# How do you find the critical points for  f(x)=3cos2(x-pi/4)+1?

Apr 18, 2016

Use trigonometry to simplify the expression for $f \left(x\right)$.

#### Explanation:

$\cos 2 \left(x - \frac{\pi}{4}\right) = \cos \left(2 x - \frac{\pi}{2}\right) = \sin 2 x$

(If you don't remember the identity above, use the difference formula for
$\cos \left(2 x - \frac{\pi}{2}\right) = \cos \left(2 x\right) \cos \left(\frac{\pi}{2}\right) + \sin \left(2 x\right) \sin \left(\frac{\pi}{2}\right) = \sin \left(2 x\right)$)

So $f \left(x\right) = 3 \sin \left(2 x\right) + 1$

and $f ' \left(x\right) = 6 \cos \left(2 x\right)$ which exists for all $x$ and is $0$ when

$\cos \left(2 x\right) = 0$. Which is true exactly when

$2 x = \frac{\pi}{2} + \pi k$ for integer $k$. Or,

$x = \frac{\pi}{4} + \frac{\pi}{2} k$ for integer $k$.

Without rewriting we can get the same answer (of course).

$f \left(x\right) = 3 \cos 2 \left(x - \frac{\pi}{4}\right) + 1$

f'(x) = -3sin2(x-pi/4)*[d/dx(2(x-pi/4)] = -3sin2(x-pi/4)*[2(1)]

$f ' \left(x\right) = - 6 \sin 2 \left(x - \frac{\pi}{4}\right)$

F'(x) is never undefined and it is $0$ for

$\sin 2 \left(x - \frac{\pi}{4}\right) = 0$ Which is true exactly when

$2 \left(x - \frac{\pi}{4}\right) = \pi k$ for integer $k$. Or,

$x - \frac{\pi}{4} = \frac{\pi}{2} k$ for integer $k$. So we need

$x = \frac{\pi}{4} + \frac{\pi}{2} k$ for integer $k$.