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\[f(x) = x^2e^{-x}\]
\[f'(x) =e^{-x}(2x-x^2)\]
\[f'' =e^{-x}(x-2)^2\] or
\[f'' =e^{-x}(x^2-4x+2)\]

\(f'' =e^{-x}(x^2-4x+2)\) is correct
\(f'' =e^{-x}(x-2)^2\) is incorrect

x^2-4x+2 will get u to correct points

well how to solve f"=0 that way

those 2 values of x will be your inflection points.

do i need to use the quadratic formula? beacuse it simplifies to (x-2)(x-2) so 2 will make it zero

@Jusaquikie plz observe that \(f'' =e^{-x}(x-2)^2\) IS INCORRECT !

and you use quadratic formula for \(x^2-4x+2=0\)

or u could complete the square

ok i see now that for (x-2) to work it would have to be X^2 - 4x +4

yes!

thanks for the help @hartnn

thats correct.

welcome ^_^

now i can sleep

good night and sweet dreams :)