+0

# Given that $f(2)=5$ and $f^{-1}(x+4)=2f^{-1}(x)+1$ for all $x$, find $f^{-1}(17)$.

+2
39
1
+193

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

#1
+18564
+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:
$$\begin{array}{|rcll|} \hline y &=&f(x) \\ x &=& f^{-1}(y) \\ \hline \end{array}$$

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

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

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

heureka  Aug 30, 2017
Sort:

#1
+18564
+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:
$$\begin{array}{|rcll|} \hline y &=&f(x) \\ x &=& f^{-1}(y) \\ \hline \end{array}$$

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

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

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

heureka  Aug 30, 2017

### 8 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details