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Increasing and Decreasing Intervals

  1. Jan 14, 2005 #1
    Hello all

    If you are given the function [tex] y = x - 3e^-x^2 [/tex] and you want to find the intervals where the function is increasing and decreasing, concavity, inflection points and any local extreme values, would I first find the derivative?

    My work

    If [tex] f(x) = x - 3e^-x^2 [/tex] then [tex] f'(x) = 1+6e^(-x^2)xln(e) [/tex[. Then I set this equal to 0 But I get [tex] {e = e, x = RootOf(`.`(1+6*exp(-ln(e)_Z^2)_Zln(e) = 0, _Z))} [/tex]

    How would you determine concavity and any local extrema? I know that to get inflection points you take the second derivative and set it equal to 0.

    Thanks a lot
     
  2. jcsd
  3. Jan 14, 2005 #2

    dextercioby

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    It's this the function??
    [tex] y=y(x)=x-3e^{-x^{2}} [/tex]

    If so,review its differentiation.

    Daniel.
     
  4. Jan 15, 2005 #3

    HallsofIvy

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    Assuming that [itex] y(x)= x- 3e^{-x^2}[/itex] then [itex]y'= 1+ 6xe^{-x^2}[/itex].
    (You really have it right: ln(e)= 1).

    A function is increasing where its derivative is positive, decreasing where its derivative is negative. Of course, those intervals are separated by points where the derivative is equal to 0 so the first thing you would do is determine where the derivative is 0: where [/itex]1+ 6xe^{-x^2}= 0[/itex].

    A function is "concave upward" if its slope is getting greater: i.e. where its derivative is increasing so its second derivative is positive. A function is concave downward where its second derivative is negative and has an inflection point where its second derivative changes sign. Note that the second derivative must be 0 at an inflection point but that is not sufficient! (y= x4 does not have an inflection point at x= 0.)
     
  5. Jan 15, 2005 #4
    thanks a lot Halls and dexter
     
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