Question 7 of 8 Complete the statements for the given function and its inverse. \[ f(x)=x^{2}+17, x \geq 0 \] The inverse function of \( f \) is \( f^{-1}(x)=\sqrt{x-17} \). The graph of \( f \) and the graph of \( f^{-1} \) are symmetric with respect to the \( \square \) line \( y=x \). The domain of \( f^{-1} \) is \( \{x \mid x \geq 17\} \), and the range of \( f^{-1} \) is \( \square \) \[ \{y \mid y \in R\} . \] \[ \{y \mid y \leq 17\} . \] \[ \{y \mid y \leq 0\} \] \[ |y| y \geq 01 . \] \[ \{y \mid y \geq 17\} . \]

To complete the statements regarding the function \( f(x) = x^2 + 17 \) and its inverse \( f^{-1}(x) = \sqrt{x - 17} \), let's analyze the properties of both functions step by step. 1. **Symmetry with respect to the line \( y = x \)**: - The graphs of a function and its inverse are symmetric with respect to the line \( y = x \). Therefore, we can fill in the first blank: \[ \text{The graph of } f \text{ and the graph of } f^{-1} \text{ are symmetric with respect to the } \textbf{line } y = x. \] 2. **Domain of \( f^{-1} \)**: - The inverse function \( f^{-1}(x) = \sqrt{x - 17} \) is defined when the expression under the square root is non-negative. Thus, we have: \[ x - 17 \geq 0 \implies x \geq 17. \] - Therefore, the domain of \( f^{-1} \) is \( \{x \mid x \geq 17\} \), which is already given. 3. **Range of \( f^{-1} \)**: - The output of the inverse function \( f^{-1}(x) = \sqrt{x - 17} \) will yield non-negative values since the square root function produces non-negative results. Therefore, the range of \( f^{-1} \) is: \[ \{y \mid y \geq 0\}. \] Now, let's summarize the completed statements: - The graph of \( f \) and the graph of \( f^{-1} \) are symmetric with respect to the **line** \( y = x \). - The domain of \( f^{-1} \) is \( \{x \mid x \geq 17\} \), and the range of \( f^{-1} \) is \( \{y \mid y \geq 0\} \). Thus, the correct options for the blanks are: - For the first blank: **line** - For the second blank: **\{y \mid y \geq 0\}**.