How do you find the domain and range of #f(x) = x^2/(1 + x^2)#?

1 Answer
Aug 20, 2017

Answer:

Domain: all real numbers
Range: all numbers y such that #0 <= y < 1#

Explanation:

The domain of a function is the set of all values x such that the function is well defined.

That means that any value of x that would cause you to divide by zero is excluded.

There is NO such value of x for this function. X can be any value, positive or negative, because there's no real value of X that, when squared, equals -1. (which would cause a divide by zero).

So the domain of this function is all real numbers X.

The range of a function is the set of all values that can be produced by the function.

We can see that the function has a minimum value of 0 when x = 0.

To determine the upper bound of the range of this function requires a concept that is a little beyond algebra (but only by just a little bit): a limit.

So, we want to know what the limit is, as x goes to infinity, of F(X).

If you take calculus you'll get into this in more detail.

For now, note that, for bigger and bigger values of x, F(x) gets closer and closer to one.

#F(100) = 10000/10001# is very close to one. Bigger values of x give a value of F(x) that is even closer to one. You can get as close as you want to a value of F(x) = 1 by just using a big enough value of X.
(Though you can never quite get EXACTLY F(x) = 1).

As I say, if you take Calculus, you'll get this all laid out in much greater detail.

#same thing on the other side: F(-100) = F(100) and is close to one.

So, the range of the function is all values y in the interval:

#0 <= y < 1#

Note the #<=# and the #<#. F(x) exactly equals 0 when x = 0, but is always less than one for any other value of x.

Here's the graph of the function, as a "sanity check".

graph{x^2/(1 + x^2) [-10, 10, -5, 5]}