Fig. 2

An example where \(g_e(x)=-0.125x^2+2x\). We want to find a \((1+\epsilon )\)-approximation for \(g_e(x)\) everywhere in the range [0, 8] where we set \(\epsilon =0.1\). The last linear piece \(f_3\) has a zero slope and passes through (8, 8), thus \(f_3(x)=8\). The first linear piece \(f_1\) is tangent to \(g_e\) at the origin, hence \(f_1(x)=2x\). At the point \(x=1.4545\), \(f_1(x)=(1+\epsilon )g_e(x)\), meaning that \(f_1\) provides a qualified approximation for \(g_e(x)\) in the range [0, 1.4545]. The second linear piece \(f_2\) passes through (1.4545, 2.909) and is tangent to \(g_e\) at (2.909, 4.7602), namely \(f_2(x)=1.2727x+1.0579\). At \(x=5.2894\), \(f_2(x)=(1+\epsilon )g_e(x)\), hence \(f_2\) provides a qualified approximation for \(g_e(x)\) in the range [1.4545, 5.2894]. The rest of the points in the range [5.2894, 8] have been covered by \(f_3\)