# Question #03083

Aug 24, 2016

Assuming you mean $f \left(x\right) = \frac{x}{2 + {x}^{2}}$, the range is $- 1 < y < 1$.

#### Explanation:

The range is the set of all $y$ values that the function encompasses. In other words, it's the places where the function actually exists on the $y$-axis.

To find it, we just need to locate the places where $f \left(x\right)$ does not exist. This involves trying some $x$-values and see if they correspond to an $f \left(x\right)$ value.

The first point we should try is $x = 0$, because sometimes there's trouble with that point in rational functions:
$f \left(x\right) = \frac{x}{2 + {x}^{2}}$
$f \left(0\right) = \frac{0}{2 + 0}$
$f \left(0\right) = \frac{0}{2} = 0$

Everything's fine here. Now let's try $x = 1$ and $x = - 1$:
$f \left(1\right) = \frac{1}{2 + {1}^{2}} = \frac{1}{3}$
$f \left(- 1\right) = - \frac{1}{2 + {\left(- 1\right)}^{2}} = - \frac{1}{3}$

So far, so good - it's looking like the range is all $y$-values. But before we jump to this conclusion, let's try a big number, say $x = 100$:
$f \left(100\right) = \frac{100}{2 + {100}^{2}} = \frac{100}{10002}$

We can see that $f \left(x\right)$ is getting smaller and smaller as $x$ is getting bigger and bigger, which forces us to ask if $f \left(x\right)$ ever is greater than $1$. We can find that out by setting $f \left(x\right) = 1$ and solving for $x$:
$1 = \frac{x}{2 + {x}^{2}}$
$2 + {x}^{2} = x$
${x}^{2} - x + 2 = 0$

This is a quadratic equation with no real solutions, which means there are no values of $x$ that allow $f \left(x\right)$ to equal $1$. In other words, no value of $x$ will make $f \left(x\right) = 1$, so the range is restricted. A similar test with $f \left(x\right) = - 1$ shows the same thing, so the range is $- 1 < y < 1$. We can confirm this by looking at the graph.
graph{x/(2+x^2) [-10, 10, -5, 5]}