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Consider the function f(x)= square root of x-1+9 for the domain [1, infinity)

Find f^-1(x) where f^-1 is the inverse of f
Also state the domain of f^-1 in interval notation.

User Gieun
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Final answer:

To find the inverse function of f(x) = \sqrt{x-1} + 9, we swap x and y, isolate the square root, square both sides, and solve for y, leading to f^{-1}(x) = (x - 9)^2 + 1. The domain of the inverse function is [9, \infty).

Step-by-step explanation:

To find the inverse of the function f(x) = \sqrt{x-1} + 9 for the domain [1, \infty), we need to follow several steps. First, swap x and y to begin solving for y. Our new equation to solve is x = \sqrt{y-1} + 9. Next, we isolate the square root term by subtracting 9 from both sides, getting x - 9 = \sqrt{y-1}. Afterward, we square both sides of the equation to remove the square root, resulting in (x - 9)^2 = y - 1. Finally, add 1 to both sides to solve for y, yielding f^{-1}(x) = (x - 9)^2 + 1.

The domain of the inverse function corresponds to the range of the original function. Since \sqrt{x} is always non-negative, and we add 9, the smallest value f(x) can take is 9. Thus, the domain of f^{-1}(x) will start at 9 and continue to infinity, represented as [9, \infty) in interval notation.

User Kalls
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