Find the range of `f(x)=(log)_2((sinx-cosx+3sqrt(2))/(sqrt(2)))`

Explanation

To find the range of the function \( f(x) = \log_2\left(\frac{\sin x - \cos x + 3\sqrt{2}}{\sqrt{2}}\right) \), we will follow these steps:

### Step 1: Determine the range of \( \sin x - \cos x \)

The expression \( \sin x - \cos x \) can be rewritten using the identity \( \sin x - \cos x = \sqrt{2}\sin\left(x - \frac{\pi}{4}\right) \). The range of \( \sin\left(x - \frac{\pi}{4}\right) \) is from -1 to 1. Therefore, the range of \( \sin x - \cos x \) is:

\[ -\sqrt{2} \leq \sin x - \cos x \leq \sqrt{2} \]

### Step 2: Shift the range by adding \( 3\sqrt{2} \)

Next, we add \( 3\sqrt{2} \) to the entire range:

\[ -\sqrt{2} + 3\sqrt{2} \leq \sin x - \cos x + 3\sqrt{2} \leq \sqrt{2} + 3\sqrt{2} \]

This simplifies to:

\[ 2\sqrt{2} \leq \sin x - \cos x + 3\sqrt{2} \leq 4\sqrt{2} \]

### Step 3: Divide the entire range by \( \sqrt{2} \)

Now, we divide the entire inequality by \( \sqrt{2} \):

\[ \frac{2\sqrt{2}}{\sqrt{2}} \leq \frac{\sin x - \cos x + 3\sqrt{2}}{\sqrt{2}} \leq \frac{4\sqrt{2}}{\sqrt{2}} \]

This simplifies to:

\[ 2 \leq \frac{\sin x - \cos x + 3\sqrt{2}}{\sqrt{2}} \leq 4 \]

### Step 4: Apply the logarithm

Now we take the logarithm base 2 of the entire inequality:

\[ \log_2(2) \leq \log_2\left(\frac{\sin x - \cos x + 3\sqrt{2}}{\sqrt{2}}\right) \leq \log_2(4) \]

This gives us:

\[ 1 \leq f(x) \leq 2 \]

### Conclusion: Range of \( f(x) \)

Thus, the range of \( f(x) \) is:

\[ [1, 2] \]