Question #03083

1 Answer
Aug 24, 2016

Assuming you mean f(x)=x2+x2, 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(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)=x2+x2
f(0)=02+0
f(0)=02=0

Everything's fine here. Now let's try x=1 and x=1:
f(1)=12+12=13
f(1)=12+(1)2=13

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)=1002+1002=10010002

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=x2+x2
2+x2=x
x2x+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]}