+0  
 
+2
331
1
avatar+598 

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

michaelcai  Aug 29, 2017

Best Answer 

 #1
avatar+19639 
+1

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:
\(\begin{array}{|rcll|} \hline y &=&f(x) \\ x &=& f^{-1}(y) \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline f(2) &=& 5 \quad & | \quad x= 2 \qquad y = 5 \\ x &=& f^{-1}(y) \\ 2 &=& f^{-1}(5) \\\\ \mathbf{f^{-1}(5)} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1) & f^{-1}(17) &=& 2f^{-1}(13) + 1 \quad & | \quad 17 = x + 4 \qquad x = 13 \\ (2) & f^{-1}(13) &=& 2f^{-1}(9) + 1 \quad & | \quad 13 = x + 4 \qquad x = 9 \\ (3) & f^{-1}(9) &=& 2f^{-1}(5) + 1 \quad & | \quad 9 = x + 4 \qquad x = 5 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline (3) & f^{-1}(9) &=& 2f^{-1}(5) + 1 \quad & | \quad \mathbf{f^{-1}(5)= 2} \\ & f^{-1}(9) &=& 2\cdot 2 + 1 \\ & f^{-1}(9) &=& 5 \\\\ (2) & f^{-1}(13) &=& 2f^{-1}(9) + 1 \quad & | \quad \mathbf{f^{-1}(9)= 5} \\ & f^{-1}(13) &=& 2\cdot 5 + 1 \\ & f^{-1}(13) &=& 11 \\\\ (1) & f^{-1}(17) &=& 2f^{-1}(13) + 1 \quad & | \quad \mathbf{f^{-1}(13)= 11} \\ & f^{-1}(17) &=& 2\cdot 11 + 1 \\ & \mathbf{f^{-1}(17)} &\mathbf{=}& \mathbf{23} \\ \hline \end{array}\)

 

laugh

heureka  Aug 30, 2017
 #1
avatar+19639 
+1
Best Answer

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:
\(\begin{array}{|rcll|} \hline y &=&f(x) \\ x &=& f^{-1}(y) \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline f(2) &=& 5 \quad & | \quad x= 2 \qquad y = 5 \\ x &=& f^{-1}(y) \\ 2 &=& f^{-1}(5) \\\\ \mathbf{f^{-1}(5)} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1) & f^{-1}(17) &=& 2f^{-1}(13) + 1 \quad & | \quad 17 = x + 4 \qquad x = 13 \\ (2) & f^{-1}(13) &=& 2f^{-1}(9) + 1 \quad & | \quad 13 = x + 4 \qquad x = 9 \\ (3) & f^{-1}(9) &=& 2f^{-1}(5) + 1 \quad & | \quad 9 = x + 4 \qquad x = 5 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline (3) & f^{-1}(9) &=& 2f^{-1}(5) + 1 \quad & | \quad \mathbf{f^{-1}(5)= 2} \\ & f^{-1}(9) &=& 2\cdot 2 + 1 \\ & f^{-1}(9) &=& 5 \\\\ (2) & f^{-1}(13) &=& 2f^{-1}(9) + 1 \quad & | \quad \mathbf{f^{-1}(9)= 5} \\ & f^{-1}(13) &=& 2\cdot 5 + 1 \\ & f^{-1}(13) &=& 11 \\\\ (1) & f^{-1}(17) &=& 2f^{-1}(13) + 1 \quad & | \quad \mathbf{f^{-1}(13)= 11} \\ & f^{-1}(17) &=& 2\cdot 11 + 1 \\ & \mathbf{f^{-1}(17)} &\mathbf{=}& \mathbf{23} \\ \hline \end{array}\)

 

laugh

heureka  Aug 30, 2017

17 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.