Exercise 6.2 · Q17: $\log|\cos x|$ falls on $\left(0,\frac\pi2\right)$ and rises on $\left(\frac{3\pi}2,2\pi\right)$

Basics you need

Solution

1. Where is $f$ defined?

   $\log|\cos x|$ needs $|\cos x| \gt 0$⁠, that is, $\cos x \ne 0$⁠. Cosine is zero only at odd multiples of $\frac{\pi}{2}$⁠; between $0$ and $2\pi$ these are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$⁠. Neither lies inside $\left(0, \frac{\pi}{2}\right)$ or $\left(\frac{3\pi}{2}, 2\pi\right)$ (the intervals are open), so $f$ is defined on both intervals.

   The absolute value lets the logarithm accept negative values of $\cos x$ too, so the domain is larger than for $\log\cos x$⁠. Only the zeros of $\cos x$ are excluded.
   $\cos x = 0$ exactly when $x$ is an odd multiple of $\frac{\pi}{2}$⁠. Basics: signs of sin and cos

2. Remove the absolute value on the two intervals

   For $0 \lt x \lt \frac{\pi}{2}$ (first quadrant) and for $\frac{3\pi}{2} \lt x \lt 2\pi$ (fourth quadrant), $\cos x \gt 0$⁠. So on both intervals

   $$|\cos x| = \cos x, \qquad f(x) = \log\cos x.$$
   An absolute value is awkward to differentiate directly. On each interval we only need the formula that holds there, and on these two intervals $\cos x$ never changes sign.
   $|u| = u$ for $u \gt 0$⁠. Basics: sign of a product

3. Differentiate with the chain rule

   Inner function $u = \cos x$ ($\frac{du}{dx} = -\sin x$⁠), outer function $\log u$ (derivative $\frac1u$⁠), valid since $u \gt 0$ here:

   $$\begin{aligned} f'(x) &= \frac{1}{\cos x}\cdot(-\sin x)\\ &= -\frac{\sin x}{\cos x} = -\tan x. \end{aligned}$$
   This formula holds on both intervals. (In fact $\frac{d}{dx}\log|u| = \frac{u'}{u}$ wherever $u \ne 0$⁠, so $f'(x) = -\tan x$ on the whole domain of $f$⁠; we do not need that here.)
   Chain rule (Theorem 4, Section 5.3.1); $\frac{d}{dx}\log x = \frac1x$ (Theorem 5, Section 5.4); $\frac{d}{dx}\cos x = -\sin x$⁠. Basics: the chain rule

4. Sign of $f' = -\tan x$ on each interval

   • $0 \lt x \lt \frac{\pi}{2}$⁠: $\sin x \gt 0$⁠, $\cos x \gt 0$⁠, so $\tan x \gt 0$ and $f'(x) = -\tan x \lt 0$⁠.
   • $\frac{3\pi}{2} \lt x \lt 2\pi$⁠: $\sin x \lt 0$⁠, $\cos x \gt 0$⁠, so $\tan x \lt 0$ and $f'(x) = -\tan x \gt 0$⁠.
   The minus sign in front of $\tan x$ is the whole point: $\log\cos x$ falls where $\cos x$ falls (first quadrant) and rises where $\cos x$ rises (fourth quadrant).
   $\tan x = \frac{\sin x}{\cos x}$⁠; read the signs of top and bottom. Basics: signs of sin and cos

5. Apply Theorem 1 on each interval

   On $\left(0, \frac{\pi}{2}\right)$⁠. For any $x_1 \lt x_2$ in it, $f$ is continuous on $[x_1, x_2]$⁠, differentiable inside, and $f' \lt 0$ there. Theorem 1 (b) gives $f(x_1) \ge f(x_2)$⁠, and the proof gives $f(x_1) \gt f(x_2)$⁠. So $f$ is decreasing (strictly) on $\left(0, \frac{\pi}{2}\right)$.

   On $\left(\frac{3\pi}{2}, 2\pi\right)$⁠. For any $x_1 \lt x_2$ in it, $f' \gt 0$ between them, so Theorem 1 (a) and its proof give $f(x_1) \lt f(x_2)$⁠. So $f$ is increasing (strictly) on $\left(\frac{3\pi}{2}, 2\pi\right)$.

   Theorem 1 is applied on a closed interval $[x_1, x_2]$ that sits inside the open interval, so the excluded points $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ (where $f$ is undefined) are never involved.
   Theorem 1 (a), (b), Section 6.3. Lesson: increasing and decreasing functions

Every fact used, and where it comes from