Results 1 to 3 of 3

Math Help - function 2--------------------------

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    6

    function 2--------------------------

    the length of a longest interval in which the function :
    3sinx-4(sinx)^3 is increasing
    a) pi/3
    b)pi/2
    c)3pi/2
    d) pi
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318
    Your function is clearly f(x)=sin3x

    f'(x)=3cos3x>0 for 3x\in(-\frac{\pi}{2},\frac{\pi}{2}) i.e for x\in(-\frac{\pi}{6},\frac{\pi}{6})

    Therefore the length of the largest interval in which f(x) is increasing is =\frac{\pi}{6}-(-\frac{\pi}{6}) = \frac{\pi}{3}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    Quote Originally Posted by torquepad View Post
    the length of a longest interval in which the function 3sinx-4(sinx)^3 is increasing
    A "function" should have a name attached. (What has been posted is just an "expression".) So let's insert the name they omitted when they composed this exercise:

    . . . . . f(x)\, =\, 3\sin{(x)}\, -\, 4\sin^3{(x)}

    As you saw in the previous reply, the customary way to answer questions of this sort is to differentiate:

    . . . . . f'(x)\, =\, 3\cos{(x)}\, -\, 12\sin^2{(x)}\cos{(x)}

    . . . . . . . . . =\, 3\cos{(x)}\left(1\, -\, 4\sin^2{(x)}\right)

    Then one would set the derivative equal to zero, and solve for the critical points:

    . . . . . 3\cos{(x)}\, =\, 0\, \mbox{ or }\, 1\, -\, 4\sin^2{(x)}\, =\, 0

    . . . . . \cos{(x)}\, =\, 0\, \mbox{ or }\, \sin^2{(x)}\, =\, \frac{1}{4}

    ...and so forth.

    However, you posted this exercise to the pre-calculus category, so you probably haven't learned anything about derivatives yet. Please reply with an explanation of whatever method you are supposed to be using to find the answer.

    When you reply, please include a clear listing of your steps and reasoning so far. Thank you!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 20
    Last Post: November 27th 2012, 06:28 AM
  2. Replies: 0
    Last Post: October 19th 2011, 05:49 AM
  3. Replies: 4
    Last Post: October 27th 2010, 06:41 AM
  4. Replies: 3
    Last Post: September 14th 2010, 03:46 PM
  5. Replies: 2
    Last Post: September 2nd 2010, 11:28 AM

Search Tags


/mathhelpforum @mathhelpforum