Q.A small town is analysing the pattern of a new street light installation. The lights are set up such that the intensity of light at any point $x$ metres from the start of the street can be modelled by $f(x) = e^{x}\sin x$, where $x$ is in metres.

Concept: Derivative Sign Analysis

We find $f'(x)$ and test its sign in $[0, \pi]$.

Step 1: Differentiate

Step 2: Critical points in $[0, \pi]$

Set $f'(x) = 0$. Since $e^x > 0$, we need $\sin x + \cos x = 0 \Rightarrow \tan x = -1$.

In $[0, \pi]$, this gives $x = \frac{3\pi}{4}$.

Step 3: Sign of $f'(x)$ on intervals

• For $0 \le x < \frac{3\pi}{4}$: $\sin x + \cos x > 0$ (e.g., at $x=0$, value $=1$), so $f'(x) > 0$ → $f$ is increasing.
• For $\frac{3\pi}{4} < x \le \pi$: $\sin x + \cos x < 0$ (e.g., at $x=\pi$, value $=-1$), so $f'(x) < 0$ → $f$ is decreasing. …

Why derivative sign analysis works

When you want to know where a function is rising or falling, you look at its slope — the derivative. If $f'(x) > 0$, the function is climbing; if $f'(x) < 0$, it's descending. The points where $f'(x) = 0$ (critical points) are where the function might pause and change direction — these are candidates for local maxima, minima, or inflection points.

For $f(x) = e^x \sin x$, the exponential factor $e^x$ is always positive, so the sign of $f'(x)$ depends entirely on the trigonometric part. That makes the analysis cleaner than it first appears.

1. Find the derivative

   Using the product rule:

Since $e^x > 0$ for all real $x$, the sign of $f'(x)$ is the same as the sign of $\sin x + \cos x$.

2. Solve $f'(x) = 0$ on $[0, \pi]$

   Set $\sin x + \cos x = 0$, which gives $\tan x = -1$.

   On $[0, \pi]$, the solution is:

This is the only critical point in the interval.

3. Test the sign of $f'(x)$ on either side

   Pick a test point in each subinterval:

   • For $x \in (0, \frac{3\pi}{4})$, try $x = \frac{\pi}{2}$:

     $\sin\frac{\pi}{2} + \cos\frac{\pi}{2} = 1 + 0 = 1 > 0$

     So $f'(x) > 0$ → $f$ is increasing on $(0, \frac{3\pi}{4})$.

   • For $x \in (\frac{3\pi}{4}, \pi)$, try $x = \pi$:

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

     So $f'(x) < 0$ → $f$ is decreasing on $(\frac{3\pi}{4}, \pi)$.

   Watch out

   A common mistake is to forget that $e^x$ is always positive and try to factor its sign into the analysis. Since $e^x > 0$, it never flips the sign — you can safely ignore it when determining monotonicity.

4. Classify the critical point $x = \frac{3\pi}{4}$

   The function changes from increasing to decreasing at this point. That is the classic behaviour of a local maximum. …