Question

trying to find an infliction point. the rest i'll make neat.

Answer 1

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

Answer 2

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

Answer 3

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 \[\sqrt{2}+2 \]

Answer 4

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

Answer 5

i believe this has something to do with it being squared but i don't understand why beacuse 0 squared is still 0

Answer 6

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

Answer 7

well how to solve f"=0 that way

Answer 8

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 ?

Answer 9

those 2 values of x will be your inflection points.

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

or u could complete the square

Answer 14

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

Answer 15

yes!

Answer 16

it's just late and my mind is jumping to the easiest answer after computing that beastly second derivatives.

Answer 17

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

Answer 18

thanks for the help @hartnn

Answer 19

thats correct.

Answer 20

welcome ^_^

Answer 21

now i can sleep

Answer 22

good night and sweet dreams :)