Answer
\[\left( {{f}^{-1}} \right)'\left( 7 \right)=-\frac{1}{2}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=-{{x}^{2}}+8;\text{ }\left( 7,1 \right) \\
& \text{We have the point }\left( {{x}_{0}},{{y}_{0}} \right)\text{ and }f\left( x \right)=-{{x}^{2}}+8 \\
& \text{Use the formula of the theorem 3}\text{.23} \\
& \left( {{f}^{-1}} \right)'\left( {{y}_{0}} \right)=\frac{1}{f'\left( {{x}_{0}} \right)} \\
& \text{Where }{{x}_{0}}={{f}^{-1}}\left( {{y}_{0}} \right)\text{ for the point }\left( 7,1 \right)\Rightarrow {{x}_{0}}=1\text{ and }{{y}_{0}}=7 \\
& \left( {{f}^{-1}} \right)'\left( 7 \right)=\frac{1}{f'\left( 1 \right)} \\
& \text{Calculate }f'\left( 1 \right) \\
& f\left( x \right)=-{{x}^{2}}+8 \\
& f'\left( x \right)=-2x \\
& f'\left( 1 \right)=-2\left( 1 \right) \\
& f'\left( 1 \right)=-2 \\
& \text{Therefore,} \\
& \left( {{f}^{-1}} \right)'\left( 7 \right)=\frac{1}{f'\left( 1 \right)} \\
& \left( {{f}^{-1}} \right)'\left( 7 \right)=-\frac{1}{2} \\
\end{align}\]
