# What is the range of 8/(x^2+2)?

Jun 15, 2015

${x}^{2} + 2$ has range $\left[2 , \infty\right)$, so $\frac{8}{{x}^{2} + 2}$ has range $\left(0 , 4\right]$

#### Explanation:

$f \left(x\right) = \frac{8}{{x}^{2} + 2}$

$f \left(0\right) = \frac{8}{2} = 4$

$f \left(- x\right) = f \left(x\right)$

As $x \to \infty$ we have $f \left(x\right) \to 0$

$f \left(x\right) > 0$ for all $x \in \mathbb{R}$

So the range of $f \left(x\right)$ is at least a subset of $\left(0 , 4\right]$

If $y \in \left(0 , 4\right]$ then $\frac{8}{y} \ge 2$ and $\frac{8}{y} - 2 \ge 0$

so ${x}_{1} = \sqrt{\frac{8}{y} - 2}$ is defined and $f \left({x}_{1}\right) = y$.

So the range of $f \left(x\right)$ is the whole of $\left(0 , 4\right]$