Thread: Max

**DivideBy0** · October 10th 2007, 06:27 AM

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

I tried differentiating to find the turning point and got

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

Plugging in x=0 i got 9... but the answer says the maximum is 2?

**topsquark** · October 10th 2007, 06:38 AM

Originally Posted by DivideBy0

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

I tried differentiating to find the turning point and got

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

Plugging in x=0 i got 9... but the answer says the maximum is 2?

The derivative of a constant is 0, so
$f^{\prime}(x) = 12 cos(3x) - 9sin(3x)$

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

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

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

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

$x \approx 0.309098$

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

-Dan

**DivideBy0** · October 10th 2007, 06:48 AM

Oops i made 2 mistakes, thanks topsquark

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