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]
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]
