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Homework Help: Find second derivative

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A curve has equation y=e^2x-x^2+x-3 , find value of x for which d^2y/dx^2=0.


    2. Relevant equations



    3. The attempt at a solution
    well. i started by finding out the 1st and 2nd derivative:
    y=e^2x-x^2+x-3
    dy/dx= 2e2^x-2x+1 and d2y/dx2=4e^2x-2 = 0

    dy/dx =>2e^2x=2x-1
    =>e^2x = 2x/2 -1/2
    e2x= x-1/2 (1)

    sub. value (1) into: d2y/dx2.
    4e^2x-2= 0
    e^2x = 1/2
    e^2x = 1/2 (x-1/2)
    no idea.. on what to do now =/.
    Thanks in adv.
     
  2. jcsd
  3. Dec 8, 2009 #2

    ideasrule

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    I don't know what you did here (why is 2e^2x=2x-1? dy/dx doesn't have to equal 0), but you already got 4e^2x-2 = 0 in the previous step. Just solve for x and you're done.
     
  4. Dec 8, 2009 #3

    Dick

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    How did you go from e^(2x)=1/2, which looks ok, to e^(2x)=(1/2)*(x-1/2) which does not look ok? If it's after 4AM there, I suggest you take a nap.
     
  5. Dec 8, 2009 #4
    Oh k, yea. i can barely hold my eyes lol, alright i guess i will sleep now, thanks for the replies and helping me out :), i will be back tomorrow.Good night for now.
     
  6. Dec 8, 2009 #5
    .....
    try to use the inverse operation of the exponential
     
    Last edited: Dec 8, 2009
  7. Dec 8, 2009 #6

    Dick

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    The policy of the forum is not to present solutions for problems, ok? Just give hints. Never do that again, ok? I'm not going to hit the Report button. But I will next time. ibysaiyan could have gotten this on his own. Don't you see the value in that?
     
    Last edited: Dec 8, 2009
  8. Dec 8, 2009 #7
    :[ sorry sorry, i won't do it again
    ...is it better now ? :D
     
  9. Dec 8, 2009 #8

    Dick

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    Much better, thanks!
     
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