Suppose that $f(x)=\frac{1}{2x+b}$. For what value of $b$ does $f^{-1}(x)=\frac{1-2x}{2x}$?
I need to find the inverse function of $f(x)$ and then compare it with the given expression for $f^{-1}(x)$. To find the inverse function, I can switch the roles of $x$ and $y$ and then solve for $y$. So, if $y=f(x)=\frac{1}{2x+b}$, then $x=\frac{1}{2y+b}$. To solve for $y$, I can cross-multiply and get $x(2y+b)=1$. Then, I can expand and move the terms with $y$ to one side and the terms without $y$ to the other side. I get $2xy+bx=1$ and then $2xy=1-bx$. Next, I can divide both sides by $2x$ to get $y=\frac{1-bx}{2x}$. This is the inverse function of $f(x)$. Now, I can compare it with the given expression for $f^{-1}(x)$ and see what value of $b$ makes them equal. I have $\frac{1-bx}{2x}=\frac{1-2x}{2x}$. To simplify this equation, I can multiply both sides by $2x$ and get $1-bx=1-2x$. Then, I can subtract $1$ from both sides and get $-bx=-2x$. Next, I can divide both sides by $-x$ and get $b=2$. This is the value of $b$ that makes $f^{-1}(x)$ match the given expression. 