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Bad question on a take home quiz

  1. Apr 1, 2004 #1
    I don't know... I've worked through this problem about ten times, but each time, I get the same answer. It's a kind of easy problem too. I don't konw what my problem with this is.

    If f(x) = sin (x/2), then there exists a number c in the interval pi/2 < x < (3*pi / 2 that satisfies the conclusion of the Mean Value Theorem. Which of teh following could be c?

    Here's my work...

    The mean value theorem is f(c) = 1/(b-a) * int (f(x), x, a, b).

    the indefinate integral of sin (x/2) is -2 cos (x/2) + c

    c = -2 (cos ((3 * pi / 2) / 2) - cos ((pi / 2) / 2) / pi

    c = -2 (cos (3 * pi / 4) - cos (pi / 4)) / pi

    c = -2 ( - sqrt (2) / 2 - sqrt (2) / 2) / pi

    c = -2 (- 2 sqrt (2) / 2) / pi

    c = -2 (- sqrt (2)) / pi

    c = 2 * sqrt (2) / pi

    but, the answer sheet only gives answers with pi in the numerator (so i know i must be wrong). Can anyone help?
     
  2. jcsd
  3. Apr 1, 2004 #2

    matt grime

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    You've dropped the f from f(c), as well as used c for two different things (the constant of the integration).
     
  4. Apr 1, 2004 #3
    As far as I know, the constant of integration only applies to taking the indefinate integral. Since I'm taking the definate integral, the constant of integration cancels out. Expalin the rest though...
     
  5. Apr 1, 2004 #4
    Your first equation was:

    [tex]f(c) = \frac{\int_a^b f(x) \, dx}{b-a}[/tex]

    Notice the left-hand side of this equation.

    Now look at the left-hand side of all your work. See something missing?

    cookiemonster
     
  6. Apr 2, 2004 #5

    matt grime

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    Using c for two things was just a note on being careful with your notation. And you spell definite with is not an a.
     
  7. Apr 3, 2004 #6
    soooo... what you guys are saying is that I should set (2 * sqrt (2)) / pi equal to my function and solve for x...

    2 * sqrt (2) / pi = sin (x / 2)
    x / 2 = arcsin (2 * sqrt (2) / pi)
    x = 2 * arcsin (2 * sqrt (2) / pi)

    But this still isn't any of my answers... I'm still doing something wrong. I dont' know what it is.
     
  8. Apr 3, 2004 #7

    matt grime

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    You say which of the choices might be c in the original post. Is there one that is closest?
     
  9. Apr 3, 2004 #8
    The five answers are (A) 2 pi / 3, (B) 3 pi / 4, (C) 5 pi / 6 (D) pi, (E) 3 pi / 2

    I can't figure out a way to simplify arcsin (2 sqrt (2) / pi), so I don't think it's any of these answers.
     
  10. Apr 3, 2004 #9

    HallsofIvy

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    The correct answer is 2arcsin(2&radic;(2)/pi) which lies between A and B.

    That's assuming you mean the "integral" mean value theorem. Normally when I see just "mean value theorem", I think "differential" mean value theorem. The value of c for that is quite easily pi.
     
  11. Apr 3, 2004 #10
    ohhhh, you're right! wow! I guess I got too caught up in the integral mean value theorum. But there's the much easier to do differential one.

    f'(c) = (sin (3 pi / 4) - sin (pi / 4)) / pi
    f'(c) = (sqrt (2) / 2 - sqrt (2) / 2) / pi
    f'(c) = 0 / pi = 0

    sin c / 2 = 0
    c / 2 = arcsin 0
    c / 2 = 0
    c = 0

    Huh, still, I get the wrong answer. Any more help?
     
  12. Apr 3, 2004 #11
    In the first step, don't you mean sin(3pi/2), not sin(3pi/4)?

    cookiemonster
     
  13. Apr 3, 2004 #12
    no, i meant sin (3 pi / 4) because it's sin (x / 2).

    if x = 3 pi / 2, then it would be sin (3 pi / 2 / 2) = sin (3 pi / 4)
     
  14. Apr 3, 2004 #13
    Ah, you're right. Sorry.

    Your error is in f'(c).

    cookiemonster
     
  15. Apr 3, 2004 #14
    OHHHHH! I GOT IT! Maybe.... Maybe... I forgot to take into account my domain. I'm going from pi / 2 to 3 pi / 2. But when i solved for the arcsin 0, i thought it was a 0 radians, which is out of my domain.

    arcsin 0 = pi, not 0

    sin (c / 2) = 0
    c / 2 = arcsin 0
    c / 2 = pi
    c = 2 * pi....

    noooooo the stupid rotten 2... i thought i almost had it.
     
  16. Apr 3, 2004 #15
    f'(c)... As in d/dx(f(x))|c... or maybe d/dx(sin(x/2))|c...

    cookiemonster
     
  17. Apr 4, 2004 #16
    Mean value theorem
    [tex]
    f'(c)= \frac{f( \frac{3 \pi}{2}) - f( \frac{ \pi}{2})}{ \pi}
    [/tex]
    u can see that

    cos(c/2)= 0

    therefore
    c = pi
     
  18. Apr 4, 2004 #17
    wow, finally i see the right answer. Now, I wonder if I can get my calc teacher with that problem. I don't think he'll be able to do it. Maybe he will though.
     
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