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Homework Help: Extrema of a complicated function.

  1. Oct 30, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the absolute maximum and minimum values of f(x) = e[tex]^{}-x[/tex] - e[tex]^{}-2x[/tex] on [0,1].

    f '(x) = -e^-x + 2e^-2x


    2. Relevant equations



    3. The attempt at a solution
    f '(x) = -e^-x + 2e^-2x
    0 = -e^-x + 2e^-2x
    e^-x = 2e^-2x
    ln(e^-x) = ln(2e^-2x)
    -x = ?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 30, 2007 #2
    ln(2e^-2x) - that's a product of two terms, right?

    Another way to do this would be to rewrite the equation (0 = -e^-x + 2e^-2x) as

    [tex]2\left(e^{-x}\right)^2 - e^{-x} = 0[/tex]
     
  4. Oct 30, 2007 #3
    so you can factor out the e^-x and your left with:
    e^-x(2(1)^2-1) = 0, correct?
     
  5. Oct 30, 2007 #4
    No. If it makes any easier, substitute y in the place of e^(-x), so that

    (2y^2 - y) = 0 becomes y(2y-1) = 0.
     
  6. Oct 30, 2007 #5
    ln(e^-x) = ln(2e^-2x)

    I am not sure that he know the rules: ln(A*B)=ln(A)+ln(B)

    Also ln is the inverse of e, which means ln(e^C)=C
     
  7. Oct 30, 2007 #6
    ok, so after substituting y for e^-x i saw that f has two critical points, on at x=0, and the other where e^-x=(1/2) . But how do you figure out x?
     
  8. Oct 30, 2007 #7
    y=0, actually. e^(-x) = 0.

    Use the log function now.
     
  9. Oct 30, 2007 #8
    ok so you take ln(e^-x) = ln(1/2)? What about the other critical number, you can't take ln(0) so how do you solve that function? Or is it a critical number because it's not defined? This is very confusing to me.
     
  10. Oct 30, 2007 #9
    what I think I have it. ln(e^-x) = -x, so -x = ln(1/2), or x = -ln(1/2). But what about the zero?
     
  11. Oct 31, 2007 #10
    Or
    x = -ln(1/2) = -(ln(1)-ln(2)) = ln(2) (It just looks better imo. :) )

    Can you find a real x such that e^(-x) = 0 let alone one within [0,1]?
     
    Last edited: Oct 31, 2007
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