To solve the given problem, we need to analyze the function \( f(x) = \sin x - \cos x + 3\sqrt{2} \) and determine its range, as well as the corresponding domain for its inverse function. Here’s a step-by-step solution:
### Step 1: Identify the function and its components
The function is given as:
\[
f(x) = \sin x - \cos x + 3\sqrt{2}
\]
We need to find the range of \( f(x) \).
### Step 2: Find the range of \( \sin x - \cos x \)
The expression \( \sin x - \cos x \) can be rewritten using the formula for the range of \( a \sin x + b \cos x \):
\[
\text{Range of } (a \sin x + b \cos x) = \left[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}\right]
\]
Here, \( a = 1 \) and \( b = -1 \):
\[
\text{Range of } \sin x - \cos x = \left[-\sqrt{1^2 + (-1)^2}, \sqrt{1^2 + (-1)^2}\right] = \left[-\sqrt{2}, \sqrt{2}\right]
\]
### Step 3: Adjust the range for the entire function \( f(x) \)
Now, we add \( 3\sqrt{2} \) to the range of \( \sin x - \cos x \):
\[
\text{Range of } f(x) = \left[-\sqrt{2} + 3\sqrt{2}, \sqrt{2} + 3\sqrt{2}\right]
\]
Calculating the endpoints:
- Lower bound: \( -\sqrt{2} + 3\sqrt{2} = 2\sqrt{2} \)
- Upper bound: \( \sqrt{2} + 3\sqrt{2} = 4\sqrt{2} \)
Thus, the range of \( f(x) \) is:
\[
\text{Range of } f(x) = [2\sqrt{2}, 4\sqrt{2}]
\]
### Step 4: Determine the domain of the inverse function
Since \( f \) is an invertible function, the range of \( f(x) \) becomes the domain of the inverse function \( f^{-1}(x) \):
\[
\text{Domain of } f^{-1}(x) = [2\sqrt{2}, 4\sqrt{2}]
\]
### Step 5: Find the range of the inverse function
To find the range of \( f^{-1}(x) \), we need to analyze the original function \( f(x) \) and its behavior:
1. The domain of \( f(x) \) is all real numbers \( \mathbb{R} \).
2. The range of \( f^{-1}(x) \) corresponds to the domain of \( f(x) \).
### Step 6: Determine the range of \( f^{-1}(x) \)
The function \( f(x) \) is continuous and differentiable, and since it is invertible, we can find the critical points to determine the range. The range of \( f^{-1}(x) \) will be the interval where \( x \) takes values in the domain of \( f \).
### Step 7: Finalize the ranges
The range of \( f^{-1}(x) \) is determined to be:
\[
\text{Range of } f^{-1}(x) = \left(-\frac{\pi}{4}, \frac{3\pi}{4}\right)
\]
### Conclusion
Thus, we conclude:
- The range of \( f(x) \) is \( [2\sqrt{2}, 4\sqrt{2}] \).
- The range of \( f^{-1}(x) \) is \( \left(-\frac{\pi}{4}, \frac{3\pi}{4}\right) \).
