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Critical points of a trig function

  1. Dec 30, 2011 #1
    1. The problem statement, all variables and given/known data

    g(x)=x - sin (pi*x) at [0,2]

    2. Relevant equations

    all values where g'(x)= 0 (or non-existant) are critical points

    3. The attempt at a solution

    g'(x)=1 - ╥(cos(╥X))
    1/╥=cos(╥X)
    cos^(-1) (1/╥) = ╥X
    X=(cos^(-1) (1/╥))/╥
    x=.397 and x=? 2nd point
     
  2. jcsd
  3. Dec 30, 2011 #2
    Remember your unit circle. Taking the inverse cosine of (1/pi) points to one of two possible triangles that lie on this circle. From the argument, we know that the x-component of this circle will be 1; the hypotenuse will be pi. Remember that the hypotenuse will always be positive but the x- or y-components can be either negative or positive. So, since the x-component is positive, we can have either a positive or a negative y-value to realize our two triangles. So, the angle is +/- 1.25 radians.

    x = +/- 0.397
     
  4. Dec 30, 2011 #3

    SammyS

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    Hello colonelAP. Welcome to PF !

    To help us respond to questions, you should include the complete problem in the text of your post, even if you wrote a portion of what you're looking for in the title of your thread.

    I gather that you are to find all the critical points in the interval [0. 2] for the function
    [itex]f(x)=x-\sin(\pi x)\,.[/itex]​

    What is the period of [itex]\cos(\pi x)\,?[/itex]

    Is [itex]\cos(\pi x)[/itex] an even function, or is it an odd function?
     
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