AP Calculus AB Mean Value Theorem Problem with Solution Today I will provide a solution for yesterday’s AP Calculus AB Mean Value Theorem Problem. Once again here is the question: Let f(x) = –3x2 + x – 5. A value of c that satisfies the conclusion of the Mean Value Theorem for f on the interval [–2,2] is (A) –2 (B) –1/2 (C) –1/6 (D) 0 Solution: f ‘ (x)= –6x + 1, so that f ‘ (c) = –6c + 1. f(2) = -3(2)2 + 2 – 5 = –3(4) – 3 = –12 – 3 = –15. f(–2) = –3(–2)2 – 2 – 5 = –3(4) – 7 = –12 – 7 = –19. We now solve the equation f ‘ (c) = (f(2) – f(–2)) / (2 – (–2)) to get –6c + 1 = (–15 – (–19)) / (2 + 2) = (–15 + 19) / 4 = 4 / 4 = 1 So –6c = 1 – 1 = 0, and c=0, choice (D). Notes: (1) The Mean Value Theorem says that if f is a function that is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there is a real number c with a < c < b such that f ‘ (c) = (f(b) – f(a)) / (b – a). In this problem a = –2, b = 2, and f(x) = –3x2 + x – 5. (2) Note that the function f(x) = –3x2 + x – 5 is a polynomial. It is therefore continuous and differentiable everywhere. In particular, it is continuous on [–2,2] and differentiable on (–2,2). Therefore by the Mean Value Theorem, there is a c between –2 and 2 such that f ‘ (c) = (f(2) – f(–2)) / (2 – (–2)). (3) Geometrically, f ‘ (c) is the slope of the tangent line to the graph of the function f at c, and (f(b) – f(a)) / (b – a) is the slope of the secant line passing through the points A(a,f(a)), and B(b,f(b)). The Mean Value Theorem says that the slopes of these two lines are the same. In other words, they are parallel. More AP Calculus Problems with Explanations If you are preparing for one of the AP Calculus exams, you may want to take a look at one of the following books. And if you liked this article, please share it with your Facebook friends: Comments comments