Restrict the domain of the function f so that the function is one-to-one and has an in- verse function. Then find the inverse function $f ^ { - 1 }$ . State the domains and ranges of f and $f ^ { - 1 }$ . $f ( x ) = ( x - 2 ) ^ { 2 }$
A) $f ^ { - 1 } ( x ) = \sqrt { x } - 2$
The domain of $f$ and the range of $f ^ { - 1 }$ are all real numbers $x$ such that $x \geq 2$.
The domain of $f ^ { - 1 }$ and the range of $f$ are all real numbers $x$ such that $x \geq 0$.
B) $f ^ { - 1 } ( x ) = \sqrt { x } + 2$
The domain of $f$ and the range of $f ^ { - 1 }$ are all real numbers $x$ such that $x \geq 0$.
The domain of $f ^ { - 1 }$ and the range of $f$ are all real numbers $x$ such that $x \geq - 2$.
C) $f ^ { - 1 } ( x ) = \sqrt { x } + 2$
The domain of $f$ and the range of $f ^ { - 1 }$ are all real numbers $x$ such that $x \geq 2$.
The domain of $f ^ { - 1 }$ and the range of $f$ are all real numbers $x$ such that $x \geq 0$.
D) $f ^ { - 1 } ( x ) = \sqrt { x } + 2$
The domain of $f$ and the range of $f ^ { - 1 }$ are all real numbers $x$ such that $x \geq 0$
The domain of $f ^ { - 1 }$ and the range of $f$ are all real numbers $x$ such that $x \geq 2$.
E) $f ^ { - 1 } ( x ) = \sqrt { x } - 2$
The domain of $f$ and the range of $f ^ { - 1 }$ are all real numbers $x$ such that $x \geq - 2$.
The domain of $f ^ { - 1 }$ and the range of $f$ are all real numbers $x$ such that $x \geq 0$.

Question

Quizplus · Accepted Answer

To find the inverse function, we set $f(x) = y$ and solve for $x$. 
\begin{align*}
y &= (x-2)^2 \
\sqrt{y} &= |x-2| \
x &= 2 + \sqrt{y} \text{ or } x = 2 - \sqrt{y}
\end{align*}
Since we want the inverse function to be one-to-one, we only choose one of the possible values for $x$. Therefore, 
$$f^{-1}(x) = \sqrt{x} + 2$$ 
The domain of $f$ is all real numbers, because we can plug in any number into the function. The range of $f^{-1}$ is all real numbers greater than or equal to 2, because the domain of $f$ was restricted to be greater than or equal to 2. 
The domain of $f^{-1}$ is all real numbers greater than or equal to 0, because the range of $f$ was restricted to be greater than or equal to 2. The range of $f$ is all real numbers greater than or equal to 0, because the function is a square, which can never be negative. 
Therefore, the best choice is C.

Restrict the Domain of the Function F So That the Function