The inflection points of $f(x)=(x^2-4x+1)e^{-x}$ I got the function $f(x)=(x^2-4x+1)e^{-x}$. The task is to find the inflection points. The correct answer is $x=4-\sqrt{5}$ and $ x=4+\sqrt{5} $. I got the second derivative to $f(x)$.
But when I equal it to zero I got $x=2+\sqrt{5}$ and $ x=2 - \sqrt{5} $. 
What am I doing wrong?

We have:
$$f(x) = P(x) e^x,$$ 
so $f'' = P'' e^x + 2 P' e^x + e^x P = e^x (2 + 4 x - 8 + x^2- 4x +1  ) = (x^2- 5) e^x$. Since $e^x > 0 $ for  $x \in (-\infty,\infty)$, the solution of $f''(x) = 0$ is given by $x^2 -5 =0 $, so the inflection points are $x_i = \pm \sqrt{5}$.
Hope this helps.
Cheers!
Edit: have a look of the graph of $f(x)$ here.
