`f(x)=sinx+cosx+3`. find the range of f(x).

Answer

To find the range of the function \( f(x) = \sin x + \cos x + 3 \), we can follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \sin x + \cos x + 3 \]

### Step 2: Find the maximum and minimum of \( \sin x + \cos x \)
To find the range of \( f(x) \), we first need to determine the range of \( \sin x + \cos x \). We can rewrite \( \sin x + \cos x \) using the identity:
\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]
This transformation helps us understand the amplitude of the function.

### Step 3: Determine the maximum and minimum values
The maximum value of \( \sin\left(x + \frac{\pi}{4}\right) \) is \( 1 \) and the minimum value is \( -1 \). Therefore:
\[ \text{Maximum of } \sin x + \cos x = \sqrt{2} \cdot 1 = \sqrt{2} \]
\[ \text{Minimum of } \sin x + \cos x = \sqrt{2} \cdot (-1) = -\sqrt{2} \]

### Step 4: Add 3 to the maximum and minimum
Now, we add 3 to both the maximum and minimum values of \( \sin x + \cos x \):
\[ \text{Maximum of } f(x) = \sqrt{2} + 3 \]
\[ \text{Minimum of } f(x) = -\sqrt{2} + 3 \]

### Step 5: Write the range of \( f(x) \)
Thus, the range of \( f(x) \) is:
\[ [-\sqrt{2} + 3, \sqrt{2} + 3] \]

### Final Answer
The range of \( f(x) = \sin x + \cos x + 3 \) is:
\[ [3 - \sqrt{2}, 3 + \sqrt{2}] \]

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