Let Q = (0, 2) and R = (9, 11) be given points in the plane. We want to find the point P = (x, 0) on the positive x-axis such that the sum of distances PQ + PR is as small as possible. (Before proceeding with this problem, draw a picture!) To solve this problem, we need to minimize the following function of x: f(z) = (x2+4)+(x2-18x + 202 X We find that f (x) has only one critical number in the interval at c = X 331 where f(x) has value 2 X Since this is smaller than the values of f(x) at the two endpoints, we conclude that this is the minimal sum of distances. Question Help: Message instructor