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Derivative of exponential function

  1. Dec 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Find extrema and points of inflection (then graph it).

    f(x) = x2e-x

    2. Relevant equations



    3. The attempt at a solution

    So, for f'(x) = xe-x(2-x)

    Critical point(s):

    f'(x) = 0

    2-x = 0
    x= 2

    I have a question before continuing. Will my critical point include 0 besides 2?

    I still have doubts about whether 0 is always part of the critical points because my prof., when working on a similar problem, stated that only the number found was the critical point and didn't use zero. He said that e-x cannot be equal to zero. Meaning he used 2 to find the extrema. What he did was using test numbers between 2 to see where it was increasing and decreasing (THAT if I didn't miss any step while copying it - I tend to rush a little when copying the notes on the board since he writes a lot).

    However, with this problem, my book seems to use 0 as the other point to find extrema. For instance, when substituting 2 and 0 back into the original function, I'll have (2, 4e-2) and (0,0). Being the first , the MAX and the last, the MIN extrema.

    Does that mean that 0 was the other critical point?
    Meaning, will 0 ALWAYS be part of the critical point and be used to find the MIN/MAX extrema?
     
    Last edited: Dec 18, 2012
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  3. Dec 18, 2012 #2

    Dick

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    If f'(0)=0 (and it is here) then x=0 is a critical point. The equation you want to solve is x(2-x)=0. It has two roots. If your prof missed x=0 that was a mistake.
     
  4. Dec 18, 2012 #3
    So I'll have zero for x= 0 and (2-x)= 0.

    What if I had e^-x(2-x)? Without the x? Then, would my critical point be only 2?
     
  5. Dec 18, 2012 #4

    Dick

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    Absolutely. 0 isn't ALWAYS a critical point. It only is when f'(0)=0.
     
  6. Dec 18, 2012 #5
    Thanks!
     
    Last edited: Dec 18, 2012
  7. Dec 18, 2012 #6
    Just a curiosity: if I had only e-x(2-x), then only 2 would be my critical point. Then, how would I know whether the point found after substituting it back into the function is MAX or MIN, considering that you can't compare it with any other point?
     
  8. Dec 18, 2012 #7

    Dick

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    You can compare it with points that aren't critical. Like x=0,1,3 etc.
     
  9. Dec 18, 2012 #8
    Thank you for clarifying that for me. I was having doubts about it. So I'll continue the problem:

    Since the problem is long, I will write the answers I found and will only solve here the step which I am having trouble with.

    So, again:

    f'(x) = xe-x(2-x)
    f'(x) = 0
    x= 0 and (2-x) = 0; x=2

    Critical point(s): 2 and 0
    Substituting it back into the original function, I will have:
    (2,4e-2) and (0,0).

    Extrema MAX: (2,4e-2)
    Extrema MIN:(0,0)

    Now, to find the point of inflection (the part I am having trouble with):

    f''(x) = xe-x(2-x)
    I figure that doing the distributive would be better (maybe that's where I went wrong?):

    f''(x) = 2xe-x -x2e-x

    f''(x) = [(2x)(e-x)' + (e-x)(2x)'] + [(-x2)(e-x)'+(e-x)(-x2)']

    Until now, am I correct?
     
  10. Dec 18, 2012 #9

    Dick

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    That looks ok so far.
     
  11. Dec 18, 2012 #10
    f''(x) = [(2x)(e-x)' + (e-x)(2x)'] + [(-x2)(e-x)'+(e-x)(-x2)']

    f''(x) = [(2x)(e-x)(-1) + (e-x)(2)] + [ (-x2)(e-x)(-1) + (e-x)(-2x)]

    f''(x) = -2xe-x + 2e-x+x2e-x - 2xe-x

    f''(x) = e-x(-2x + 2 + x2 - 2x)
    f''(x) = e-x(-4x + 2 + x2)

    How is it so far?
     
    Last edited: Dec 18, 2012
  12. Dec 18, 2012 #11
    Thanks a lot! The help you, mentors, from this forum, provide is of tremendous relevance when I have doubts/can't understand something.
     
  13. Dec 18, 2012 #12

    Dick

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    It agrees with what I got.
     
  14. Dec 18, 2012 #13
    Now, I am having trouble with how to proceed from here.


    f'(x)=0
    x2-4x + 2=0

    I was expecting to get a "proper" number.
     
  15. Dec 18, 2012 #14

    Dick

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    Use the quadratic equation. Irrationals are numbers too.
     
  16. Dec 18, 2012 #15
    I can't factor this out.
    What should I do?

    x(-4+x)=-2
    x=-2/(-4+x) ?
     
  17. Dec 18, 2012 #16

    Dick

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  18. Dec 18, 2012 #17

    Mark44

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    The two equations above are incorrect since you haven't found the 2nd derivative yet. What's on the right side is actually f'(x).

    The corrected second equation would be
    f''(x) = d/dx(2xe-x -x2e-x)
     
  19. Dec 18, 2012 #18

    Dick

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    Yeah, I'll usually gloss over bad notation on the preceding lines if the last one comes out correct. Good to correct that though for writing up the final problem.
     
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