# How do you find the domain and range of f(x) = - sqrt2 / (x² - 16)?

Jul 11, 2017

Domain is all value of $x$ other than $- 4$ and $4$.

Range is all values except those between $0$ and $\frac{\sqrt{2}}{16}$ i.e. $\left(0 , \frac{\sqrt{2}}{16}\right)$

#### Explanation:

AS $f \left(x\right) = - \frac{\sqrt{2}}{{x}^{2} - 16} = - \frac{\sqrt{2}}{\left(x + 4\right) \left(x - 4\right)}$,

it is quite apparent that $x$ cannot take values $- 4$ and $4$.

Hence domain of $x$ is all values other than $- 4$ and $4$.

Further, when $x$ lies between $- 4$ and $4$ i.e. $| x | < 4$,

the denominator is always negative i.e. $f \left(x\right)$ is positive and maximum value of denominator is when $x = 0$, and then $f \left(x\right) = \frac{\sqrt{2}}{16}$. Hence this is minimum value of $f \left(x\right)$ in this range.

When $| x | > 4$, $f \left(x\right)$ is always negative and when $x \to \infty$,

$f \left(x\right) = - \frac{\sqrt{2}}{{x}^{2} - 16} = - \frac{\frac{\sqrt{2}}{x} ^ 2}{1 - \frac{16}{x} ^ 2}$ i.e. $f \left(x\right) \to 0$

Hence range of $f \left(x\right)$ is all values except those between $0$ and $\frac{\sqrt{2}}{16}$

graph{-sqrt2/(x^2-16) [-5, 5, -2.5, 2.5]}