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Trig derivatives

  1. Oct 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the maximum value and minimum value of the function [tex] f(x) = cos 2x + 2sinx[/tex]

    [tex] 0 \leq x \leq \frac{3\pi}{4}[/tex]


    2. Relevant equations

    [tex] f(x) = cos 2x + 2sinx[/tex]

    3. The attempt at a solution

    [tex] f(x) = cos 2x + 2sinx[/tex]

    [tex] f'(x) = -2sin2x + 2cosx[/tex]

    [tex] 0=-2sin2x + 2 cosx[/tex]

    [tex] 2sin2x=2cosx[/tex]

    [tex] sin2x = cosx[/tex]

    [tex] 2sinxcosx = cosx[/tex]

    [tex] 2sinx = 1 [/tex]

    [tex] sin x = \frac{1}{2}[/tex]

    [tex] x = \frac{\pi}{6}[/tex]

    [tex] f''(x) = -4cos2x - 2sinx [/tex]
    If I check, it is a max. But everything(in the domain) I check is max. What's up with this? Is my derivative wrong?
     
  2. jcsd
  3. Oct 16, 2009 #2
    You have several values you did not check. You must check the endpoints of the interval, x=0 and [itex]x=\frac{3\pi}{4}[/itex]. they may or may not be turning points, but their values may yet be higher or lower than any turning point in the interior of the interval.
    You also divided by cos(x) in one step, which means you must go back and see what happens to f'(x) when cos(x) = 0, as your solution from that point onwards assumes that cos(x) is not 0. First, check to see whether cos(x) is ever 0 in the interval given. If so, you must check that point(s) as well. This, in fact, gives you an additional critical point where sin(2x) = 0.
    Your derivatives are correct. Why are you checking every point in the domain? Remember, the second derivative only gives you concavity/curvature. It does not, by itself, tell you whether a point is a maximum or minimum unless that point is also a critical point.
     
  4. Oct 16, 2009 #3

    Mark44

    Staff: Mentor

    In the line below, you have divided both sides by cosx, effectively discarding cos x = 0 as a solution to your original equation.

    You should have changed the equation above to this:
    2sinxcosx - cos x = 0
    ==> cosx(2sinx - 1) = 0
    ==> cos x = 0 or sinx = 1/2
    I don't understand what you're saying when everything in the domain is a max. On the restricted domain there are a local minimum and a local maximum. Since the domain is restricted, you should be able to find an absolute min and absolute max.
     
  5. Oct 16, 2009 #4
    Ah, I should've worded it better. I checked [tex]x=\frac{\pi}{6}[/tex] with the second derivative. I got -3 which makes it a local max.
     
  6. Oct 16, 2009 #5
    That's correct. However, there are 3 other potential maximums or minimums in the interval to check as well. :)
     
  7. Oct 16, 2009 #6
    :eek: what! I checked 0, 3pi/4 and pi/2 also. 0 and 3pi/4 I got local maximums for that. But a local minimum at pi/2 :/
     
  8. Oct 16, 2009 #7
    Yes! I knew I did that part right. But my teacher just said it was right. By dividing both sides by cos, I would lose a x value - which is a huge mistake. This is how I got pi/2 the first time.
     
  9. Oct 16, 2009 #8
    Right. Now the only thing left to do is compare the values of f at those points to find the absolute maximum and the absolute minimum (since your original post has the singular forms of these words, I assume they mean absolute).
     
  10. Oct 16, 2009 #9
    Whew, you scared me there. I thought you meant that there's more! Yes, these are absolute values (as it is in a restricted domain). The absolute max is at x= pi/6 and absolute min is at x = pi/2.

    However, f(0) = 1 and f(pi/2) = 1. But I checked with 2nd derivative and f(pi/2) is the only min. Is this enough proof?
     
  11. Oct 16, 2009 #10
    No. Both 0 and pi/2 are absolute minima, with minimum value 1.
    Again, the second derivative only tells you the concavity of a region; it does not tell you whether a value is an absolute minimum or not. It gives only enough information to tell whether a critical point is a local minimum or a local maximum, or possibly neither. It is even possible for the 2nd derivative to fail to give you this information, ie., consider the function f(x) = x4 for x = 0. It is obviously a local (and global) minimum, but the second derivative test is inconclusive.
     
  12. Oct 16, 2009 #11
    Hmm. I must've done the 2nd derivative(the calculator part :p) wrong since I got it as a -4 :/
     
  13. Oct 16, 2009 #12
    Damn it. No one told me this :(/ How would I go proving that there is 2 minimas? I know that there is only 1 max.
     
  14. Oct 16, 2009 #13
    The second derivative at x=0 is -4, meaning the graph has negative concavity there (it opens downwards). This does not tell you much about the relation of this point to the values of points in the closed interval under consideration, since this point is not a turning point. Draw a picture.
     
  15. Oct 16, 2009 #14
    You are correct. f(o) = f(pi/2) = 1 but since I know that f"(pi/2) = 2 > 0 it's a min. Logically, f(o) is a min too.
     
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