# absolute max and mins

• October 26th 2009, 04:38 PM
hazecraze
absolute max and mins
Find the absolute max and mins of f on the given interval.
$f(x)=2cos(x) + sin(2x), [0, \frac{pi}{2}]$

$f'(x)=-2sin(x) + 2cos(2x)=0$
$cos(2x)=sin(x)
$

I got stuck after this.
• October 26th 2009, 04:50 PM
Debsta
Remember that cos (2x) = 1 - 2 (sin x)^2
So use that on the LHS, then you'll have a quadratic to solve fo sin x.