Question

What is the range of the function `f(x) = x/(1+|x|)`?

Answer 1

(-1,1)

Explanation:

First split it into 2 functions, when `x>=0` let `f^1(x)=x/(1+x)`
and when `x<0` let `f^2(x)=x/(1-x)` now we have 2 functions that are continuous on their domains.

Take the `lim_(x->oo) x/(1+x)` Substitution give `(oo)/(oo)` therefore we can use L' Hopitale to show the function approaches 1 at infinity.

Take the `lim_(x->-oo) x/(1-x)` Substitution gives `-(oo)/(oo)`
therefore we can use L'Hopitale once more to show the function approaches -1 at `-oo` therefore adding the range of our two continous function shows the original `f(x) in (-1,1)`

Answer 2

`(-1,+1)` over the non-infinite real domain

Explanation:

First search for minima and maxima of the function. We do this by setting its first derivative to zero. But there's a catch with this function - the modulus of the variable that is in the denominator of the fraction.

By the quotient rule,
`d/dx[x/(1+|x|)]=((1+|x|)-xd/dx[1+|x|])/(1+|x|)^2`

But what is `d/dx|x|`?

We split the function `g(x)=|x|` into two parts, around its change of behaviour at `x=0`:
If `x>=0`, then `g(x)=x`
If `x<0`, then `g(x)=-x`
Note that at `x=0`, both apply: `g(x)=x=-x`.

We can differentiate each of these separately to obtain a derivative for each part of the function (note that we'll leave `x=0` out for now):
If `x>0`, then `g'(x)=1`
If `x<0`, then `g'(x)=-1`
As there is a sharp kink in the function `|x|` at `x=0`, it is not differentiable there and there is no immediately obvious value to assign to its derivative at that point.

There are a couple of ways of writing `g(x)` as a single expression. We can write it without introducing new notation by writing `g(x)=|x|/x` (or equivalently `x/|x|`), but this is a bit cumbersome when it is basically a piecewise constant function. A better solution is to use the sign function (also known as signum), defined by
If `x>0`, then sgn`(x)=1`
If `x=0`, then sgn`(x)=0`
If `x<0`, then sgn`(x)=-1`

The choice of value for `x=0` is largely immaterial, fits the concept of the function, and has the advantage of symmetry. But let's not forget that we've made a choice here.

Return to problem

We'd deduced that
`d/dx[x/(1+|x|)]=((1+|x|)-xd/dx[1+|x|])/(1+|x|)^2`

Now we can complete this differentiation and set it to zero:
`((1+|x|)-xsgn(x))/(1+|x|)^2=0`

We know that for any real number `x` the denominator will never be `<1`, so there are no nasty infinite blow-ups lurking. We simplify:
`1+|x|-x`sgn`(x)=0`
`1+|x|=x`sgn`(x)`

Let's split this into pieces again:
If `x>0`, then we want `1+x=x`, i.e. `1=0`
If `x=0`, then we want `1=0` (notice that the chosen value doesn't matter)
If `x<0`, then we want `1-x=-x`, i.e. `1=0`

All three produce an immediate contradiction, so there are no solutions over the real numbers to `(df)/dx=0`. Thus `f` has no turning points - it is monotonic over the whole range `(-oo,oo)`. So its two real infinite limits give us the two extremes of the range, and we can now solve the problem.

Real infinite limits

The positive infinite limit is 1: `|x|` dominates the constant in the denominator for large absolute values, and is equal to the `x` numerator for positive values. Hence `lim_(x->+oo)x/(1+|x|)=lim_(x->+oo)x/|x|=1`

The negative infinite limit is -1: `|x|` is equal to `-x` for negative values. `lim_(x->-oo)x/(1+|x|)=lim_(x->-oo)x/|x|=-1`

So the range of our function `f(x)` with the domain taken as the real axis (not including infinities) is `(-1,+1)`. If we do include infinities, we add the end points of the open interval to make it closed: `[-1,+1]`.

Sanity check our solution by plotting the graph of the function along with the asymptotes that we've deduced:
graph{(y-x/(1+|x|))(y-1)(y+1)=0 [-100, 100, -2, 2]}