We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
419
1
avatar

Suppose that $f(x)=\frac{1}{2x+b}$. For what value of $b$ does $f^{-1}(x)=\frac{1-2x}{2x}$?

 Mar 21, 2018
 #1
avatar+22152 
+1

Suppose that $f(x)=\frac{1}{2x+b}$. For what value of $b$ does $f^{-1}(x)=\frac{1-2x}{2x}$?

 

\(\begin{array}{|rcll|} \hline f(x) &=& \dfrac{1}{2x+b} \\ \hline y &=& \dfrac{1}{2x+b} \\\\ \dfrac{1}{y} &=& 2x+b \\\\ 2x &=& \dfrac{1}{y}-b \\\\ x &=& \dfrac{\dfrac{1}{y}-b}{2} \quad & | \quad x \leftrightarrow y \\\\ y &=& \dfrac{\dfrac{1}{x}-b}{2} \\\\ f^{-1}(x) &=& \dfrac{\dfrac{1}{x}-b}{2} \\ \hline f^{-1}(x) = \dfrac{\dfrac{1}{x}-b}{2} &=& \dfrac{1-2x}{2x} \\\\ \dfrac{\dfrac{1}{x}-b}{2} &=& \dfrac{1-2x}{2x} \\\\ \dfrac{1}{x}-b &=& \dfrac{1-2x}{x} \\\\ \dfrac{1}{x}-b &=& \dfrac{1}{x}-2 \\\\ -b &=& -2 \\\\ \mathbf{b} & \mathbf{=}& \mathbf{2} \\ \hline \end{array}\)

 

 

laugh

 Mar 21, 2018

34 Online Users

avatar
avatar
avatar
avatar