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Math Help - Find the absolute maximum and absolute minimum values of f on the given interval.

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    Find the absolute maximum and absolute minimum values of f on the given interval.

    Find the absolute maximum and absolute minimum values of f on the given interval. (Round all answers to two decimal places.)
    f(t) = 2 cos t + sin 2t
    [0, ( pi)/2]

    so i know the deriv is -2(sint)+cos(2t), but i don't understand how to find a value which makes this zero

    Use calculus to find the absolute maximum and minimum values of the function. (Round all answers to three decimal places.)
    f(x) = x + 2cos(x)
    [0, 2]

    again, I know the derivative is 1+-2sinx, but i don't understand how to get a value for x which makes the equation equal to zero.
    Last edited by mr fantastic; May 23rd 2010 at 07:08 PM. Reason: Re-titled.
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    Quote Originally Posted by vjelmy View Post
    Find the absolute maximum and absolute minimum values of f on the given interval. (Round all answers to two decimal places.)
    f(t) = 2 cos t + sin 2t
    [0, ( pi)/2]

    so i know the deriv is -2(sint)+cos(2t), but i don't understand how to find a value which makes this zero
    Actually it's

    f'(t) = -2\sin{t} + 2\cos{2t}

     = -2\sin{t} + 2(\cos^2{t} - \sin^2{t})

     = -2\sin{t} + 2(1 - \sin^2{t} - \sin^2{t}) from the Pythagorean Identity

     = -2\sin{t} + 2(1 - 2\sin^2{t})

     = -2\sin{t} + 2 - 2\sin^2{t}

     = -2x^2 - 2x + 2, where x = \sin{t}.


    This is now a quadratic. So setting the derivative = 0 gives:

    -2x^2 - 2x + 2 = 0

    x^2 + x - 1 = 0

    x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}

     = \frac{-1 \pm \sqrt{5}}{2}.


    So that means:

    \sin{t} = \frac{-1 + \sqrt{5}}{2} or \sin{t} = \frac{-1 - \sqrt{5}}{2}.

    Solve these for t.
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    Quote Originally Posted by vjelmy View Post
    Find the absolute maximum and absolute minimum values of f on the given interval. (Round all answers to two decimal places.)
    f(t) = 2 cos t + sin 2t
    [0, ( pi)/2]

    so i know the deriv is -2(sint)+cos(2t), but i don't understand how to find a value which makes this zero

    Use calculus to find the absolute maximum and minimum values of the function. (Round all answers to three decimal places.)
    f(x) = x + 2cos(x)
    [0, 2]

    again, I know the derivative is 1+-2sinx, but i don't understand how to get a value for x which makes the equation equal to zero.
    The derivative is not 1 \pm 2\sin{x}, it's...

    f'(x) = 1 - 2\sin{x}.

    If this is 0, then

    0 = 1 - 2\sin{x}

    2\sin{x} = 1

    \sin{x} = \frac{1}{2}

    x = \left\{\frac{\pi}{6}, \pi - \frac{\pi}{6}\right\} + 2\pi n where n \in \mathbf{Z}

    x = \left\{\frac{\pi}{6}, \frac{5\pi}{6}\right\} + 2\pi n.

    Check which ones are in the region x \in [0, 2]...


    Now use the second derivative test to find which of these are maximums and which are minimums.
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    im sorry

    but i ammm sooooo confused
    the first problem, i don't even understand a little
    why is the deriv of sin(2t)) = 2cos2t? the deriv of sinx=cosx. right?

    and the second problem, i understand that i can only use pi/6 or 5pi/6
    and i know that f" =-2cosx
    do i just plug in these values?
    I am extremely lost
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    Quote Originally Posted by vjelmy View Post
    but i ammm sooooo confused
    the first problem, i don't even understand a little
    why is the deriv of sin(2t)) = 2cos2t? the deriv of sinx=cosx. right?

    and the second problem, i understand that i can only use pi/6 or 5pi/6
    and i know that f" =-2cosx
    do i just plug in these values?
    I am extremely lost
    Using the chain rule:

    If y = \sin{2t}

    Let u = 2t so that y = \sin{u}.


    \frac{du}{dt} = 2


    \frac{dy}{du} = -\cos{u} = -\cos{2t}.


    So \frac{dy}{dt} = \frac{du}{dt}\,\frac{dy}{du}

     = -2\cos{2t}.



    For the second problem, the second derivative test states that if x=a is a critical point, then...

    If f''(a) < 0 then a is a maximum.

    If f''(a) > 0 then a is a minimum.

    If f''(a) = 0 then the result is inconclusive.
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    is a = t?
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    and how do you get


    where
    ?

    the only way i see solving for x is x = arcsin(1/2)
    howwwwwwww do you do that?
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    Quote Originally Posted by vjelmy View Post
    and how do you get


    where
    ?

    the only way i see solving for x is x = arcsin(1/2)
    howwwwwwww do you do that?
    Special Triangles
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    Quote Originally Posted by vjelmy View Post
    and how do you get


    where
    ?

    the only way i see solving for x is x = arcsin(1/2)
    howwwwwwww do you do that?

    In the unit circle, \sin{\theta} is positive in the first and second quadrants, that means that two solutions are

    \arcsin{\frac{1}{2}} = \frac{\pi}{6} (Quadrant 1)

    and

    \pi - \arcsin{\frac{1}{2}} = \frac{5\pi}{6} (Quadrant 2).


    But are these the only solutions? Of course not. It depends on how many times you go around the unit circle. So adding an integer multiple of 2\pi specifies that you have gone around the unit circle a certain number of times.
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