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Math Help - Maximum Value of a function on a closed interval

  1. #1
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    Maximum Value of a function on a closed interval

    On the closed interval [0,2pi], the maximum value of the function f(x)=4sinx-3cosx is:

    A.)3
    B.)4
    C.)24/5
    D.)5
    E.)None of these

    Explanations appreciated.
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  2. #2
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    Quote Originally Posted by Manabu View Post
    On the closed interval [0,2pi], the maximum value of the function f(x)=4sinx-3cosx is:

    A.)3
    B.)4
    C.)24/5
    D.)5
    E.)None of these

    Explanations appreciated.
    f(x)=4\sin x-3\cos x

    f'(x)=4\cos x+3\sin x

    for max let f'(x)=0

    0=4\cos x+3\sin x

    4\cos x=-3\sin x

    \frac{-4}{3}=\frac{\sin x}{\cos x}

    \tan x =\frac{-4}{3}

    Can you finish this?
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  3. #3
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    Quote Originally Posted by Manabu View Post
    On the closed interval [0,2pi], the maximum value of the function f(x)=4sinx-3cosx is:

    A.)3
    B.)4
    C.)24/5
    D.)5
    E.)None of these

    Explanations appreciated.
    A couple of suggestions:
    Have you learnt to write that function in the form sin (a+b) or cos(a+b)?
    If so, that'd be the easiest way.
    2. Calculus??
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  4. #4
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    Quote Originally Posted by pickslides View Post
    f(x)=4\sin x-3\cos x

    f'(x)=4\cos x+3\sin x

    for max let f'(x)=0

    0=4\cos x+3\sin x

    4\cos x=-3\sin x

    \frac{-4}{3}=\frac{\sin x}{\cos x}

    \tan x =\frac{-4}{3}

    Can you finish this?
    Not really. I understand how you got to that point, but I'm not 100% sure how to calculate maximum values.
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  5. #5
    Super Member Failure's Avatar
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    Quote Originally Posted by Manabu View Post
    On the closed interval [0,2pi], the maximum value of the function f(x)=4sinx-3cosx is:

    A.)3
    B.)4
    C.)24/5
    D.)5
    E.)None of these

    Explanations appreciated.
    D. is the correct answer, and it can be found without calculus (but with some trigonometry), because

    4\sin x-3\cos x=5\cdot\left(\frac{4}{5}\cdot \sin x-\frac{3}{5}\cdot\cos x\right)=-5\cos(x+\varphi)
    for \varphi := \arccos(3/5). Now, since x varies over a whole period of the function, the maximum value of f(x) must be 5.
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