Letx)=x2-2x on [-5, 1]. Use the IVT to determine if there is a solution to x) = 15 in the interval between -5 and 1. If 50, nd the value of c in the interval such that c) = 15. x) is continuous on [- 5, 1] O 15is between f(-5)=35 and H1): -1 c= -3 x) is continuous on [- 5, 1] O 15 is not between -5 and 1 so the IVT does not apply x) is continuous on [- 5, 1] O 15is between f(-5)=35 and H1): -1 c=5 x) is continuous on [- 5, 1] O 15is between f(-5)=35 and f(1)= -1 = - 3, c = 5