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!

What are all the functions that satisfy (Calculus II)

  1. Mar 4, 2008 #1
    The problem:
    [tex](\int f(x)dx)(\int \frac{1}{f(x)} dx) = -1[/tex]
    solve for f(x)

    The attempt at a solution

    divide both sides by integral(f(x)dx):
    [tex]\int \frac{1}{f(x)} dx = \frac{-1}{\int f(x)dx}[/tex]

    took derivative of both sides:
    [tex]\frac{1}{f(x)} = \frac{f(x)}{(\int f(x) dx)^2}[/tex]

    multiplied both sides by f(x):
    [tex]1 = \frac{f(x)^2}{(\int f(x) dx)^2}[/tex]

    square root:
    [tex]1 = \frac{f(x)}{\int f(x) dx}[/tex]

    Now, here I applied L'Hopital's incorrectly (this isn't even a limit I know!) by taking derivative of top and bottom on the right side; it seemed to work later that's why I tried to stick with it:
    [tex]1 = \frac{f'(x)}{f(x)}[/tex]

    took antiderivative of both sides...
    [tex]x = ln|f(x)|[/tex]

    [tex]e^x = |f(x)|[/tex]

    [tex]f(x) = e^x and -e^x[/tex]

    when plugging those two answers into the original equation it seems to work...
    but something isn't right and anyway there is no way that only those two functions satisfy that. help?

    (first time trying to use the tex tags so i have my fingers crossed that I did not make a mistake in formatting..)
     
    Last edited: Mar 4, 2008
  2. jcsd
  3. Mar 4, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    When you took an antiderivative there is a constant of integration possible. So ln(|f(x)|)=x+C. Now what? Also when you took the square root, the square root of 1 could also be -1. That gives you more solutions.
     
    Last edited: Mar 4, 2008
  4. Mar 4, 2008 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    And actually the use of l'Hopital is completely incorrect. It's not even a limit. Do it this way. When you get to integral(f(x))^2=f(x)^2, just differentiate both sides and cancel the f(x). Then differentiate again. Much simpler. Now you just have the differential equation d^2(f(x))/dx^2=f(x).
     
  5. Mar 4, 2008 #4
    However, I am unsure how a -1 will give me more answers. when taking the square root of both sides, it would become:

    [tex]\frac{|f(x)|}{|\int f(x) dx|} = (negative?)1[/tex]

    sorry, but can you elaborate, by including -e^x in the answer (from the absolute value another step or two down), how is that excluding anything?

    As far as the constants go, we could bring them both to the right side, bringing us [tex]e^{ln|f(x)| + c} = e^c * |f(x)|[/tex] we could just call that constant [tex]e^c = k[/tex] bring it to the left side thus making the answer [tex]ke^x and -ke^x[/tex]
     
    Last edited: Mar 4, 2008
  6. Mar 4, 2008 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Um, how about e^(-x)?? You are missing stuff related to those. Like I said, if you reduce the problem to f''=f there are ODE type techniques to handle that.
     
  7. Mar 4, 2008 #6
    ODE type techniques?
     
  8. Mar 5, 2008 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ordinary Differential Equations. More systematic ways of doing the same thing you've been doing.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?