x}^{2} - 7 = 0$

$\sqrt{{x}^{2}} = \sqrt{7}$

$x = \pm \sqrt{7}$

This means that the domain of the function cannot include these two values, $x = - \sqrt{7}$ and $\sqrt{7}$. No other restrictions exist for the values $x$ can take, so the domain of the function will be $\mathbb{R} - \left\{\pm \sqrt{7}\right\}$, or $\left(- \infty , - \sqrt{7}\right) \cup \left(- \sqrt{7} , \sqrt{7}\right) \cup \left(\sqrt{7} , + \infty\right)$.

The range of the function will also be affected by the domain restriction. Basically, the graph will have two vertical asymptotes at $x = - \sqrt{7}$ and $x = \sqrt{7}$.

For values of $x$ located in the interval $\left(- \sqrt{7} , \sqrt{7}\right)$, the expression ${x}^{2} - 7$ is maximum for $x = 0$.

$f \left(0\right) = \frac{10}{{0}^{2} - 7} = - \frac{10}{7}$

This means that the range of the function will be $\left(- \infty , - \frac{10}{7}\right) \cup \left(0 , + \infty\right)$.

graph{10/(x^2-7) [-10, 10, -5, 5]}