# Thread: Help with finding Critical Numbers

1. ## Help with finding Critical Numbers

Ok so the equation is
$\displaystyle f(x) = cos^2(x) - 2sin(x)$

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

I get the derivative which is

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

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

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

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

I get the derivative which is

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

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

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

then the first derivative (use chain rule!) is

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

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

I leave the rest for you.

3. Thanks I can do the rest but what does the V sign stand for?

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

$\displaystyle \vee$ means or

and

$\displaystyle \wedge$ means and