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

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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
+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
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### 1+0 Answers

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

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

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

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

heureka  Aug 30, 2017

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