## Precalculus (6th Edition) Blitzer

The value of x which satisfies the equation $8+{{f}^{-1}}\left( x-1 \right)=10$ is $x=7$.
$8+{{f}^{-1}}\left( x-1 \right)=10$ Solve the equation further, \begin{align} & {{f}^{-1}}\left( x-1 \right)=10-8 \\ & {{f}^{-1}}\left( x-1 \right)=2 \\ \end{align} (I) Also, $f\left( 2 \right)=6$ Take the inverse of the above function ${{f}^{-1}}\left( f\left( 2 \right) \right)={{f}^{-1}}\left( 6 \right)$ As ${{f}^{-1}}\left( f\left( x \right) \right)=x$, therefore${{f}^{-1}}\left( f\left( 2 \right) \right)=2$ And, $2={{f}^{-1}}\left( 6 \right)$ From (1), ${{f}^{-1}}\left( x-1 \right)=2$ And, $2={{f}^{-1}}\left( 6 \right)$ So, ${{f}^{-1}}\left( x-1 \right)={{f}^{-1}}\left( 6 \right)$ Now compare both sides, \begin{align} & x-1=6 \\ & x=1+6 \\ & x=7 \end{align} Thus, the value of x = 7 satisfies $8+{{f}^{-1}}\left( x-1 \right)=10$.