Sue is a runner. At time t = 0 she runs down a steep hill, turns around and runs back up it. At time t = 1 she is in the same place she started. Assume that her elevation (let the bottom of the hill be elevation 0), is f(x) = (x - 0.5|, [0, 1]. Can Rolle's Theorem be applied to show that at some point between t = 0 and t = 1, Sue has achieved a maximum and/or minimum speed? O The conditions of Rolle's Theorem are met and Rolle's Theorem does not state that you can find min or max points. O The conditions of Rolle's Theorem are not met but Rolle's Theorem does not state that you can find min or max points. O The conditions of Rolle's Theorem are not met. The conditions of Rolle's Theorem are met and Rolle's Theorem does state that you can find min or max points.Let f(ac) = 23 - 4x. Using Rolle's Theorem, what are all the values c in the interval [-2, 2] such that f'(c) = 0? O c= = 2V3 3 O c = 0.5 and c = -0.5 O c= 12V3 O c = 1 and c = 2For f(a) = - Vac + 2, find all the values of c on the interval [2, 7] that satisfy the Mean Value Theorem. O C=-4.75 O C=-4.25 O c = 4.25 O C = 4.75For f(ac ) = 2ac2 - ac + 1, find all the values of c on the interval [0, 1] that satisfy the Mean Value Theorem. O C= -0.5 O C= 0.375 O C = 0.5 O C= 0.68For at) 2 (c2 + 3 |:r:|does the Mean Value Theorem apply on the interval [-2, 2] and why? 0 The Mean Value Theorem does not apply because f[x) is not continuous on the closed interval [-2, 2]. O The Mean Value Theorem does not apply because f(x) is not differentiable on the closed interval [-2, 2]. O The Mean Value Theorem applies because f(x) is continuous on [-2, 2]. O The Mean Value Theorem applies because, f(x) is differentiable on the closed interval [-2, 2]