## Jusaquikie trying to find an infliction point. the rest i'll make neat. one year ago one year ago

1. Jusaquikie

$f(x) = x^2e^{-x}$ $f'(x) =e^{-x}(2x-x^2)$ $f'' =e^{-x}(x-2)^2$ or $f'' =e^{-x}(x^2-4x+2)$

2. hartnn

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

3. Jusaquikie

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 $\sqrt{2}+2$

4. hartnn

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

5. Jusaquikie

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

6. Jusaquikie

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

7. Jusaquikie

well how to solve f"=0 that way

8. hartnn

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 ?

9. hartnn

those 2 values of x will be your inflection points.

10. Jusaquikie

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

11. hartnn

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

12. hartnn

and you use quadratic formula for $$x^2-4x+2=0$$

13. hartnn

or u could complete the square

14. Jusaquikie

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

15. hartnn

yes!

16. Jusaquikie

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

17. Jusaquikie

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

18. Jusaquikie

thanks for the help @hartnn

19. hartnn

thats correct.

20. hartnn

welcome ^_^

21. Jusaquikie

now i can sleep

22. hartnn

good night and sweet dreams :)