1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

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

    If so,review its differentiation.

  4. Jan 15, 2005 #3


    User Avatar
    Science Advisor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook