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Define the function $g(x)=3x+2$. If $g(x)=2f^{-1}(x)$ and $f^{-1}(x)$ is the inverse of the function $f(x)=ax+b$, find $\dfrac{a+b}{2}$.

 Mar 21, 2018

Best Answer 

 #1
avatar+20857 
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Define the function $g(x)=3x+2$. If $g(x)=2f^{-1}(x)$ and $f^{-1}(x)$ is the inverse of the function $f(x)=ax+b$, find $\dfrac{a+b}{2}$.

\(\begin{array}{|rcll|} \hline f(x) &=& ax+b \\ \hline y &=& ax+b \\\\ ax &=& y-b \\\\ x &=& \dfrac{y-b}{a} \quad & | \quad x \leftrightarrow y \\\\ y &=& \dfrac{x-b}{a} \\\\ \mathbf{f^{-1}(x)} &\mathbf{=}& \mathbf{\dfrac{x-b}{a}} \\ \hline g(x) &=& 3x+2 \\ g(x) &=& 2f^{-1}(x) \\ 3x+2 &=& 2\left( \dfrac{x-b}{a} \right) \\\\ \dfrac{3x+2}{2} &=& \dfrac{x-b}{a} \\\\ \mathbf{\dfrac{3}{2}x+1} &\mathbf{=}& \mathbf{\dfrac{1}{a}x-\dfrac{b}{a}} \quad & | \quad \text{compare} \\\\ \hline \dfrac{3}{2} &=& \dfrac{1}{a} \\ \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{2}{3}} \\ \hline 1 &=& -\dfrac{b}{a} \\ \dfrac{b}{a} &=& -1 \\ b &=& -a \\ \mathbf{b} &\mathbf{=}& \mathbf{-\dfrac{2}{3}} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \dfrac{a+b}{2} &=& \dfrac{ \dfrac{2}{3}-\dfrac{2}{3} }{2}\\\\ \mathbf{\dfrac{a+b}{2}} &\mathbf{=}& \mathbf{0} \\ \hline \end{array}\)

 

laugh

 Mar 22, 2018
 #1
avatar+20857 
+2
Best Answer

Define the function $g(x)=3x+2$. If $g(x)=2f^{-1}(x)$ and $f^{-1}(x)$ is the inverse of the function $f(x)=ax+b$, find $\dfrac{a+b}{2}$.

\(\begin{array}{|rcll|} \hline f(x) &=& ax+b \\ \hline y &=& ax+b \\\\ ax &=& y-b \\\\ x &=& \dfrac{y-b}{a} \quad & | \quad x \leftrightarrow y \\\\ y &=& \dfrac{x-b}{a} \\\\ \mathbf{f^{-1}(x)} &\mathbf{=}& \mathbf{\dfrac{x-b}{a}} \\ \hline g(x) &=& 3x+2 \\ g(x) &=& 2f^{-1}(x) \\ 3x+2 &=& 2\left( \dfrac{x-b}{a} \right) \\\\ \dfrac{3x+2}{2} &=& \dfrac{x-b}{a} \\\\ \mathbf{\dfrac{3}{2}x+1} &\mathbf{=}& \mathbf{\dfrac{1}{a}x-\dfrac{b}{a}} \quad & | \quad \text{compare} \\\\ \hline \dfrac{3}{2} &=& \dfrac{1}{a} \\ \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{2}{3}} \\ \hline 1 &=& -\dfrac{b}{a} \\ \dfrac{b}{a} &=& -1 \\ b &=& -a \\ \mathbf{b} &\mathbf{=}& \mathbf{-\dfrac{2}{3}} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \dfrac{a+b}{2} &=& \dfrac{ \dfrac{2}{3}-\dfrac{2}{3} }{2}\\\\ \mathbf{\dfrac{a+b}{2}} &\mathbf{=}& \mathbf{0} \\ \hline \end{array}\)

 

laugh

heureka Mar 22, 2018

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