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

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Answer 1 · 2015-06-15T16:57:07

$x^2+2$ has range $[2, oo)$ , so $8/(x^2+2)$ has range $(0,4]$

$f(x) = 8/(x^2+2)$

$f(0) = 8/2 = 4$

$f(-x) = f(x)$

As $x->oo$ we have $f(x)->0$

$f(x) > 0$ for all $x in RR$

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

If $y in (0, 4]$ then $8/y >= 2$ and $8/y - 2 >= 0$

so $x_1 = sqrt(8/y - 2)$ is defined and $f(x_1) = y$ .

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

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