I don't understand how they proved the double root:

Consider r(x) = - sin x - 5x cos x over (-5, 5). Its derivative is r'(x) = 5 cosx - 5 (cos x - x sin x) = 5x sin x. Solving r'(x) = xsin x =0 in the interval (-5, 5) gives three critical points -T, 0, T. In fact, the derivative r' (x) = >x sin x has simple root at x = -7 and 7, and a double y =r(x) root at x = 0. The later is because -5 5 r' (x) sin x -1 lim N/- lim 22 = 370. Since sin x is positive for -5