Let's work through the problem step-by-step to find the inverse function [tex]\( f^{-1}(x) \)[/tex] of the function [tex]\( f(x) = \sqrt{3 - x} + 1 \)[/tex] and also determine its domain.
### Step 1: Set up the initial equation
We start with the function:
[tex]\[ y = \sqrt{3 - x} + 1 \][/tex]
To find the inverse, we need to solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex].
### Step 2: Isolate the square root
First, we subtract 1 from both sides:
[tex]\[ y - 1 = \sqrt{3 - x} \][/tex]
### Step 3: Square both sides
Square both sides to eliminate the square root:
[tex]\[ (y - 1)^2 = 3 - x \][/tex]
### Step 4: Solve for [tex]\( x \)[/tex]
Rearrange the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ x = 3 - (y - 1)^2 \][/tex]
Thus, the inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ f^{-1}(x) = 3 - (x - 1)^2 \][/tex]
### Step 5: Determine the domain of the inverse function
The domain of the inverse function [tex]\( f^{-1}(x) \)[/tex] is derived from the range of the original function [tex]\( f(x) \)[/tex]. Since,
[tex]\[ f(x) = \sqrt{3 - x} + 1 \][/tex]
For [tex]\( f(x) \)[/tex] to be real, the argument of the square root must be non-negative:
[tex]\[ 3 - x \geq 0 \implies x \leq 3 \][/tex]
This tells us that the range of [tex]\( f(x) \)[/tex] is [tex]\( [1, \infty) \)[/tex] because:
[tex]\[ y = \sqrt{3 - x} + 1 \][/tex]
reaches a minimum value of 1 when [tex]\( x = 3 \)[/tex] and increases without bound as [tex]\( x \)[/tex] decreases.
Therefore, the domain of [tex]\( f^{-1}(x) \)[/tex] is [tex]\( [1, \infty) \)[/tex].
### Final Answer
Given this step-by-step process, we find that:
[tex]\[ f^{-1}(x) = 3 - (x - 1)^2 \][/tex]
and the domain of [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ [1, \infty) \][/tex]
Therefore:
[tex]\[ f^{-1}(x) = 3 - (x - 1)^2 \ \text{for the domain} \ [1, \infty) \][/tex]
Here is the formatted answer as requested:
[tex]\[ f^{-1}(x) = 3 - (x - 1)^2 \ \text{for the domain} \ [1, \infty) \][/tex]
