Question

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

Answer 1

The domain of `f` is `RR`,
and the range is `{f(x) in RR:-1/2<=f(x)<=1/2}`.

Explanation:

Solving for the domain of `f`, we will observe that the denominator is always positive, regardless of `x`, and indeed is least when `x=0`. And because `x^2>=0`, no value of `x` can give us `x^2=-1` and we can therefore rid ourself of the fear of the denominator ever equalling naught. By this reasoning, the domain of `f` is all real numbers.

By contemplating the output of our function, we will notice that, from the right the function is decreasing until the point `x=-1`, after which the function steadily increases. From the left, it is the opposite: the function is increasing until the point `x=1`, after which the function steadily decreases.

From either direction, `f` cannot ever equal `0` except at `x=0` because for no number `x>0 or x<0` can `f(x)=0`.

Therefore the highest point on our graph is `f(x)=1/2` and the lowest point is `f(x)=-1/2`. `f` can equal all numbers in between though, so the range is given by all real numbers in between `f(x)=1/2` and `f(x)=-1/2`.

Answer 2

The domain is `x in RR`. The range is `y in [-1/2, 1/2]`

Explanation:

The denominator is

`1+x^2>0, AA x in RR`

The domain is `x in RR`

To find, the range procced as follows :

Let `y=x/(x^2+1)`

`y(x^2+1)=x`

`yx^2-x+y=0`

In order for this quadratic equation to have solutions, the discriminant `Delta >=0`

Therefore,

`(-1)^2-4*y*y>=0`

`1-4y^2>=0`

The solution to this inequality is

`y in [-1/2, 1/2]`

The range is `y in [-1/2, 1/2]`

graph{x/(x^2+1) [-3, 3.93, -1.47, 1.992]}