Miscellaneous · Q3: Where $\frac{4\sin x-2x-x\cos x}{2+\cos x}$ increases and decreases

Basics you need

Solution

1. Domain: every real number

   Since $-1 \le \cos x \le 1$⁠, the denominator satisfies

   $$2 + \cos x \ge 2 - 1 = 1 > 0.$$

   It is never zero, so $f$ is defined, continuous and differentiable at every real $x$⁠.

   Theorem 1 needs continuity on closed intervals and differentiability inside them. Checking the denominator once settles this for every interval we will use.

2. Simplify: $f(x) = \frac{4\sin x}{2 + \cos x} - x$

   Group the last two terms of the numerator: $-2x - x\cos x = -x(2 + \cos x)$⁠. So

   $$\begin{aligned} f(x) &= \frac{4\sin x - x(2 + \cos x)}{2 + \cos x} \\ &= \frac{4\sin x}{2 + \cos x} - x. \end{aligned}$$
   Splitting the fraction is allowed because $2 + \cos x \ne 0$ (Step 1). Differentiating the original fraction would also work, but it creates a long numerator; the simplification makes the derivative short and easy to factorise.

3. Differentiate and factorise

   For $\frac{4\sin x}{2 + \cos x}$ use the quotient rule with $u = 4\sin x$⁠, $u' = 4\cos x$⁠, $v = 2 + \cos x$⁠, $v' = -\sin x$⁠:

   $$\begin{aligned} \frac{u'v - uv'}{v^2} &= \frac{4\cos x(2 + \cos x) + 4\sin^2 x}{(2 + \cos x)^2} \\ &= \frac{8\cos x + 4(\cos^2 x + \sin^2 x)}{(2 + \cos x)^2} \\ &= \frac{8\cos x + 4}{(2 + \cos x)^2}. \end{aligned}$$

   The derivative of $-x$ is $-1$⁠. Writing $1 = \frac{(2 + \cos x)^2}{(2 + \cos x)^2}$ and expanding $(2 + \cos x)^2 = 4 + 4\cos x + \cos^2 x$⁠:

   $$\begin{aligned} f'(x) &= \frac{8\cos x + 4 - 4 - 4\cos x - \cos^2 x}{(2 + \cos x)^2} \\ &= \frac{4\cos x - \cos^2 x}{(2 + \cos x)^2} \\ &= \frac{\cos x\,(4 - \cos x)}{(2 + \cos x)^2}. \end{aligned}$$
   $-uv' = -4\sin x\cdot(-\sin x) = +4\sin^2 x$⁠: the two minus signs cancel. Then $\cos^2 x + \sin^2 x = 1$ collapses the numerator. Putting everything over one denominator lets us take out the common factor $\cos x$⁠.
   Quotient rule, Section 5.3 rule (3); derivatives of $\sin$ and $\cos$⁠, Table 5.3; $\sin^2 x + \cos^2 x = 1$ (Basic 3).

4. Only $\cos x$ can change sign

   • $4 - \cos x \ge 4 - 1 = 3 > 0$ for every $x$⁠;
   • $(2 + \cos x)^2 \ge 1 > 0$ for every $x$ (Step 1).

   So $f'(x)$ has the same sign as $\cos x$⁠, and $f'(x) = 0$ exactly when $\cos x = 0$⁠, at the odd multiples of $\frac\pi2$⁠.

   A product of a positive quantity with $\cos x$ is positive, negative or zero exactly when $\cos x$ is. The whole problem has become "where is $\cos x$ positive?".
   Sign of a product (Basic 5). Basics: sign of a product

5. Sign chart on one period, $[0, 2\pi]$

   The question names no interval. Since $f'$ depends on $x$ only through $\cos x$⁠, its sign pattern repeats every $2\pi$⁠, so we first work on $[0, 2\pi]$ (as the answer key does). There $\cos x = 0$ at $x = \frac\pi2$ and $x = \frac{3\pi}2$⁠, and

