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Finding a Point of Inflexion for f(x)

  1. Apr 29, 2015 #1
    1. The problem statement, all variables and given/known data
    The function ##f## is defined on the domain ##x≥0## by ##f(x) = \frac{x^2}{e^x}##.
    (a) Find the maximum value of ##f(x)##, and justify that it is a maximum.
    (b) Find the ##x## coordinates of the points of inflexion on the graph of ##f##.

    I believe that I did (a) correctly, but (b) is where I got stuck.
    2. Relevant equations
    Point of Inflexion Criteria:
    • Is 0 at ##f''(x)##
    • Changes Sign at ##f''(x) = 0##
    I believe that's all that's really relevant, and of course differentiation, power/quotient rules.

    3. The attempt at a solution
    (a)
    ##f(x) = \frac{x^2}{e^x}##
    ##f'(x) = \frac{2xe^x - x^2e^x}{e^{2x}}##
    ∴ ##2xe^x - x^2e^x = 0##
    ##2xe^x = x^2e^x##
    ##2x = x^2##
    ##x=2##
    Since the exponentially growing ##e^x## will increase at a much greater pace than the ##x^2## term, in ##f(x) = \frac{x^2}{e^x}##, there can be no other maxima.

    (b)
    This is the part I had trouble on. I figured the second derivative to be:
    ##f''(x) = \frac{3x^2-4x+2}{e^x}##
    ∴ ##3x^2-4x+2 = 0##
    ##x = \frac{4±\sqrt{40}}{6} \rightarrow x ≥ 0##
    ∴ ##x = \frac{4+\sqrt{40}}{6}##

    I've probably made some mistake somewhere in my algebra, or more likely that annoying differentiation, but when I plug values on either side of that resulting x back into ##f''(x)##, there is no change in sign. Am I missing something? Have I made a mistake somewhere in my algebra or calculus? I don't believe there could be any other points that equal zero in the equation I've generated, so where'd I go wrong?

    Thanks to anyone who can help me with this problem!
     
  2. jcsd
  3. Apr 29, 2015 #2

    SteamKing

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    Using the quotient rule to take the first and second derivatives is the hard way to go.

    Re-write f(x) = x2 * e-x instead and take both derivatives again. This can serve as a check on your original work, which I don't think is entirely correct, especially f"(x). Note: f'(x) can be simplified from your result.
     
  4. Apr 29, 2015 #3
    I've just double checked. First derivative is (presumably) correct, and simplified down to:
    ##f'(x)=\frac{2x-x^2}{e^x}##
    Second derivative I did indeed get a different result for:
    ##f''(x)=\frac{x^2-3x}{e^x}##

    After that I figured ##x=3## and it all worked out. Thanks for your suggestion! I don't know why I didn't go for the power rule in the first place, but anyways, thanks for your assistance!
     
  5. Apr 29, 2015 #4

    Ray Vickson

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    Please: never again write ##1/e^x##; always convert it to ##e^{-x}##. Believe it or not, that will simplify your life a lot!
     
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