$f(x) = \ln(xe^x)$. Find all $a >$ 0 such that two functions $f′(x)$ and $1 − f′′(x)$ with [2] domain $(a, ∞)$ have the same range.

Question

f(x) = ln(xe^x). Find all a > 0 such that two functions f′(x) and 1 − f′′(x) with [2] domain (a, ∞) have the same range.

ln(xe^x) = lnx + le^x lne=1; therefore ln(xe^x) = lnx+1 f'=1/x +1 f" = -1/(x^2) but i do not know the next step to this question.

Answer 1

$f'(x)=1+\frac 1x$ and $1-f''(x)=1+\frac 1 {x^{2}}$. On $(a,\infty)$ these functions have range $(1,1+\frac 1 a)$ and $(1,1+\frac 1 {a^{2}})$. The ranges are equal iff $a=a^{2}$ which gives $a=1$.