Q.Find the intervals in which the function $f$ given by $f(x) = \sin x + \cos x,\ 0 \le x \le 2\pi$ is increasing or decreasing.

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

Step 1: Differentiate.

$f'(x) = \cos x - \sin x$.

Step 2: Set $f'(x) = 0$ to find critical points.

$\cos x - \sin x = 0 \implies \tan x = 1$.

In $[0, 2\pi]$, this gives $x = \frac{\pi}{4},\ \frac{5\pi}{4}$.

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

• For $0 \le x < \frac{\pi}{4}$: $\cos x > \sin x \implies f'(x) > 0$ → increasing.
• For $\frac{\pi}{4} < x < \frac{5\pi}{4}$: $\cos x < \sin x \implies f'(x) < 0$ → decreasing. …

To decide where a function is increasing or decreasing, we look at the sign of its derivative. If $f'(x) > 0$, the function is rising; if $f'(x) < 0$, it is falling. The key here is to find where $f'(x)$ changes sign, which happens at its zeros.

1. Find the derivative.

   $f(x) = \sin x + \cos x$

   $f'(x) = \cos x - \sin x$

2. Set the derivative to zero to find critical points.

   $\cos x - \sin x = 0 \implies \cos x = \sin x$

   This happens when $\tan x = 1$. Within $0 \le x \le 2\pi$, the solutions are:

   $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.

   These two points divide the interval $[0, 2\pi]$ into three subintervals:

   $[0, \frac{\pi}{4})$, $(\frac{\pi}{4}, \frac{5\pi}{4})$, and $(\frac{5\pi}{4}, 2\pi]$.

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

   Pick a convenient test point inside each interval and evaluate $f'(x) = \cos x - \sin x$.

   • Interval $[0, \frac{\pi}{4})$: Choose $x = 0$.

     $f'(0) = \cos 0 - \sin 0 = 1 - 0 = 1 > 0$

     So $f$ is increasing here.

   • Interval $(\frac{\pi}{4}, \frac{5\pi}{4})$: Choose $x = \pi$.

     $f'(\pi) = \cos \pi - \sin \pi = -1 - 0 = -1 < 0$

     So $f$ is decreasing here.

   • Interval $(\frac{5\pi}{4}, 2\pi]$: Choose $x = 2\pi$.

     $f'(2\pi) = \cos 2\pi - \sin 2\pi = 1 - 0 = 1 > 0$

     So $f$ is increasing here.

Method: Finding Intervals of Increase/Decrease for a Sum of Sine and Cosine

Applies to any $f(x) = a\sin x + b\cos x$ type function on a given interval.

Steps

Step 1: Differentiate term by term.

Step 2: Solve $f'(x) = 0$ by converting to a single trig ratio.

Rearranging $a\cos x = b \sin x$ gives $\tan x = a/b$ (when $b \ne 0$) — remember that $\tan x = c$ has two solutions in any interval of length $2\pi$ (one in each of two "opposite" quadrants where tangent has the matching sign), not just one. Solve for all of them within the given domain.

Step 3: Use the critical points to split the domain into sub-intervals, and test the sign of $f'(x)$ in each. …

Common Mistakes

Mistake 1: Finding only one solution of $\tan x = 1$ in $[0, 2\pi]$ instead of both.

Why it's wrong: $\tan x = 1$ has two solutions in any full period of $2\pi$ — $x=\pi/4$ and $x = 5\pi/4$ — because tangent has the same sign in two opposite quadrants; stopping at $\pi/4$ alone misses an entire increasing sub-interval at the end of the domain. Correct approach: always find every solution of a trig equation within the given domain, remembering the periodicity of $\tan x$ is $\pi$, not $2\pi$.

Mistake 2: Mis-locating $\pi/4$ and $5\pi/4$ relative to $\pi$ when choosing test points. …