+0  
 
0
37
2
avatar

\(Suppose that $f(x)$ is a linear function satisfying the equation $f(x) = 4f^{-1}(x) + 6$. Given that $f(1) = 4$, find $f(2)$.\)

Guest Apr 10, 2018

Best Answer 

 #1
avatar+19207 
+1

Suppose that $f(x)$ is a linear function satisfying the equation $f(x) = 4f^{-1}(x) + 6$. Given that $f(1) = 4$, find $f(2)$.

 

 

\(\text{1. $f(x)$ is a linear function}\)

\(\begin{array}{|lrcll|} \hline f(x) =& ax +b && \quad & | \quad f(1) = 4 \\\\ f(1) =& a\cdot1 +b &=& 4 \\ & \mathbf{a +b} &\mathbf{=}& \mathbf{4 } \qquad &(1) \\ & \mathbf{a } &\mathbf{=}& \mathbf{4-b } \\ \hline \end{array} \)

 

\(\text{2. $f^{-1}(x)=\ ? $ }\)

\(\begin{array}{|rcll|} \hline f(x) &=& ax +b \\ y &=& ax +b \\ ax &=& y-b \quad & | \quad : a \\ x &=& \dfrac{y-b}{a} \quad & | \quad x \leftrightarrow y \\ y &=& \dfrac{x-b}{a} \\ \mathbf{f^{-1}(x)} &\mathbf{=}& \mathbf{\dfrac{x-b}{a}} \\ \hline \end{array}\)

 

 

\(\text{3. $f(x) = 4f^{-1}(x) + 6$ }\)

\(\begin{array}{|rcll|} \hline f(x) &=& 4f^{-1}(x) + 6 \quad & | \quad x = 1 \\\\ f(1) &=& 4f^{-1}(1) + 6 \quad & | \quad f(1) = 4 \qquad f^{-1}(1) = \dfrac{1-b}{a} \\ 4 &=& 4 \left(\dfrac{1-b}{a} \right) + 6 \quad & | \quad - 6 \\ -2 &=& 4 \left(\dfrac{1-b}{a} \right) \quad & | \quad \cdot a \\ -2a &=& 4 (1-b) \\ -2a &=& 4-4b \quad & | \quad +4b \\ 4b-2a &=& 4 \quad & | \quad : 2 \\ \mathbf{2b-a} & \mathbf{=}& \mathbf{2} \qquad &(2) \\ \hline \end{array}\)

 

 

\(\text{4. $a=\ ? \qquad b=\ ?$ }\)

\(\begin{array}{|rcll|} \hline 2b-a & = & 2 \quad & | \quad a = 4-b \\ 2b-(4-b) & = & 2 \\ 2b-4+b & = & 2 \\ 3b-4 & = & 2 \quad & | \quad +4 \\ 3b & = & 6 \quad & | \quad :3 \\ \mathbf{ b} & \mathbf{=} & \mathbf{2} \\\\ a &=& 4-b \\ a &=& 4-2 \\ \mathbf{ a} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

 

 

\(\text{5. $f(x)=\ ?$ }\)

\(\begin{array}{|rcll|} \hline f(x) &=& ax +b \quad & | \quad a=2 \qquad b = 2 \\\\ \mathbf{ f(x)} & \mathbf{=} & \mathbf{2x +2} \\ \hline \end{array}\)

 

 

\(\text{6. $f(2)=\ ?$ } \)

\(\begin{array}{|rcll|} \hline \mathbf{ f(x)} & \mathbf{=} & \mathbf{2x +2} \quad & | \quad x = 2 \\\\ f(2) & = & 2\cdot 2 +2 \\ \mathbf{ f(2)} & \mathbf{=} & \mathbf{6} \\ \hline \end{array}\)

 

 

laugh

heureka  Apr 11, 2018
Sort: 

2+0 Answers

 #1
avatar+19207 
+1
Best Answer

Suppose that $f(x)$ is a linear function satisfying the equation $f(x) = 4f^{-1}(x) + 6$. Given that $f(1) = 4$, find $f(2)$.

 

 

\(\text{1. $f(x)$ is a linear function}\)

\(\begin{array}{|lrcll|} \hline f(x) =& ax +b && \quad & | \quad f(1) = 4 \\\\ f(1) =& a\cdot1 +b &=& 4 \\ & \mathbf{a +b} &\mathbf{=}& \mathbf{4 } \qquad &(1) \\ & \mathbf{a } &\mathbf{=}& \mathbf{4-b } \\ \hline \end{array} \)

 

\(\text{2. $f^{-1}(x)=\ ? $ }\)

\(\begin{array}{|rcll|} \hline f(x) &=& ax +b \\ y &=& ax +b \\ ax &=& y-b \quad & | \quad : a \\ x &=& \dfrac{y-b}{a} \quad & | \quad x \leftrightarrow y \\ y &=& \dfrac{x-b}{a} \\ \mathbf{f^{-1}(x)} &\mathbf{=}& \mathbf{\dfrac{x-b}{a}} \\ \hline \end{array}\)

 

 

\(\text{3. $f(x) = 4f^{-1}(x) + 6$ }\)

\(\begin{array}{|rcll|} \hline f(x) &=& 4f^{-1}(x) + 6 \quad & | \quad x = 1 \\\\ f(1) &=& 4f^{-1}(1) + 6 \quad & | \quad f(1) = 4 \qquad f^{-1}(1) = \dfrac{1-b}{a} \\ 4 &=& 4 \left(\dfrac{1-b}{a} \right) + 6 \quad & | \quad - 6 \\ -2 &=& 4 \left(\dfrac{1-b}{a} \right) \quad & | \quad \cdot a \\ -2a &=& 4 (1-b) \\ -2a &=& 4-4b \quad & | \quad +4b \\ 4b-2a &=& 4 \quad & | \quad : 2 \\ \mathbf{2b-a} & \mathbf{=}& \mathbf{2} \qquad &(2) \\ \hline \end{array}\)

 

 

\(\text{4. $a=\ ? \qquad b=\ ?$ }\)

\(\begin{array}{|rcll|} \hline 2b-a & = & 2 \quad & | \quad a = 4-b \\ 2b-(4-b) & = & 2 \\ 2b-4+b & = & 2 \\ 3b-4 & = & 2 \quad & | \quad +4 \\ 3b & = & 6 \quad & | \quad :3 \\ \mathbf{ b} & \mathbf{=} & \mathbf{2} \\\\ a &=& 4-b \\ a &=& 4-2 \\ \mathbf{ a} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

 

 

\(\text{5. $f(x)=\ ?$ }\)

\(\begin{array}{|rcll|} \hline f(x) &=& ax +b \quad & | \quad a=2 \qquad b = 2 \\\\ \mathbf{ f(x)} & \mathbf{=} & \mathbf{2x +2} \\ \hline \end{array}\)

 

 

\(\text{6. $f(2)=\ ?$ } \)

\(\begin{array}{|rcll|} \hline \mathbf{ f(x)} & \mathbf{=} & \mathbf{2x +2} \quad & | \quad x = 2 \\\\ f(2) & = & 2\cdot 2 +2 \\ \mathbf{ f(2)} & \mathbf{=} & \mathbf{6} \\ \hline \end{array}\)

 

 

laugh

heureka  Apr 11, 2018
 #2
avatar+85726 
+1

Nice, heureka !!!

 

cool cool cool

CPhill  Apr 11, 2018

16 Online Users

avatar
avatar
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