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Finding slope locations

  1. Mar 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Let f(x) = xe^(2x)

    A) Find the values of x for which the slope of the cruve y = f(x) is 0
    B) Explain the meaning of your answer to part (a) in terms of the graph f


    2. Relevant equations



    3. The attempt at a solution

    I tried setting xe^(2x) to 0 by making it 0 = xe^(2x). I dont know what to do from there on. Am i supposed to take the derivative of xe^(2x)?
     
  2. jcsd
  3. Mar 23, 2012 #2
    A derivative is a fancy way of saying "instantaneous slope." So if you need to find at which values of x the slope is instantaneously 0, you need to set the derivative to zero, not the function itself.
     
  4. Mar 23, 2012 #3

    So i get 2xe^(2x) = 0
    The answers in the back of the book say that x = 1/2 but i just cant seem to find a way to that answer
     
  5. Mar 23, 2012 #4
    Product rule.
     
  6. Mar 23, 2012 #5
    Im sorry i just cant seem to do this right..so far i got

    x(2e^(2x))+e^(2x)
    =2e^(2x)x+e^(2x)
     
  7. Mar 23, 2012 #6
    Yes, that looks right. Now set that equal to zero and solve for x. For the record, as you've written f(x), I believe the answer in the book is incorrect.
    [tex] 2xe^{2x}+e^{2x} = 0[/tex]

    Can you solve it from here? What answer do you find? I will tell you if I find the same answer.
     
  8. Mar 23, 2012 #7
    I thinking that you can factor out a e^(2x).

    So it would be e^(2x)(x+1).

    Then subtract 1 to the other side, then divide by...I have no clue
     
  9. Mar 23, 2012 #8
    if A(x)B(x) = 0, what must be true about at least A(x) or B(x)? Think back to algebra I where you solved

    [tex]x^2-1=(x+1)(x-1)=A(x)B(x) = 0[/tex]
    [tex]A(x)=x+1[/tex]
    [tex]B(x)=x-1[/tex]

    Another method to arriving to the same conclusion is:
    What does
    [tex] \frac{0}{A}[/tex]
    equal if A is nonzero? Is an exponential ever zero? A/A = 1 if A is nonzero too. Can you think of a single step to remove the exponential?
     
    Last edited: Mar 23, 2012
  10. Mar 23, 2012 #9
    Im sorry i dont understand
     
  11. Mar 23, 2012 #10
    Divide both sides by the exponential or you can simply say if A(x)B(x) = 0, A(X) = 0 or B(x) = 0.
    [tex] A(x) = e^{2x}[/tex]
    [tex] B(x) = 2x+1[/tex]
    A(x) cannot reach zero for a finite x. B(x) can. So solve for when B(x) = 0 for the solution. Note, you end up solving for B(x) = 0 if you divide both sides by A(x) (the exponential).
     
    Last edited: Mar 23, 2012
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