Q.Find the maximum and minimum values of $x + \sin 2x$ on $[0, 2\pi]$.

Idea: on the closed interval $[0,2\pi]$, compare $f$ at its interior critical points and its endpoints.

Let $f(x)=x+\sin 2x$, so

Set $f'(x)=0:\ \cos 2x=-\tfrac12$. With $2x\in[0,4\pi]$,

Evaluate $f$ at these and the endpoints $0,2\pi$:

$f\!\left(\tfrac{\pi}{3}\right)=\tfrac{\pi}{3}+\tfrac{\sqrt3}{2}\approx1.91,\quad f\!\left(\tfrac{2\pi}{3}\right)=\tfrac{2\pi}{3}-\tfrac{\sqrt3}{2}\approx1.23,$ …

The plan

$f$ is continuous on the closed interval $[0,2\pi]$, so it attains both a greatest and a least value there, and these can only occur at an interior critical point ($f'=0$) or at an endpoint. We find all such $x$, evaluate $f$, and compare.

Step 1 — Critical points

As $x$ runs over $[0,2\pi]$, the angle $2x$ runs over $[0,4\pi]$, where $\cos 2x=-\tfrac12$ at

Hence

Step 2 — Evaluate $f$ at every candidate

Endpoints:

Interior critical points (using $\sin\tfrac{2\pi}{3}=\tfrac{\sqrt3}{2}$, $\sin\tfrac{4\pi}{3}=-\tfrac{\sqrt3}{2}$, and their periodic repeats):

$f\!\left(\tfrac{\pi}{3}\right)=\tfrac{\pi}{3}+\tfrac{\sqrt3}{2}\approx1.91,\qquad f\!\left(\tfrac{2\pi}{3}\right)=\tfrac{2\pi}{3}-\tfrac{\sqrt3}{2}\approx1.23,$ …

Method: Closed-Interval Extreme Value Method (Critical Points + Endpoints)

This method finds the absolute maximum and minimum of a continuous function on a closed interval $[a, b]$ — used whenever a function combines a polynomial-type term with a periodic (trigonometric) term.

Steps

Step 1: Confirm continuity and differentiability

Since the function is continuous on the closed interval $[a,b]$, the Extreme Value Theorem guarantees it attains both an absolute maximum and an absolute minimum somewhere on $[a,b]$ — and these can only occur at an interior critical point ($f'(x)=0$ or undefined) or at one of the two endpoints.

Step 2: Find every critical point inside the interval

Solve $f'(x) = 0$. When the equation involves a trigonometric term like $\cos(kx)$, remember the argument $kx$ sweeps through a wider range than $x$ itself as $x$ runs over $[a,b]$ — so list every solution of $\cos(kx) = \text{value}$ in that wider range, not just the first one, then convert back to $x$. …

Common Mistakes

Mistake 1: Missing some of the critical points because the trig argument's range was not extended

Why it's wrong: here $2x$ ranges over $[0, 4\pi]$ as $x$ ranges over $[0, 2\pi]$ — so $\cos 2x = -\tfrac12$ has four solutions for $2x$ in that range, not the usual two you'd get from a basic reference-angle lookup. Missing solutions means missing candidate points, which can silently give the wrong max or min. Correct approach: always convert the interval for the original variable into the corresponding interval for the trig argument before listing solutions.

Mistake 2: Sign errors evaluating $\sin$ at second- and fourth-quadrant angles …