Consider the function [tex]f(x)=\sqrt{x-5}+1[/tex] for the domain [tex][5, \infty)[/tex].

1. Find [tex]f^{-1}(x)[/tex], where [tex]f^{-1}[/tex] is the inverse of [tex]f[/tex].
2. State the domain of [tex]f^{-1}[/tex] in interval notation.

[tex]f^{-1}(x) = \square[/tex] for the domain [tex]\square[/tex].



Answer :

To find the inverse of the function [tex]\( f(x) = \sqrt{x-5} + 1 \)[/tex], we will follow a detailed, step-by-step approach:

1. Express the function in terms of [tex]\( y \)[/tex] instead of [tex]\( f(x) \)[/tex]:
[tex]\[ y = \sqrt{x-5} + 1 \][/tex]

2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to begin solving for the inverse function:
[tex]\[ x = \sqrt{y-5} + 1 \][/tex]

3. Isolate the square root term:
[tex]\[ x - 1 = \sqrt{y-5} \][/tex]

4. Square both sides to remove the square root:
[tex]\[ (x - 1)^2 = y - 5 \][/tex]

5. Solve for [tex]\( y \)[/tex]:
[tex]\[ y = (x - 1)^2 + 5 \][/tex]

Therefore, the inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ f^{-1}(x) = (x - 1)^2 + 5 \][/tex]

Next, we need to state the domain of the inverse function. The given function [tex]\( f(x) = \sqrt{x-5} + 1 \)[/tex] is defined for [tex]\( x \geq 5 \)[/tex]. This means that for [tex]\( f^{-1}(x) \)[/tex] to be valid, the output of [tex]\( f(x) \)[/tex], which becomes the input for [tex]\( f^{-1}(x) \)[/tex], must be in the range of [tex]\( f \)[/tex].

Since [tex]\( \sqrt{x-5} \geq 0 \)[/tex], adding 1 implies that:
[tex]\[ f(x) = \sqrt{x-5} + 1 \geq 1 \][/tex]

Thus, the domain of the inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ [1, \infty) \][/tex]

In summary:
[tex]\[ f^{-1}(x) = (x - 1)^2 + 5 \quad \text{for the domain} \quad [1, \infty) \][/tex]