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Calculus I - max/min, concavity

  1. Nov 18, 2006 #1
    This problem is really frustrating me. Any help is appreciated!

    Answer the following questions for the function f(x) = [tex]\sin^{2}{\frac{x}{6}}[/tex] defined on the interval [-6π + 0.8, 6π/4 - 0.3].

    A. f(x) is concave down on the region ____ to ____.
    B. A global minimum for this function occurs at _____.
    C. A local maximum for this function which is not a global maximum occurs at ____.
    D. The function is increasing on ____ to ____ and on _____ to ____.

    So I found the first derivative, [tex]2\sin{\frac{x}{6}}\cos{(\frac{x}{6})}\frac{1}{6}[/tex] which I thought I could simplify to [tex]\frac{1}{6}\sin{\frac{2x}{6}}[/tex]

    Then taking the second derivative gives, [tex]\frac{1}{18}\cos{\frac{2x}{6}}[/tex]

    Trig has never been my strong suit, and I guess the major problem I'm having here is the domain. I can't look at it and understand clearly what it means. It seems to me that there are several minima and maxima in that domain, but I don't remember sine or cosine graphs having local extrema that aren't also global extrema. I'm hazarding a guess that this means the graph isn't along the x-axis... like, it's either tilted toward the positive y values or the negative y values? Does that mean that the global minimum occurs at one of the end points? The phrasing of the question "A global minimum" suggests there is more than one of those, too, though. Confusing.

    And for the local maximum... I think there's going to be more than one, so maybe I can just set the first derivative to zero and choose any of them.

    As for concavity, I know I use the second derivative to find that, and the first derivative to find increasing/decreasing, but... I just don't know what to do in this case because I don't fully understand the domain. I tried looking at the unit circle but since the values for the domain aren't actually on it, it didn't really help.
    Last edited: Nov 18, 2006
  2. jcsd
  3. Nov 18, 2006 #2
    If you're having trouble visualizing trigonometric domains, write an inequality with your second derivative and then solve it for the value of x (i.e. set your second derivative less than zero). You can determine domains of concavity more easily from there.
  4. Nov 18, 2006 #3
    I suggest you graph the function, it is quite helpful. Rewrite your derivative as [tex]\frac{1}{6}sin(\frac{x}{3})[/tex] just so you understand easier what the trig function is doing when you set it to 0.

    Secondly, remember that on a closed interval you must consider the end points for maxima or minima. This will be important.

    When you set the first derivative to 0 you pick all of them to evaluate for maxima or minima. Giving a maxima is not sufficient. You should get 2 maximas and 2 minimas.
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