\[ f(x)=60.89-x^{2} \] The lead engineer is Ashton. This is a quiz or exam question and is © 2024 MF-LM-JMM. Sharing, posting, copying, saving (et problem in any way or sending/sharing it to any app/website/Al-tool is a violation of the Honor Co A) Follow all course work guidelines and note any other specific directions below. (6090) What did Ashton find to be the inverse? (FZHI) You must show all steps of your algebraic work to receive full credit. Answer: \( f^{-1}(x)=\sqrt{60.89-x} \)

When tackling the function \( f(x)=60.89-x^{2} \) to find its inverse, you start by replacing \( f(x) \) with \( y \), giving you \( y = 60.89 - x^{2} \). Next, rearranging the equation for \( x \) involves solving for \( x \) in terms of \( y \). You first isolate the \( x^{2} \) term: \( x^{2} = 60.89 - y \). Then, by taking the square root of both sides, you get an expression for \( x \): \( x = \sqrt{60.89 - y} \). Thus, the final inverse function is expressed as \( f^{-1}(x)=\sqrt{60.89-x} \). When finding the inverse, pay close attention to the domain and range. Since \( f(x) \) is a downward-opening parabola, its range is limited to \( (-\infty, 60.89] \). Therefore, the inverse \( f^{-1}(x) \) is only valid when \( x \) is in \( [0, 60.89] \) because of the square root. A common mistake is neglecting to specify these conditions, which can lead to confusion or incorrect conclusions about the behavior of the function and its inverse. So remember to always clarify the domains when dealing with square roots!