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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- Oct 10th 2007, 06:27 AM #1

- Oct 10th 2007, 06:38 AM #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

- Oct 10th 2007, 06:48 AM #3