Answer
\[\frac{1}{4}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)={{x}^{3}}+x+1 \\
& \text{Differentiate} \\
& f'\left( x \right)=3{{x}^{2}}+1 \\
& \text{Use the Derivative of the Inverse Function}\left( \text{Theorem 3}\text{.23} \right) \\
& \left( {{f}^{-1}} \right)'\left( {{y}_{0}} \right)=\frac{1}{f'\left( {{x}_{0}} \right)},\text{ where }{{y}_{0}}=f\left( {{x}_{0}} \right) \\
& \text{We need to find }\left( {{f}^{-1}} \right)'\left( 3 \right)\Rightarrow {{y}_{0}}=3,\text{ therefore} \\
& f\left( {{x}_{0}} \right)=3 \\
& \text{Evaluating }f\left( 1 \right) \\
& f\left( 1 \right)={{1}^{3}}+1+1=3\Rightarrow {{x}_{0}}=1 \\
& \left( {{f}^{-1}} \right)'\left( {{y}_{0}} \right)=\frac{1}{f'\left( {{x}_{0}} \right)}\Rightarrow \left( {{f}^{-1}} \right)'\left( 3 \right)=\frac{1}{f'\left( 1 \right)} \\
& \left( {{f}^{-1}} \right)'\left( 3 \right)=\frac{1}{3{{\left( 1 \right)}^{2}}+1} \\
& \left( {{f}^{-1}} \right)'\left( 3 \right)=\frac{1}{4} \\
\end{align}\]
