The maximum value of for all real values of x is

1/2 1 2 3

3

Let $S = S’ (x) = –2 = –2$ $S’ (x) = 0 x = 1$. Since $S’ (x) > 0$ for $x < -1$ and $S’ (x) < 0$ for $–1 < x < 1$ and $S’ (x) > 0$ for $x > 1$. So S is maximum when $x = – 1$. Hence $S_{max} = 1 + 2/1 = 3$.