Let \(f: \mathbb{R} \to \mathbb{R}\) be the function \(f(x) = x^3 - x\) . By restricting the domain and range of \(f\) appropriately, obtain from \(f\) a bijective function \(g\) . Draw the graphs of \(g\) and \(g^{-1}\) . (There are several possible choices for \(g\) .)
Solution
Put \(c=1/\sqrt3\), so that \(f(c)=c^3-c=\frac1{3\sqrt3}-\frac1{\sqrt3}=-\frac2{3\sqrt3}\).
1. \(f\) is strictly increasing on \([c,\infty)\).
\(f'(x)=3x^2-1>0\) for \(x>c\).
2. \(f([c,\infty))=[-\frac2{3\sqrt3},\infty)\).
By step 1, \(f(x)\ge f(c)\) for \(x\ge c\). Since \(f\) is continuous and \(f(x)\to\infty\) as \(x\to\infty\), the intermediate value theorem gives every value \(y\ge f(c)\).
3. Q.E.D.
By steps 1 and 2, \[ \boxed{g\colon[1/\sqrt3,\infty)\to[-2/(3\sqrt3),\infty),\quad g(x)=x^3-x} \] is injective and surjective. Its graph is the part of the cubic \(y=x^3-x\) to the right of the local minimum \((c,f(c))\), and the graph of \(g^{-1}\) is its reflection in the line \(y=x\).