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(x) does not exist. This involves trying some x-values and see if they correspond to an f(x) value.
The first point we should try is x=0, because sometimes there's trouble with that point in rational functions:
f(x)=x/(2+x^2)
f(0)=0/(2+0)
f(0)=0/2=0
Everything's fine here. Now let's try x=1 and x=-1:
f(1)=1/(2+1^2)=1/3
f(-1)=-1/(2+(-1)^2)=-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(100)=100/(2+100^2)=100/(10002)
We can see that f(x) is getting smaller and smaller as x is getting bigger and bigger, which forces us to ask if f(x) ever is greater than 1. We can find that out by setting f(x)=1 and solving for x:
1=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(x) to equal 1. In other words, no value of x will make f(x)=1, so the range is restricted. A similar test with f(x)=-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]}
