\[
f(x)=60.89-x^{2}
\]
The lead engineer is Ashton.
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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: \( f^{-1}(x)= \) \( \square \)
B) Follow all course work guidelines and note any other specific directions below. (6089)
You must include work and/or an explanation of why you chose your answer in order to receive ful
According to Ashton, which of the following must be true for \( f(x) \) to have an inverse?
\( x \geq 60.89 \)
\( x \geq 0 \)
\( x \leq 60.89 \)
\( -60.89
Ask by Ellis Tran. in the United States
Jan 23,2025
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The Deep Dive
To find the inverse of the function \( f(x) = 60.89 - x^2 \), we need to switch \( x \) and \( y \) in the equation and solve for \( y \). Start by rewriting the equation as \( y = 60.89 - x^2 \). Now swap \( x \) and \( y \) to get \( x = 60.89 - y^2 \). Rearranging gives \( y^2 = 60.89 - x \), and taking the square root results in \( y = \sqrt{60.89 - x} \) (considering the positive root since the domain must be restricted). Thus, the inverse is \( f^{-1}(x) = \sqrt{60.89 - x} \). For a function to have an inverse, it must be one-to-one, meaning it must not repeat any values. In this case, since \( f(x) = 60.89 - x^2 \) is a downward-opening parabola, it is only one-to-one when \( x \) is restricted to either half: either \( x \leq \sqrt{60.89} \) or \( x \geq -\sqrt{60.89} \). However, a more conventional choice is \( x \leq \sqrt{60.89} \) to ensure all values are included without repeating, isolating the upper portion of the parabola for the inverse.