The maximum value of f(x) `=2 sin x + sin 2x`, in the interval `[0, (3)/(2)pi]` is

Options

A. `sqrt2 + 1`
B. `2 sqrt3`
C. `(3 sqrt3)/(2)`
D. `sqrt3`

Answer

To find the maximum value of the function \( f(x) = 2 \sin x + \sin 2x \) in the interval \([0, \frac{3\pi}{2}]\), we will follow these steps:

### Step 1: Find the derivative of \( f(x) \)
We start by differentiating the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(2 \sin x + \sin 2x) \]
Using the derivative rules, we have:
\[ f'(x) = 2 \cos x + 2 \cos 2x \]
Using the double angle formula, \( \cos 2x = 2 \cos^2 x - 1 \), we can rewrite \( f'(x) \):
\[ f'(x) = 2 \cos x + 2(2 \cos^2 x - 1) = 2 \cos x + 4 \cos^2 x - 2 \]
Thus, we can simplify this to:
\[ f'(x) = 4 \cos^2 x + 2 \cos x - 2 \]

### Step 2: Set the derivative equal to zero
To find critical points, we set \( f'(x) = 0 \):
\[ 4 \cos^2 x + 2 \cos x - 2 = 0 \]
Dividing the entire equation by 2 gives:
\[ 2 \cos^2 x + \cos x - 1 = 0 \]
Let \( u = \cos x \). The equation becomes:
\[ 2u^2 + u - 1 = 0 \]

### Step 3: Solve the quadratic equation
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ u = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \]
This gives us two solutions:
\[ u = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad u = \frac{-4}{4} = -1 \]

### Step 4: Find the corresponding \( x \) values
1. For \( \cos x = \frac{1}{2} \):
\[ x = \frac{\pi}{3}, \quad \text{(in the interval [0, \( \frac{3\pi}{2} \)])} \]
2. For \( \cos x = -1 \):
\[ x = \pi \]

### Step 5: Evaluate \( f(x) \) at critical points and endpoints
Now we will evaluate \( f(x) \) at the critical points and the endpoints of the interval:
1. \( f(0) = 2 \sin 0 + \sin 0 = 0 \)
2. \( f\left(\frac{\pi}{3}\right) = 2 \sin\left(\frac{\pi}{3}\right) + \sin\left(\frac{2\pi}{3}\right) = 2 \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = 3\frac{\sqrt{3}}{2} \)
3. \( f(\pi) = 2 \sin \pi + \sin 2\pi = 0 \)
4. \( f\left(\frac{3\pi}{2}\right) = 2 \sin\left(\frac{3\pi}{2}\right) + \sin(3\pi) = 2(-1) + 0 = -2 \)

### Step 6: Determine the maximum value
Now we compare the values:
- \( f(0) = 0 \)
- \( f\left(\frac{\pi}{3}\right) = 3\frac{\sqrt{3}}{2} \)
- \( f(\pi) = 0 \)
- \( f\left(\frac{3\pi}{2}\right) = -2 \)

The maximum value occurs at \( x = \frac{\pi}{3} \):
\[ \text{Maximum value} = 3\frac{\sqrt{3}}{2} \]

### Final Answer
The maximum value of \( f(x) \) in the interval \([0, \frac{3\pi}{2}]\) is \( \frac{3\sqrt{3}}{2} \).