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Jusaquikie

  • 3 years ago

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

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  1. Jusaquikie
    • 3 years ago
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    \[f(x) = x^2e^{-x}\] \[f'(x) =e^{-x}(2x-x^2)\] \[f'' =e^{-x}(x-2)^2\] or \[f'' =e^{-x}(x^2-4x+2)\]

  2. hartnn
    • 3 years ago
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    \(f'' =e^{-x}(x^2-4x+2)\) is correct \(f'' =e^{-x}(x-2)^2\) is incorrect

  3. Jusaquikie
    • 3 years ago
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    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
    • 3 years ago
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    x^2-4x+2 will get u to correct points

  5. Jusaquikie
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    well how to solve f"=0 that way

  8. hartnn
    • 3 years ago
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    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
    • 3 years ago
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    those 2 values of x will be your inflection points.

  10. Jusaquikie
    • 3 years ago
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    do i need to use the quadratic formula? beacuse it simplifies to (x-2)(x-2) so 2 will make it zero

  11. hartnn
    • 3 years ago
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    @Jusaquikie plz observe that \(f'' =e^{-x}(x-2)^2\) IS INCORRECT !

  12. hartnn
    • 3 years ago
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    and you use quadratic formula for \(x^2-4x+2=0\)

  13. hartnn
    • 3 years ago
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    or u could complete the square

  14. Jusaquikie
    • 3 years ago
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    ok i see now that for (x-2) to work it would have to be X^2 - 4x +4

  15. hartnn
    • 3 years ago
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    yes!

  16. Jusaquikie
    • 3 years ago
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    it's just late and my mind is jumping to the easiest answer after computing that beastly second derivatives.

  17. Jusaquikie
    • 3 years ago
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    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
    • 3 years ago
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    thanks for the help @hartnn

  19. hartnn
    • 3 years ago
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    thats correct.

  20. hartnn
    • 3 years ago
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    welcome ^_^

  21. Jusaquikie
    • 3 years ago
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    now i can sleep

  22. hartnn
    • 3 years ago
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    good night and sweet dreams :)

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