Draw the graph of `y=|x|`.

Explanation

We have `y=f(x) = xe^(x)`
Clearly, the domain of the function is R.
`f(0) =-0`
and `f(x)=0` when `x=0`
`f^(')(x) = e^(x)+xe^(x) = e^(x)(1+x)`
`f^(')(x) gt 0` for `x lt -1`, where `f(x)` increases
`f^(')(x) lt 0` for `x lt -1`, where `f(x)` decreases
`f^(')(x) =0 therefore x=-1`
Also at `x=-1, (dy)/(dx),` Changes sign from negative to positive, hence `x=-1` is the point of minima,
`f(-1)=-e^(-1)=-1//e`
When `x to infty, f(x) to infty`
Also `underset(x to -infty) "lim" xe^(x) = underset(xto -infty) x/(e^(-x)) = underset(x to -infty) 1/(-e^(-x)) =0`
So x-axis is an asymptote.
Thus, in `(-infty,-1), f(x)` decreases from `0^(+) to -1//e`
In `(1, infty), f(x)` increases, intersecting the x-axis at x=0
`f^('')(x) =e^(x)(2+x)`
`f^('')(x)=0 therefore x=-2`
So `x=-2` is the point of inflection.
From the above discussion, the graph of the function is as follows:

Range of the function is `[f(-1), infty)` or `[-e^(-1),infty)`.