Question

Show that \(f\) and \(g\) are inverse functions (a analytically and (b) graphically. $$ f(x)=\frac{1}{1+x}, \quad x \geq 0 ; \quad g(x)=\frac{1-x}{x}, \quad 0

Short answer

By plugging \(g(x)\) into \(f(x)\) and \(f(x)\) into \(g(x)\), and simplifying both to get \(x\), we showed analytically that \(f(x)\) and \(g(x)\) are inverses of each other. Visually, the graph of \(f(x)\) and \(g(x)\) would be mirror images about the line \(f(x) = g(x)\).

Step-by-step solution

Analytically Verify \(f(g(x)) = x\)

First, substitute \(g(x)\) into \(f(x)\), so that we get \(f(g(x)) = f(\frac{1-x}{x})\). If we substitute \(g(x)\) into \(f(x)\), we get \(f(\frac{1-x}{x})= \frac{1}{1+((1-x)/x)} = \frac{x}{1+1-x} = x\), which simplifies to \(x\).

Analytically Verify \(g(f(x)) = x\)

Now, substitute \(f(x)\) into \(g(x)\), that is, \(g(f(x)) = g(\frac{1}{1+x})\). This gives us \(g(\frac{1}{1+x}) = \frac{1-\frac{1}{1+x}}{\frac{1}{1+x}} = \frac{1+x-1}{1} = x\), which simplifies to \(x\). Thus, since \(f(g(x)) = x\) and \(g(f(x)) = x\), we have confirmed analytically that \(f(x)\) and \(g(x)\) are inverse functions of each other.

Graphical Verification

Graphically, if \(f(x)\) and \(g(x)\) are inverse functions their graphs will be mirror images of each other about the line \(y=x\). This can be verified by using a graphing tool or software. It's crucial throughout the range of x where both functions are defined. The reflection property should hold.