Let f(x) = x-2x on [-5, 1]. Use the IVT to determine if there is a solution to f(x) = 15 in the interval between -5 and 1. If so, find the value of c in the interval such that f(c) = 15. O f(x) is continuous on [-5, 1] 15 is not between -5 and 1 so the IVT does not apply f(x) is continuous on [-5, 1] 15 is between f(-5)= 35 and f(1) = -1 c=5 f(x) is continuous on [-5, 1] 15 is between f(-5)= 35 and f(1) = -1 c=-3, c=5 O f(x) is continuous on [-5, 1] 15 is between f(-5)= 35 and f(1) = -1 c = -3