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Calculus Problem (involes critical numbers)

  1. Mar 15, 2004 #1

    Cod

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    The directions state to find any critical numbers of the function within the interval 0 < x < 2pi. The function is:

    [tex]h(x) = sin^2(x) + cos(x)[/tex]

    I've already found that the derivative is:

    [tex]2sin(x)cos(x) - sin(x)[/tex]

    I've set that equal to zero, and that's where I'm stuck. I guess this is more of an algebra question than anything. The answer I keep getting is:

    [tex]tan(x)/sin(x) = 2[/tex]

    However, I don't think its correct. So any help would be greatly appreciated.
     
  2. jcsd
  3. Mar 15, 2004 #2
    [tex]2\sin{x}\cos{x} - \sin{x} = 0[/tex]
    [tex]2\sin{x}\cos{x}= \sin{x}[/tex]
    [tex]2\cos{x}=0[/tex]
    [tex]\cos{x}=0[/tex]

    ...?

    cookiemonster
     
  4. Mar 15, 2004 #3

    Cod

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    Ohhhhh, I'm an idiot. I was adding [tex]sin x[/tex] to both sides then dividing by [tex]cos x[/tex] to get [tex]tan x[/tex]. Well, I guess I was just looking over the obvious the whole time.

    Thanks for the assistance.
     
  5. Mar 15, 2004 #4

    HallsofIvy

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    Staff Emeritus
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    cod: Don't call yourself an idiot. Let's reserve that for cookiemonster! (Hey, it's not often I can do that!!)

    NO, 2sin x cos x= cos x does NOT immediately lead to
    "2cos x= 0"- it leads to 2cos x= 1 !

    In fact, the best way to do this is to factor the original form:
    2sin x cos x- sin x= sin x(2 cos x- 1)= 0 so

    either sin x= 0 or 2 cos x-1= 0. That is, either sin x= 0
    or cos x= 1/2. You can get x itself from those.
     
  6. Feb 10, 2011 #5
    the answer for: Find any critical numbers of the function. (Enter your answers as a comma-separated list.)
    (sin(x))2 + cos(x) 0 < x < 2π

    is: Pi/3 , Pi, 5Pi/3 so you know
     
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