Draw the graph of `y=xe^(x)`. Find the range of the function. Also find the point of inflection.

Explanation

We have y=f(x)=`xe^(x)`
clearly domain of the function is R.
f(0)=0
Curve passes through origin.
Now `f(x) =e^(x)+de^(x)=e^(x)(1+x)`
f(x)=0 there x=-1
`f(x)gt0forxgt-1`, where f(x)increase
`f(x)lt 0 for xlt-1` where f(x) decreases.
x=-1 is point of minima.
`f(-1)=-e^(-1)=-1//e`
`underset(xrarr00)lim xe^(x)=00`
and `1underset(xrarr00)lim xe^(x)=underset(xrarr00)lim(x)/(-x)=underset(xrarr-00)lim (1)(-x)=0`
Thus in (-00,-1),f(x) decrease from `0^(-) to -1//e`
In (-1,00),f(x) increases interesecting x axis at x=0
Also `f(x) =e^(x)(2+x)`
f(x)=0=-2
So x =-2 is point of inflection
From the above discussion the graph of the function is as shown in the following figure.

Range of the function is `[f(-1),00] or -[.^(-1,00)]`
If `xe^(x)`=k` has two distinct real roots line y =k must interesect the graph between f(-1) and `0 -1//e` and 0
So,k in `(-1//e,0)`