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 $x \underline{quadquad}$ and $y \underline{quadquad}$
options: $-$, $+$, $x^2 + 2$, $x^2 - 2$, $-x^2 + 2$, $-x^2 - 2$, $\sqrt{-x^2 + 4}$, $\sqrt{-x^2 - 4}$, $<$, $>$, $\leq$, $\geq$, $4$, $-2$, $0$, $2$, $4$

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 \( y=\sqrt{x - 2} \), where \( x\geq2 \).

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 \( x=\sqrt{y - 2} \) to get rid of the square root. Squaring both sides, we have \( x^{2}=y - 2 \). Then, add 2 to both sides to solve for \( y \). So, \( y=x^{2}+2 \). Thus, the inverse function \( f^{-1}(x)=x^{2}+2 \).

Step4: Determine the domain and range of the inverse

For the original function \( f(x)=\sqrt{x - 2} \) with \( x\geq2 \):

• The domain of \( f(x) \) is \( x\geq2 \), so the range of \( f(x) \) is \( y\geq0 \) (since the square root of a non - negative number is non - negative and when \( x = 2 \), \( y=0 \) and as \( x \) increases, \( y \) increases).
• For the inverse function \( f^{-1}(x) \), the domain of \( f^{-1}(x) \) is the range of \( f(x) \), so the domain of \( f^{-1}(x) \) is \( x\geq0 \).
• The range of \( f^{-1}(x) \) is the domain of \( f(x) \), so the range of \( f^{-1}(x) \) is \( y\geq2 \).

Answer:

\( f^{-1}(x)=x^{2}+2 \) for \( x\geq0 \) and \( y\geq2 \)