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Thread: 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

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

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

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

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

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

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


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

    $\displaystyle -2x^2 - 2x + 2 = 0$

    $\displaystyle x^2 + x - 1 = 0$

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

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


    So that means:

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

    Solve these for $\displaystyle t$.
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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 $\displaystyle 1 \pm 2\sin{x}$, it's...

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

    If this is 0, then

    $\displaystyle 0 = 1 - 2\sin{x}$

    $\displaystyle 2\sin{x} = 1$

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

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

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

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


    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 $\displaystyle y = \sin{2t}$

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


    $\displaystyle \frac{du}{dt} = 2$


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


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

    $\displaystyle = -2\cos{2t}$.



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

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

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

    If $\displaystyle f''(a) = 0$ 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, $\displaystyle \sin{\theta}$ is positive in the first and second quadrants, that means that two solutions are

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

    and

    $\displaystyle \pi - \arcsin{\frac{1}{2}} = \frac{5\pi}{6}$ (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 $\displaystyle 2\pi$ specifies that you have gone around the unit circle a certain number of times.
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