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  1. #1
    Senior Member DivideBy0's Avatar
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    Max

    Find the maximum value of $\displaystyle 4\sin{3x}+3\cos{3x}-3$

    I tried differentiating to find the turning point and got

    $\displaystyle f'(x)=12\cos{3x}-9\sin{3x}-3$

    Plugging in x=0 i got 9... but the answer says the maximum is 2?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    Find the maximum value of $\displaystyle 4\sin{3x}+3\cos{3x}-3$

    I tried differentiating to find the turning point and got

    $\displaystyle f'(x)=12\cos{3x}-9\sin{3x}-3$

    Plugging in x=0 i got 9... but the answer says the maximum is 2?
    The derivative of a constant is 0, so
    $\displaystyle f^{\prime}(x) = 12 cos(3x) - 9sin(3x)$

    To find a local min/max you set the derivative to 0, not x. So setting this to 0 and solving I get:
    $\displaystyle 0 = 12cos(3x) - 9sin(3x)$

    $\displaystyle 9sin(3x) = 12cos(3x)$

    $\displaystyle tan(3x) = \frac{12}{9} = \frac{4}{3}$

    $\displaystyle 3x \approx 0.927295$ (There is no "nice" exact value for $\displaystyle tan^{-1}(4/3)$.)

    $\displaystyle x \approx 0.309098$

    Plugging this back into f(x) I get:
    $\displaystyle f(0.309098) \approx 2$

    -Dan
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  3. #3
    Senior Member DivideBy0's Avatar
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    Oops i made 2 mistakes, thanks topsquark
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