+619 Given that $f(2)=5$ and $f^{-1}(x+4)=2f^{-1}(x)+1$ for all $x$, find $f^{-1}(17)$.
f^(-1) is the inverse of f
+26393 Given that
\(f(2)=5\)
and \(f^{-1}(x+4)=2f^{-1}(x)+1\)
for all \(x\),
find \(f^{-1}(17)\).
\(f^{-1}\) is the inverse of f
Formula:
\(\begin{array}{|rcll|} \hline y &=&f(x) \\ x &=& f^{-1}(y) \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline f(2) &=& 5 \quad & | \quad x= 2 \qquad y = 5 \\ x &=& f^{-1}(y) \\ 2 &=& f^{-1}(5) \\\\ \mathbf{f^{-1}(5)} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline (1) & f^{-1}(17) &=& 2f^{-1}(13) + 1 \quad & | \quad 17 = x + 4 \qquad x = 13 \\ (2) & f^{-1}(13) &=& 2f^{-1}(9) + 1 \quad & | \quad 13 = x + 4 \qquad x = 9 \\ (3) & f^{-1}(9) &=& 2f^{-1}(5) + 1 \quad & | \quad 9 = x + 4 \qquad x = 5 \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline (3) & f^{-1}(9) &=& 2f^{-1}(5) + 1 \quad & | \quad \mathbf{f^{-1}(5)= 2} \\ & f^{-1}(9) &=& 2\cdot 2 + 1 \\ & f^{-1}(9) &=& 5 \\\\ (2) & f^{-1}(13) &=& 2f^{-1}(9) + 1 \quad & | \quad \mathbf{f^{-1}(9)= 5} \\ & f^{-1}(13) &=& 2\cdot 5 + 1 \\ & f^{-1}(13) &=& 11 \\\\ (1) & f^{-1}(17) &=& 2f^{-1}(13) + 1 \quad & | \quad \mathbf{f^{-1}(13)= 11} \\ & f^{-1}(17) &=& 2\cdot 11 + 1 \\ & \mathbf{f^{-1}(17)} &\mathbf{=}& \mathbf{23} \\ \hline \end{array}\)
+26393 Given that
\(f(2)=5\)
and \(f^{-1}(x+4)=2f^{-1}(x)+1\)
for all \(x\),
find \(f^{-1}(17)\).
\(f^{-1}\) is the inverse of f
Formula:
\(\begin{array}{|rcll|} \hline y &=&f(x) \\ x &=& f^{-1}(y) \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline f(2) &=& 5 \quad & | \quad x= 2 \qquad y = 5 \\ x &=& f^{-1}(y) \\ 2 &=& f^{-1}(5) \\\\ \mathbf{f^{-1}(5)} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline (1) & f^{-1}(17) &=& 2f^{-1}(13) + 1 \quad & | \quad 17 = x + 4 \qquad x = 13 \\ (2) & f^{-1}(13) &=& 2f^{-1}(9) + 1 \quad & | \quad 13 = x + 4 \qquad x = 9 \\ (3) & f^{-1}(9) &=& 2f^{-1}(5) + 1 \quad & | \quad 9 = x + 4 \qquad x = 5 \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline (3) & f^{-1}(9) &=& 2f^{-1}(5) + 1 \quad & | \quad \mathbf{f^{-1}(5)= 2} \\ & f^{-1}(9) &=& 2\cdot 2 + 1 \\ & f^{-1}(9) &=& 5 \\\\ (2) & f^{-1}(13) &=& 2f^{-1}(9) + 1 \quad & | \quad \mathbf{f^{-1}(9)= 5} \\ & f^{-1}(13) &=& 2\cdot 5 + 1 \\ & f^{-1}(13) &=& 11 \\\\ (1) & f^{-1}(17) &=& 2f^{-1}(13) + 1 \quad & | \quad \mathbf{f^{-1}(13)= 11} \\ & f^{-1}(17) &=& 2\cdot 11 + 1 \\ & \mathbf{f^{-1}(17)} &\mathbf{=}& \mathbf{23} \\ \hline \end{array}\)
