Suppose that $f(x)=\frac{1}{2x+b}$. For what value of $b$ does $f^{-1}(x)=\frac{1-2x}{2x}$?
+26367 Suppose that $f(x)=\frac{1}{2x+b}$. For what value of $b$ does $f^{-1}(x)=\frac{1-2x}{2x}$?
\(\begin{array}{|rcll|} \hline f(x) &=& \dfrac{1}{2x+b} \\ \hline y &=& \dfrac{1}{2x+b} \\\\ \dfrac{1}{y} &=& 2x+b \\\\ 2x &=& \dfrac{1}{y}-b \\\\ x &=& \dfrac{\dfrac{1}{y}-b}{2} \quad & | \quad x \leftrightarrow y \\\\ y &=& \dfrac{\dfrac{1}{x}-b}{2} \\\\ f^{-1}(x) &=& \dfrac{\dfrac{1}{x}-b}{2} \\ \hline f^{-1}(x) = \dfrac{\dfrac{1}{x}-b}{2} &=& \dfrac{1-2x}{2x} \\\\ \dfrac{\dfrac{1}{x}-b}{2} &=& \dfrac{1-2x}{2x} \\\\ \dfrac{1}{x}-b &=& \dfrac{1-2x}{x} \\\\ \dfrac{1}{x}-b &=& \dfrac{1}{x}-2 \\\\ -b &=& -2 \\\\ \mathbf{b} & \mathbf{=}& \mathbf{2} \\ \hline \end{array}\)
