Show that the equation x3 - 14x + c = 0 has at most one solution in the interval [-2, 2]. Let f(x) = x3. -Select- Now f'(r)= 14x + c for x in [-2, 2]. If f has two real solutions a and b in [-2, 2], with a < b, then ---Select-- implies that there is a number r in (a, b) such that f'(r) = 0. . Since the polynomial fis continuous on [a, b] and differentiable on (a, b), Since r is in (a, b), which is contained in [-2, 2], we have Irl < 2, so r2 < 4. It follows that 3r - 14 73-4-14=-270. This contradicts f'(r) = 0, so the gi equation cannot have two real solutions in [-2,2]. Hence, it has at most one real solution in [-2, 2].