# Critical numbers problem

• Oct 30th 2012, 04:14 PM
nubshat
Critical numbers problem
Find all the critical numbers of the function f(x) = x+ln(|cos(x)|) that lie in the interval [pi, 3pi].

• Oct 30th 2012, 04:32 PM
Soroban
Re: Critical numbers problem
Hello, nubshat!

Quote:

$\text{Find all the critical numbers of the function: }\:f(x) \:=\: x\;+\;\ln(|\cos x|)\:\text{ in the interval }[\pi, 3\pi].$

Differentiate and equate to zero: . $f'(x) \:=\:1 + \frac{-\sin x}{\cos x} \:=\:0$

We have: . $1 - \tan x \:=\:0 \quad\Rightarrow\quad \tan x \:=\:1$

Hence: . $x\:=\:\pm\tfrac{\pi}{4},\:\pm\tfrac{5\pi}{4},\:\pm \tfrac{9 \pi}{4},\:\pm\tfrac{13\pi}{4}\:\hdots$

Answers: . $x \;=\;\frac{5\pi}{4},\:\frac{9\pi}{4}$
• Oct 30th 2012, 05:36 PM
hollywood
Re: Critical numbers problem
Doesn't the derivative fail to exist at $\frac{3\pi}{2}$ and $\frac{5\pi}{2}$, so that they are critical points, too?

- Hollywood
• Oct 31st 2012, 04:35 AM
nubshat
Re: Critical numbers problem
would you also have to include the two end points at pi and 3pi?
• Oct 31st 2012, 07:56 AM
hollywood
Re: Critical numbers problem
Yes, I think both end points, too. Critical points are only used for finding minima and maxima, so assuming you're finding the minimum or maximum over the range $\pi$ to $3\pi$, then you would need to include those points, too.

I suppose it could be that you're finding the minimum or maximum over a different range, and for some reason you're temporarily limiting yourself to finding critical points in the given range. That doesn't make much sense, though.

- Hollywood