The range of the function `f(x)=3|sin x|-2|cos x|` is :

Options

A. `[-2 sqrt13]`
B. [-2,3]
C. `[3,sqrt13]`
D. `[-3,2]`

Answer

To find the range of the function \( f(x) = 3|\sin x| - 2|\cos x| \), we will analyze the function step by step.

### Step 1: Understand the behavior of the absolute values
The function consists of absolute values of sine and cosine functions. The values of \( |\sin x| \) and \( |\cos x| \) both range from 0 to 1. Therefore, we can express the function as:
\[ f(x) = 3|\sin x| - 2|\cos x| \]
where \( |\sin x| \) can take values from 0 to 1 and \( |\cos x| \) can also take values from 0 to 1.

### Step 2: Analyze the critical points
To find the maximum and minimum values of \( f(x) \), we can analyze the function at the critical points where \( |\sin x| \) and \( |\cos x| \) achieve their extreme values.

1. **When \( |\sin x| = 1 \) and \( |\cos x| = 0 \)**:
\[ f(x) = 3(1) - 2(0) = 3 \]

2. **When \( |\sin x| = 0 \) and \( |\cos x| = 1 \)**:
\[ f(x) = 3(0) - 2(1) = -2 \]

### Step 3: Consider intermediate values
Next, we need to consider the cases when \( |\sin x| \) and \( |\cos x| \) take values between 0 and 1. Since both functions are periodic, we can analyze the function over one period, say \( x \in [0, 2\pi] \).

### Step 4: Find the maximum and minimum values
From the calculations above:
- The maximum value of \( f(x) \) occurs when \( |\sin x| = 1 \) (which happens at \( x = \frac{\pi}{2} + k\pi \)), giving \( f(x) = 3 \).
- The minimum value of \( f(x) \) occurs when \( |\cos x| = 1 \) (which happens at \( x = k\pi \)), giving \( f(x) = -2 \).

### Step 5: Conclusion
Thus, the range of the function \( f(x) = 3|\sin x| - 2|\cos x| \) is:
\[ [-2, 3] \]

### Final Answer
The range of the function \( f(x) = 3|\sin x| - 2|\cos x| \) is \([-2, 3]\).

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