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Jusaquikie
trying to find an infliction point. the rest i'll make neat.
\[f(x) = x^2e^{-x}\] \[f'(x) =e^{-x}(2x-x^2)\] \[f'' =e^{-x}(x-2)^2\] 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
for my infliction points i need f"=0 i thought this should be at 2 but that is not right. when i solve f" for x on my calculator i get \[-(\sqrt{2}-2) \]and \[\sqrt{2}+2 \]
x^2-4x+2 will get u to correct points
i believe this has something to do with it being squared but i don't understand why beacuse 0 squared is still 0
i know the answer is the first of the infliction points i have listed but i just don't see how to solve the second derivitive that way
well how to solve f"=0 that way
what exactly is your doubt ? u got \(f'' =e^{-x}(x^2-4x+2)\) now f''(x)=0 will give \((x^2-4x+2)=0\) can u solve this quadratic to get 2 values of x ?
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
it's just late and my mind is jumping to the easiest answer after computing that beastly second derivatives.
so the quadratic formula would be what i'd use if they didn't come out even and if i used that i bet i'd get the right answer
thanks for the help @hartnn
good night and sweet dreams :)