Local Min Given Local Max

Any local maximum/minimum will occur when $f'(x) = 1 - \frac{a}{x^{2}} = 0$ , i.e., when $a = x^{2}$ .

Thus the local maximum occurs when $2x + 1 = -3 \longrightarrow x = -2$ . This tells us that $a = (-2)^{2} = 4$ .

This means that our function is $f(x) = x + 1 + \frac{4}{x}$ , and that any local critical points occur when $x^{2} = 4$ . Since the local maximum occurs when $x = -2$ we are left with $x = 2$ , at which $f(2) = 5$ .

This at first seems paradoxical, having the local minimum greater than the local maximum, but a quick examination of the behavior of $f(x)$ shows us that it has two branches, one entirely in the first quadrant and one entirely in the fourth quadrant. The branch in the first quadrant has a minimum at $(2,5)$ , and the branch in the fourth quadrant has a maximum at $(-2,-3)$ .

Thus with $a = 4$ and $b = 5$ we have the solution $a + b = \boxed{9}$ .