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Max min trig functions

  1. Apr 6, 2010 #1
    x in (0,pi];

    f(x) = sin(x)-x2;

    f'(x) = cos(x) - 2x;

    f'(x) = 0 ==> cos(x) - 2x = 0;

    since |cos(x)| ≤ 1,

    cos(x) - 2x ≤ 1 - 2x;

    Now 1-2x = 0 <==> x = 1/2;

    f'(1/4) = cos(1/4) - 2*(1/4) > 0 and f'(3/4) = cos(3/4) - 2*(3/4) < 0;

    ==> x = 1/2 is maximum and f'(x) ≤ 1/2;

    Is my logic correct?
     
  2. jcsd
  3. Apr 6, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    The fact that 1- 2x= 0 does NOT mean cos(x)- 2x= 0 since cos(x)- 2x< 1- 2x, if 1- 2x= 0 then cos(x)- 2x< 0. In fact, with x= 1/2, cos(1/2)- 2(1/2)= -1.22. cos(x)- 2x= 0 when cos(x)= 2x which happens when x is approximately .45.

    No, it is not. saying that cos(x)- 2x< 1- 2x= 0 does not mean cos(x)- 2x= 0.
     
  4. Apr 6, 2010 #3
    I'm trying to find the max and min of this function. Can someone give me some advice?
     
  5. Apr 6, 2010 #4

    Mark44

    Staff: Mentor

    You'll have to do this by numerical approximation, since the equation is not solvable by analytic means.

    cos (x) - 2x = 0 <==> x = .5 cos(x)

    Start with an initial x value of your choice.
    Evaluate .5*cos(x) at that value to get your next x value.
    Repeat step 2.
    When the input value and output value are "close enough" that's your approximate solution.
     
  6. Apr 6, 2010 #5
    How about just getting a bound on this function ( f'(x) ) on (0,pi]; can I do that without finding the max and min?

    f'(0) = 1 and f'(pi) = -1 - 2*pi

    is my guess, but I am unsure of any bigger fluctuations on that interval.
     
    Last edited: Apr 6, 2010
  7. Apr 6, 2010 #6
    Well let's look at the function shall we?

    sinx is always less than 1.
    the max slope of sinx occurs at zero and equals 1; thereafter it is always below the line y=x.
    at small values

    [tex]\sin x\quad \simeq x[/tex]

    since [tex]x \le {(x)^2}[/tex]

    and [tex] - {(x)^2}[/tex] decreases monotonically from 0 to [tex] - {(\pi )^2} \simeq 10[/tex]

    The function is negative once [tex]x \le {(x)^2}[/tex] is greaer than 1.

    Can you sketch it?
     
    Last edited: Apr 6, 2010
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