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
$$\eqalign{
  f(x)&=(x^2-4x+1)e^x\cr
  f'(x)&=(x^2-2x-3)e^x\cr
  f''(x)&=(x^2-5)e^x\ .\cr}$$
There are


*

*zeros at $f(x)=0$, that is, $x=2\pm\sqrt3$;

*stationary points at $f'(x)=0$, that is, $x=-1,3$;

*inflection points at $f''(x)=0$, that is, $x=\pm\sqrt5$.


I can't explain where the answer of $x=4\pm\sqrt5$ came from, are you sure you copied the question correctly?
Edit.  As we can now see, $f(x)=(x^2-4x+1)e^x$ as originally posed in the question was not the function that the OP intended.
