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Critical Points

  1. Jun 19, 2011 #1
    1. The problem statement, all variables and given/known data
    g(x) = 4x - tanx
    2. Relevant equations

    3. The attempt at a solution
    [tex]g(\theta) = 4\theta - tan\theta[/tex]
    [tex]g'(\theta)=4-sec^{2}\theta[/tex]
    [tex]g'(\theta)=\frac{4cos^{2}\theta-1}{cos^{2}\theta}[/tex]
    [tex]g'(\theta)=\frac{-4sin^{2}\theta}{cos^{2}\theta}[/tex]
    [tex]-4tan^{2}\theta=0[/tex]
    [tex]-4(tan\theta)(tan\theta)=0[/tex]
    [tex]\theta = arctan0[/tex]
    [tex]\theta = k\pi[/tex]

    So, that lists the zeros of the derivative if I am correct. I believe the other critical points are pi/2 + kpi, where tan is undefined. That is all of the critical points correct?
     
  2. jcsd
  3. Jun 19, 2011 #2
    You made an error:

    We have cos² θ + sin² θ = 1, but 4 cos² θ - 1 = 4 (1 - sin² θ) - 1 = 3 - 4 \sin² θ.

    Instead you should simply set

    4 cos² θ - 1 = 0,

    which gives you cos θ = ±½.
     
  4. Jun 19, 2011 #3
    Thank you, yeah I caught that right after I posted. Now I am stuck on another problem. I suppose this should more or less be posted in the pre-calc/trig section.

    Finding the critical points of:
    [tex]f(x)=2cos(x)+sin(2x)[/tex]
    [tex]f(x)=2cos(x)+2sin(x)cos(x)[/tex]

    [tex]f'(x)=-2sin(x)+2(cos^{2}(x)-sin^{2}(x))[/tex]
    [tex]f'(x)=-2sin(x)+2cos^{2}(x)-2sin^{2}(x)[/tex]

    I just know there is a pythag. identity in there somewhere to aid in solving for 0.
     
  5. Jun 19, 2011 #4
    Well, just express the cosine by the sine using the Pythagorean identity to get f'(x) = -2 sin x + 2 - 4 sin² x, which is a polynomial in z = sin x with roots z = -1/4 ± 3/4. Then you calculate the x corresponding to those values of sin x. That's a standard trick to solve trigonometric equations - if you have odd powers of only one trigonometric function (here sin), express the other (here cos) and you get a polynomial in the first trig function.
     
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