Answer :
To find the inverse [tex]\( f^{-1}(x) \)[/tex] of the function [tex]\( f(x) = \sqrt{x-4} + 2 \)[/tex], follow these steps:
1. Start with the function:
[tex]\[ f(x) = \sqrt{x-4} + 2 \][/tex]
2. Interchange [tex]\( x \)[/tex] and [tex]\( f(x) \)[/tex] and solve for [tex]\( x \)[/tex]:
Let [tex]\( y = f(x) \)[/tex]. Then the function becomes:
[tex]\[ y = \sqrt{x-4} + 2 \][/tex]
Interchanging [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = \sqrt{y-4} + 2 \][/tex]
3. Isolate the square root term:
[tex]\[ x - 2 = \sqrt{y-4} \][/tex]
4. Square both sides to eliminate the square root:
[tex]\[ (x - 2)^2 = y - 4 \][/tex]
5. Solve for [tex]\( y \)[/tex]:
[tex]\[ y = (x - 2)^2 + 4 \][/tex]
Therefore, the inverse function is:
[tex]\[ f^{-1}(x) = (x - 2)^2 + 4 \][/tex]
6. Determine the domain of [tex]\( f^{-1}(x) \)[/tex]:
The original function [tex]\( f(x) = \sqrt{x-4} + 2 \)[/tex] is defined for [tex]\( x \geq 4 \)[/tex]. The function [tex]\( f(x) \)[/tex] outputs values starting from 2 and increasing without bound, i.e., the range of [tex]\( f(x) \)[/tex] is [tex]\([2, \infty)\)[/tex].
Since the range of [tex]\( f(x) \)[/tex] becomes the domain of [tex]\( f^{-1}(x) \)[/tex]:
The domain of [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ [2, \infty) \][/tex]
To summarize:
[tex]\[ f^{-1}(x) = (x - 2)^2 + 4 \quad \text{with the domain} \quad [2, \infty) \][/tex]
1. Start with the function:
[tex]\[ f(x) = \sqrt{x-4} + 2 \][/tex]
2. Interchange [tex]\( x \)[/tex] and [tex]\( f(x) \)[/tex] and solve for [tex]\( x \)[/tex]:
Let [tex]\( y = f(x) \)[/tex]. Then the function becomes:
[tex]\[ y = \sqrt{x-4} + 2 \][/tex]
Interchanging [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = \sqrt{y-4} + 2 \][/tex]
3. Isolate the square root term:
[tex]\[ x - 2 = \sqrt{y-4} \][/tex]
4. Square both sides to eliminate the square root:
[tex]\[ (x - 2)^2 = y - 4 \][/tex]
5. Solve for [tex]\( y \)[/tex]:
[tex]\[ y = (x - 2)^2 + 4 \][/tex]
Therefore, the inverse function is:
[tex]\[ f^{-1}(x) = (x - 2)^2 + 4 \][/tex]
6. Determine the domain of [tex]\( f^{-1}(x) \)[/tex]:
The original function [tex]\( f(x) = \sqrt{x-4} + 2 \)[/tex] is defined for [tex]\( x \geq 4 \)[/tex]. The function [tex]\( f(x) \)[/tex] outputs values starting from 2 and increasing without bound, i.e., the range of [tex]\( f(x) \)[/tex] is [tex]\([2, \infty)\)[/tex].
Since the range of [tex]\( f(x) \)[/tex] becomes the domain of [tex]\( f^{-1}(x) \)[/tex]:
The domain of [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ [2, \infty) \][/tex]
To summarize:
[tex]\[ f^{-1}(x) = (x - 2)^2 + 4 \quad \text{with the domain} \quad [2, \infty) \][/tex]
