Question

find the inverse of the following function. also, list the domain and range restriction for the inverse.
$f(x) = \sqrt{x - 2}$, for $x \geq 2$
$f^{-1}(x) = \underline{quadquad}$ for $x \underline{quadquad}$
options: $x$, $x + $, $x^2 + 2$, $x^2 - 2$, $-x^2 + 2$, $-x^2 - 2$, $\sqrt{-x^2 + 4}$, $\sqrt{-x^2 - 4}$, $<$, $>$, $\leq$, $\geq$, $-4$, $-2$, $0$, $2$, $4$

Explanation:

Step1: Replace \( f(x) \) with \( y \)

We start with the function \( f(x)=\sqrt{x - 2} \) for \( x\geq2 \). Replace \( f(x) \) with \( y \), so we have \( y=\sqrt{x - 2} \).

Step2: Swap \( x \) and \( y \)

To find the inverse, we swap the roles of \( x \) and \( y \). This gives us \( x=\sqrt{y - 2} \).

Step3: Solve for \( y \)

First, square both sides of the equation to eliminate the square root: \( x^{2}=y - 2 \). Then, add 2 to both sides to solve for \( y \): \( y=x^{2}+2 \). So the inverse function \( f^{-1}(x)=x^{2}+2 \).

Step4: Determine domain and range of inverse

For the original function \( f(x)=\sqrt{x - 2} \) with \( x\geq2 \), the range of \( f(x) \) is \( y\geq0 \) (since the square root of a non - negative number is non - negative). The domain of the inverse function \( f^{-1}(x) \) is the range of the original function, so the domain of \( f^{-1}(x) \) is \( x\geq0 \). The range of the inverse function \( f^{-1}(x) \) is the domain of the original function, so the range of \( f^{-1}(x) \) is \( y\geq2 \).

Answer:

The inverse function \( f^{-1}(x)=x^{2}+2 \). The domain of \( f^{-1}(x) \) is \( x\geq0 \) and the range of \( f^{-1}(x) \) is \( y\geq2 \).