# Help with finding Critical Numbers

• Nov 28th 2007, 05:45 PM
FalconPUNCH!
Help with finding Critical Numbers
Ok so the equation is
$f(x) = cos^2(x) - 2sin(x)$

0 < or equal to x < or equal to 2pi

I get the derivative which is

$f'(x) = -sin^2(x) - 2cos(x)$

I was never good with Trigonometric identities so I need help with finding the critical numbers.
• Nov 28th 2007, 09:25 PM
earboth
Quote:

Originally Posted by FalconPUNCH!
Ok so the equation is
$f(x) = cos^2(x) - 2sin(x)$

0 < or equal to x < or equal to 2pi

I get the derivative which is

$f'(x) = -sin^2(x) - 2cos(x)$

I was never good with Trigonometric identities so I need help with finding the critical numbers.

Hello,

if
$f(x)=\cos^2(x)-2\sin(x)\ ,~0 \leq x \leq 2\pi$

then the first derivative (use chain rule!) is

$f'(x)=2\cos(x)\cdot (-\sin(x)) - 2\cos(x)=-2\cos(x)(\sin(x)+1)$

$f'(x)=0~\implies~-2\cos(x)=0~\vee~\sin(x)+1=0$

I leave the rest for you.
• Nov 28th 2007, 09:38 PM
FalconPUNCH!
Thanks I can do the rest but what does the V sign stand for?
• Nov 28th 2007, 09:41 PM
earboth
Quote:

Originally Posted by FalconPUNCH!
Thanks I can do the rest but what does the V sign stand for?

Hi,

$\vee$ means or

and

$\wedge$ means and