   • on $\left(0, \frac\pi2\right)$⁠: $\cos x > 0$ (test: $f'\!\left(\frac\pi3\right) = \frac{\frac12\cdot\frac72}{\left(\frac52\right)^2} = 0.28$⁠);
   • on $\left(\frac\pi2, \frac{3\pi}2\right)$⁠: $\cos x < 0$ (test: $f'(\pi) = \frac{(-1)(5)}{1} = -5$⁠);
   • on $\left(\frac{3\pi}2, 2\pi\right)$⁠: $\cos x > 0$ (test: $f'\!\left(\frac{5\pi}3\right) = 0.28$⁠).
   The test values are a check on the factor analysis, not a replacement for it: the factor argument covers every point of each interval, a test value covers one.
   Signs of $\cos$ on the unit circle (Basic 4). Basics: signs of sin and cos

6. Conclude on $[0, 2\pi]$ with Theorem 1

   $f$ is continuous everywhere (Step 1), so Theorem 1 applies on each closed piece:

   • $f' > 0$ on $\left(0, \frac\pi2\right)$⁠: $f$ is increasing on $\left[0, \frac\pi2\right]$⁠;
   • $f' < 0$ on $\left(\frac\pi2, \frac{3\pi}2\right)$⁠: $f$ is decreasing on $\left[\frac\pi2, \frac{3\pi}2\right]$⁠;
   • $f' > 0$ on $\left(\frac{3\pi}2, 2\pi\right)$⁠: $f$ is increasing on $\left[\frac{3\pi}2, 2\pi\right]$⁠.

   By the proof of Theorem 1 each of these is even strict. The endpoints may be included because $f$ is continuous there.

   Note that $f$ is increasing on each of $\left[0, \frac\pi2\right]$ and $\left[\frac{3\pi}2, 2\pi\right]$ separately, not on the two together: $\frac\pi2 < \frac{3\pi}2$ but $f\!\left(\frac\pi2\right) = 2 - \frac\pi2 \approx 0.429$ is larger than $f\!\left(\frac{3\pi}2\right) = -2 - \frac{3\pi}2 \approx -6.712$⁠.

   Theorem 1 is a statement about one interval at a time. Joining two intervals on which $f$ increases does not give an increasing function, because the drop in between can bring the values down, as the graph shows.
   Theorem 1, Section 6.3 (Basic 2); Definition 1 (Basic 1). Lesson: increasing and decreasing functions

7. All real $x$

   $\cos x > 0$ on every interval $\left(2n\pi - \frac\pi2, 2n\pi + \frac\pi2\right)$ and $\cos x < 0$ on every interval $\left(2n\pi + \frac\pi2, 2n\pi + \frac{3\pi}2\right)$⁠, $n \in \Z$⁠. By Theorem 1, for every integer $n$⁠:

   $$\begin{aligned} &\text{increasing on } \left[2n\pi - \tfrac\pi2,\ 2n\pi + \tfrac\pi2\right], \\ &\text{decreasing on } \left[2n\pi + \tfrac\pi2,\ 2n\pi + \tfrac{3\pi}2\right]. \end{aligned}$$

   With $n = 0$ and $n = 1$ these, cut down to $[0, 2\pi]$⁠, give exactly the intervals of Step 6.

   $f$ itself is not periodic (the $-x$ term keeps pulling it down), but $f'$ is, because it only involves $\cos x$⁠. That is why the increasing and decreasing intervals repeat with period $2\pi$⁠.

8. Compare with the answer key

   The key, working on $[0, 2\pi]$⁠, gives (i) $0 \le x \le \frac\pi2$ and $\frac{3\pi}2 < x < 2\pi$⁠; (ii) $\frac\pi2 < x < \frac{3\pi}2$⁠. These are our intervals with some endpoints left out. Both versions are correct: a function that is increasing on $\left[\frac{3\pi}2, 2\pi\right]$ is also increasing on the smaller interval $\left(\frac{3\pi}2, 2\pi\right)$⁠, and similarly for the others. Our version simply records the largest intervals that Theorem 1 allows.

   Whether to include an endpoint where $f' = 0$ is a matter of convention in many books. With Definition 1 and Theorem 1 of this book, the closed intervals are justified, because $f$ is continuous at the endpoints.

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