Q.At $x = \dfrac{5\pi}{6}$, $f(x) = 2\sin 3x + 3\cos 3x$ is:
(A) maximum
(B) minimum
(C) zero
(D) neither maximum nor minimum

To classify a point as a maximum or minimum, first check whether the derivative is zero there. If $f'\neq 0$, the point cannot be an extremum.

Step 1 — Differentiate. $f(x)=2\sin 3x+3\cos 3x$, so

Step 2 — Evaluate at $x=\frac{5\pi}{6}$. Here $3x=\frac{5\pi}{2}=2\pi+\frac{\pi}{2}$, so $\cos\frac{5\pi}{2}=0$ and $\sin\frac{5\pi}{2}=1$: …

The idea

A smooth function can only have a local maximum or minimum where its derivative is zero. So the first thing to test is whether $f'$ vanishes at the given point. If $f'\neq 0$ there, the graph is still climbing or falling through the point and it cannot be a turning point.

Set up

Work the steps

1. Differentiate (chain rule, since the angle is $3x$):

2. Plug in $x=\frac{5\pi}{6}$, so $3x=\frac{5\pi}{2}$. Since $\frac{5\pi}{2}=2\pi+\frac{\pi}{2}$,

Therefore

$f'\left(\tfrac{5\pi}{6}\right)=6(0)-9(1)=-9.$ …

Method: Testing Whether a Given Point Is a Local Extremum

Use this whenever a question gives a specific $x$-value and asks whether the function is at a maximum, minimum, zero, or neither there.

Steps

Step 1: Differentiate the function.

Find $f'(x)$ using the standard rules, including the chain rule for any composite arguments like $kx$.

Step 2: Evaluate the derivative at the given point.

Substitute the specified $x$-value into $f'(x)$ and simplify, using known values or periodicity of trig functions as needed to reduce the angle to a standard reference angle.

Step 3: Check whether $f'(x_0)=0$ — this is the necessary first test. …

Common Mistakes

Mistake 1: Jumping straight to classifying max/min without first checking whether the derivative is even zero.

Why it's wrong: a point can only be a local extremum where $f'(x)=0$; testing anything else at a point where $f'\ne 0$ is meaningless and leads to a wrong conclusion. Correct approach: always compute $f'(x_0)$ first and confirm it equals zero before applying any further classification test.

Mistake 2: Mis-reducing the angle when it goes beyond $2\pi$ or involves a multiple angle like $3x$.

Why it's wrong: forgetting to subtract off full rotations ($2\pi$) before evaluating $\sin$ or $\cos$ at a large angle leads to the wrong reference-angle value and a wrong derivative sign. Correct approach: always reduce the angle modulo $2\pi$ (here $\tfrac{5\pi}{2}=2\pi+\tfrac{\pi}{2}$) before reading off the standard sine/cosine value. …