Answer
\[\frac{3}{2}\]
Work Step by Step
\[\begin{align}
& \text{We have that the slope of the function }y=f\left( x \right)\text{ at the point} \\
& \left( 7,4 \right)\text{ is }\frac{2}{3},\text{ which implies that} \\
& m=f'\left( x \right)=\frac{2}{3}\text{ at the point }\left( 7,4 \right) \\
& f'\left( 7 \right)=\frac{2}{3} \\
& \text{Use the Derivative of the Inverse Function}\left( \text{Theorem 3}\text{.23} \right) \\
& \left( {{f}^{-1}} \right)'\left( {{y}_{0}} \right)=\frac{1}{f'\left( {{x}_{0}} \right)},\text{ where }{{y}_{0}}=f\left( {{x}_{0}} \right) \\
& \text{Calculating }\left( {{f}^{-1}} \right)'\left( x \right)\text{ at }\left( 4,7 \right),\text{ then }\left( {{f}^{-1}} \right)'\left( 4 \right) \\
& \left( {{f}^{-1}} \right)'\left( 4 \right)=\frac{1}{f'\left( {{x}_{0}} \right)}=\frac{1}{f'\left( 7 \right)} \\
& \text{Substituting }f'\left( 7 \right)=\frac{2}{3} \\
& \left( {{f}^{-1}} \right)'\left( 4 \right)=\frac{1}{f'\left( {{x}_{0}} \right)}=\frac{1}{2/3} \\
& \left( {{f}^{-1}} \right)'\left( 4 \right)=\frac{3}{2} \\
\end{align}\]
