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!

Is this correct?

  1. Feb 21, 2005 #1
    Find the interval in which the function [itex]y=x+\sin x\cos x[/itex] is increasing. So, first I differentiated to get [itex]y'=1+\cos 2x[/itex]. Then I set [itex]y'[/itex] equal to zero:

    [tex]1+\cos 2x=0[/tex]
    [tex]\cos 2x=-1[/tex]
    [tex]2x=\pm \arccos m+2n\pi[/tex], where [tex]n\in\mathbb{Z}[/tex]
    [tex]2x=\pm \arccos(-1)+2n\pi[/tex]
    [tex]2x=\pm\pi+2n\pi[/tex]
    [tex]x=\pm\frac{\pi}{2}+n\pi[/tex]

    So, since [itex]y'=0[/itex] is true only at certain points (because [itex]n\in\mathbb{Z}[/itex]), we know that the function is strictly increasing or decreasing. To find out which, we do this:

    [tex]x_1=-50 : y(x_1)=y_1=-49.75[/tex]
    [tex]x_2=30 : y(x_2)=y_2=29.85[/tex]

    Thus the function is strictly increasing because in the case of [itex]x_1 < x_2[/itex] we have [itex]y_1 < y_2[/itex].

    So the function is increasing on the open interval [itex]X^{\uparrow}=]-\infty;\infty[[/itex].

    Is this correct?

    - Kamataat
     
  2. jcsd
  3. Feb 21, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    True,the derivative is never negative...But you can't say that the function's increasing on R,you'll have to exclude the infinite amount of points where it is zero...:wink:

    Daniel.
     
  4. Feb 21, 2005 #3
    So it's strictly increasing on [itex]X^\uparrow=\mathbb{R}\ \backslash\ (\pm\pi/2+\pi n)[/itex], where [itex]n\in\mathbb{Z}[/itex]?

    - Kamataat
     
  5. Feb 21, 2005 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Remove the minus,if the "n" is integer,automatically the value [itex] -\frac{\pi}{2} [/itex] is assumed.

    Daniel.
     
  6. Feb 21, 2005 #5
    ok, but why is -pi/2 assumed if n is an integer? is there a mathematical reason or is it just agreed on by mathematicians?

    - Kamataat
     
  7. Feb 21, 2005 #6

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Take "n=-1" in the "plus" formula...

    Daniel.
     
  8. Feb 21, 2005 #7
    I fail to see why this should create a problem :confused: .

    - Kamataat
     
  9. Feb 21, 2005 #8

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Because mathematicians like to count the DISTINCT elements of a set ONLY ONCE...

    Daniel.
     
  10. Feb 21, 2005 #9
    yeah, i know that. but i don't get why i have to assume it's a "minus" and not a "plus". wouldn't n=1 in the "minus" formula cause the same problem as n=-1 in the "plus" formula?

    - Kamataat
     
  11. Feb 21, 2005 #10

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    You can of course use the minus version rather then the plus version. But, as Daniel said, to use both is redundant (and hence, not well liked).
    Stick to the one you like best (most would prefer the plus version).
     
  12. Feb 21, 2005 #11
    ok, that's how i understand it. i just got confused, because i thought he meant to use only the "minus" version. thanks anyway, both of you!

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

Have something to add?



Similar Discussions: Is this correct?
  1. Is this correct? (Replies: 4)

  2. Is this correct? (Replies: 3)

  3. Is this correct? (Replies: 0)

Loading...