Example 12: Where $\sin 3x$ increases and decreases on $\left[0,\frac{\pi}{2}\right]$

Basics you need

Solution

1. Differentiate

   By the chain rule (Basic 2), with outer function $\sin u$ and inner function $u = 3x$⁠:

   $$f'(x) = \cos 3x \cdot 3 = 3\cos 3x.$$
   The factor 3 comes from differentiating the inner function $3x$⁠. It is positive, so it never affects the sign: $f'(x)$ has the same sign as $\cos 3x$ everywhere.
   $\sin 3x$ is continuous and differentiable at every real $x$⁠, so in particular on $\left[0, \frac{\pi}{2}\right]$⁠, and Theorem 1 can be used on any closed piece of this interval.

2. Where is $f'(x) = 0$ in the domain?

   $f'(x) = 0 \iff \cos 3x = 0$⁠. First find the range of the angle: multiplying $0 \le x \le \frac{\pi}{2}$ by 3 gives

   $$0 \le 3x \le \frac{3\pi}{2}.$$

   The cosine is zero only at odd multiples of $\frac{\pi}{2}$⁠; those lying in $\left[0, \frac{3\pi}{2}\right]$ are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ (the next ones, $-\frac{\pi}{2}$ and $\frac{5\pi}{2}$⁠, are outside). So

   $$\begin{aligned} 3x &= \frac{\pi}{2} \ \text{ or } \ 3x = \frac{3\pi}{2}, \\ \text{so} \quad x &= \frac{\pi}{6} \ \text{ or } \ x = \frac{\pi}{2}. \end{aligned}$$

   The point $x = \frac{\pi}{2}$ is the right-hand end of the domain, so only $x = \frac{\pi}{6}$ splits the domain, into $\left[0, \frac{\pi}{6}\right)$ and $\left(\frac{\pi}{6}, \frac{\pi}{2}\right]$⁠.

   Solving $\cos 3x = 0$ without the range of $3x$ would give infinitely many answers $x = \frac{\pi}{6} + \frac{n\pi}{3}$⁠. Fixing the range of the angle first picks out exactly the ones inside the domain, and makes sure none is missed.
   Cosine is zero exactly where the unit-circle point is at the top or bottom of the circle: $\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \dots$ Basics: signs of sin and cos

3. The sign of $f'$ on each piece

   For $0 \le x < \frac{\pi}{6}$⁠: multiply by 3 to get $0 \le 3x < \frac{\pi}{2}$⁠. The angle is in the first quadrant (or at 0), where cosine is positive, so $f'(x) = 3\cos 3x > 0$⁠.

   For $\frac{\pi}{6} < x < \frac{\pi}{2}$⁠: multiply by 3 to get $\frac{\pi}{2} < 3x < \frac{3\pi}{2}$⁠. The angle is in the second or third quadrant, where cosine is negative, so $f'(x) < 0$⁠.

   Test values: $f'\!\left(\frac{\pi}{12}\right) = 3\cos\frac{\pi}{4}$ $= \frac{3}{\sqrt2} \approx 2.1213 > 0$ and $f'\!\left(\frac{\pi}{3}\right) = 3\cos\pi = -3 < 0$⁠.

   Multiplying an inequality by a positive number (here 3) keeps its direction, so the range of $x$ translates directly into a range of the angle. Then the unit circle answers the sign question for the whole range at once.
   First quadrant: $\cos > 0$⁠. Second and third quadrants: $\cos < 0$⁠. Basics: signs of sin and cos

4. Conclude with Theorem 1, end points included

   On $\left[0, \frac{\pi}{6}\right]$⁠: $f$ is continuous on this closed interval, differentiable inside it, and $f'(x) > 0$ for $0 < x < \frac{\pi}{6}$⁠. By Theorem 1(a), $f$ is increasing on $\left[0, \frac{\pi}{6}\right]$.

   On $\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$⁠: $f$ is continuous there and $f'(x) < 0$ for $\frac{\pi}{6} < x < \frac{\pi}{2}$⁠. By Theorem 1(b), $f$ is decreasing on $\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$.

   Because $f'$ is strictly positive (strictly negative) inside, the Mean Value Theorem argument of the book's proof gives $f(x_1) < f(x_2)$ (respectively $>$⁠): the monotonicity is in fact strict on each closed piece.

   Theorem 1 only asks about the sign of $f'$ inside $(a,b)$ plus continuity on $[a,b]$⁠. That is why the end points 0, $\frac{\pi}{6}$ and $\frac{\pi}{2}$ can be included even though $f'(\frac{\pi}{6}) = 0$ and $f'(\frac{\pi}{2}) = 0$⁠. The book first writes $\left[0, \frac{\pi}{6}\right)$ and $\left(\frac{\pi}{6}, \frac{\pi}{2}\right]$⁠, then uses continuity at 0 and $\frac{\pi}{6}$ to close them up, exactly as here.
   Theorem 1, Section 6.3: continuous on $[a,b]$⁠, differentiable on $(a,b)$⁠, and the sign of $f'$ on $(a,b)$ decides. Lesson: increasing and decreasing functions

5. Picture the result

   The key values are $f(0) = \sin 0 = 0$⁠, $f\!\left(\frac{\pi}{6}\right) = \sin\frac{\pi}{2} = 1$ and $f\!\left(\frac{\pi}{2}\right) = \sin\frac{3\pi}{2} = -1$⁠. So the graph climbs from 0 to its top value 1 at $x = \frac{\pi}{6}$⁠, then falls all the way to $-1$ at $x = \frac{\pi}{2}$⁠, crossing the axis at $x = \frac{\pi}{3}$ (where $3x = \pi$⁠).

   Over the domain, $3x$ sweeps three quarters of a full turn, from 0 to $\frac{3\pi}{2}$⁠. The sine of the angle rises during the first quarter turn and falls during the next two, which is the graph we see.

Every fact used, and where it comes from