Q.Find intervals in which the function given by $f(x) = \sin 3x,\ x \in \left[0, \dfrac{\pi}{2}\right]$ is

Concept: Monotonic Function Analysis — we check the sign of $f'(x)$ over the given interval.

Step 1: Differentiate.

$f'(x) = 3 \cos 3x$.

Step 2: Find critical points in $[0, \pi/2]$.

Set $\cos 3x = 0 \Rightarrow 3x = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{6}$.

Also check endpoints: $x=0$ and $x=\pi/2$.

Step 3: Test sign of $f'(x)$ in each subinterval.

• For $x \in \left[0, \frac{\pi}{6}\right)$: $3x \in [0, \pi/2) \Rightarrow \cos 3x > 0 \Rightarrow f'(x) > 0$ → increasing. …

We need to find where $\sin 3x$ rises and falls on the given closed interval. The key is to examine the sign of the derivative — a function increases where its derivative is positive and decreases where it is negative. This is a direct application of the monotonic function test from calculus.

Why this approach works

For a differentiable function, the sign of $f'(x)$ tells us the direction of motion. If $f'(x) > 0$, the function is strictly increasing; if $f'(x) < 0$, it is strictly decreasing. Here, $f(x) = \sin 3x$ is a sine wave compressed horizontally by a factor of 3, so it completes one full cycle in $2\pi/3$. On our interval $[0, \pi/2]$, which is shorter than half a cycle, we expect exactly one turning point — the peak of the sine wave.

Let’s work through it step by step.

1. Find the derivative. Using the chain rule:

The factor 3 is always positive, so the sign of $f'(x)$ is entirely determined by $\cos 3x$.

2. Find where the derivative is zero (critical points). Set $f'(x) = 0$:

The cosine function is zero at odd multiples of $\pi/2$:

So

Now restrict to $x \in [0, \pi/2]$.

• For $n = 0$: $x = \pi/6$ (inside the interval).
• For $n = 1$: $x = \pi/6 + \pi/3 = \pi/2$ (endpoint).
• For $n = -1$: $x = \pi/6 - \pi/3 = -\pi/6$ (outside).

So the only interior critical point is $x = \pi/6$. The endpoint $x = \pi/2$ also gives $\cos(3\pi/2) = 0$, but we’ll handle endpoints separately.

3. Test the sign of $f'(x)$ in each subinterval.

   The critical point $\pi/6$ splits $[0, \pi/2]$ into two intervals:

   • Interval I: $0 \le x < \pi/6$
   • Interval II: $\pi/6 < x \le \pi/2$

   Pick a test point in each:

   • Interval I: Take $x = 0$. Then $3x = 0$, and $\cos 0 = 1 > 0$. So $f'(0) = 3(1) = 3 > 0$.

     Hence $f$ is increasing on $[0, \pi/6)$.

   • Interval II: Take $x = \pi/4$ (which is $0.785$, and $\pi/6 \approx 0.524$). Then $3x = 3\pi/4$, and $\cos(3\pi/4) = -\frac{\sqrt{2}}{2} < 0$. So $f'(\pi/4) = 3(-\sqrt{2}/2) < 0$.

     Hence $f$ is decreasing on $(\pi/6, \pi/2]$. …

Method: Finding Intervals of Increase/Decrease for a Compressed/Stretched Trig Function on a Restricted Domain

Applies to any $f(x) = \sin(kx)$ or $\cos(kx)$ type function given only on a specific closed interval, not all of $\mathbb{R}$.

Steps

Step 1: Differentiate using the chain rule.

Never drop the multiplier $k$ that comes from the chain rule — even though it doesn't change the sign of the derivative here (since $k>0$), forgetting it is a real computational error that matters in other problems, such as when a question asks for the value of the derivative rather than only its sign.

Step 2: Solve $f'(x) = 0$ for the general solution, then restrict to the given domain.

Trig equations like $\cos(kx) = 0$ have infinitely many solutions ($kx = \tfrac{\pi}{2} + n\pi$); solve for $x$ in terms of the integer $n$, then keep only the values of $n$ that land inside the given closed interval — discard the rest. …

Common Mistakes

Mistake 1: Forgetting to restrict the infinite family of solutions of $\cos 3x = 0$ to the given domain.

Why it's wrong: $\cos 3x = 0$ has solutions $x = \dfrac{\pi}{6}+\dfrac{n\pi}{3}$ for every integer $n$; treating a value like $x=-\pi/6$ as a valid critical point, or mishandling the endpoint $x=\pi/2$ as an "extra" interior critical point, gives a wrong or over-complicated interval breakdown. Correct approach: solve the general equation first, then keep only the values that actually lie inside $[0, \pi/2]$.

Mistake 2: Dropping the chain-rule factor of 3 when differentiating $\sin 3x$. …