Question

Show that \(f\) and \(g\) are inverse functions (a analytically and (b) graphically. $$ f(x)=1-x^{3}, \quad g(x)=\sqrt[3]{1-x} $$

Short answer

Based on the analytical and graphical check, the functions \(f\) and \(g\) are indeed inverse to each other.

Step-by-step solution

Analytical Proof part 1: \(f(g(x))=x\)

First substitute \(g(x)\) into \(f(x)\) to find \(f(g(x))\). This comes out to be \(f(g(x))=f(\sqrt[3]{1-x})=1-(\sqrt[3]{1-x})^3\). This simplifies to \(f(g(x))=1-(1-x)=x\). So, \(f(g(x))=x\).

Analytical Proof part 2: \(g(f(x))=x\)

Now substitute \(f(x)\) into \(g(x)\) to find \(g(f(x))\). This comes out to be \(g(f(x))=g(1-x^{3})=\sqrt[3]{1-(1-x^3)}\). This simplifies to \(g(f(x))=\sqrt[3]{x^3}=x\). So, \(g(f(x))=x\).

Graphical Proof: Draw \(f(x)\) and \(g(x)\)

Draw both \(f(x)\) and \(g(x)\) on the same graph and include the line \(y = x\). Check if \(f(x)\) is a mirror image of \(g(x)\) across the line \(y = x\). If yes, then \(f\) and \(g\) are inverse functions.