1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Extrema of a complicated function.
  1. Extrema of function (Replies: 3)

Loading...