At ` x = (5 pi)/( 6)` , the function f (x) = 2 sin 3 x + 3 cos 3 x is

Options

A. maximum
B. minimum
C. zero
D. neither maximum nor minimum

Answer

To determine whether the function \( f(x) = 2 \sin(3x) + 3 \cos(3x) \) has a maximum, minimum, zero, or is neither at \( x = \frac{5\pi}{6} \), we will follow these steps:

### Step 1: Find the derivative of the function
To find the critical points, we first need to differentiate \( f(x) \).

\[ f'(x) = \frac{d}{dx}(2 \sin(3x) + 3 \cos(3x)) \]

Using the chain rule, we get:

\[ f'(x) = 2 \cdot 3 \cos(3x) - 3 \cdot 3 \sin(3x) = 6 \cos(3x) - 9 \sin(3x) \]

### Step 2: Evaluate the derivative at \( x = \frac{5\pi}{6} \)
Next, we will substitute \( x = \frac{5\pi}{6} \) into the derivative to check if it equals zero.

\[ f'\left(\frac{5\pi}{6}\right) = 6 \cos\left(3 \cdot \frac{5\pi}{6}\right) - 9 \sin\left(3 \cdot \frac{5\pi}{6}\right) \]

Calculating \( 3 \cdot \frac{5\pi}{6} = \frac{15\pi}{6} = \frac{5\pi}{2} \).

Now, we find the values of cosine and sine:

\[ \cos\left(\frac{5\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]
\[ \sin\left(\frac{5\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]

Substituting these values back into the derivative:

\[ f'\left(\frac{5\pi}{6}\right) = 6 \cdot 0 - 9 \cdot 1 = -9 \]

### Step 3: Determine the nature of the critical point
Since \( f'\left(\frac{5\pi}{6}\right) = -9 \), which is not equal to zero, \( x = \frac{5\pi}{6} \) is not a critical point.

### Conclusion
Since \( x = \frac{5\pi}{6} \) is not a critical point, it cannot be a maximum or minimum. Therefore, the answer is that \( f(x) \) at \( x = \frac{5\pi}{6} \) is neither maximum nor minimum.