1. Name: PerngTa Wu Teacher: Ms.Delacruz Date: 12/1/07 Subject: Math Period: 5-6

3. Proof The statement of the theorem Suppose f is a function which is continuous on the closed interval [a,b] and differentiable on the open interval (a,b). Suppose in addition that f(a) = 0 and f(b) = 0. Then there is a point c lying strictly between a and b such that f'(c) = 0. A continuous function on a closed interval attains a maximum and a minimum on the interval. At a point where a differentiable function has a max or a min, the derivative is zero. The plan of the proof is to break the situation up into three cases, the cases are:

4. 1. f is not positive or negative on the interval [a,b]. 2. f is positive somewhere on the interval [a,b]. 3. f is negative somewhere on the interval [a,b]. Case 1: f is not positive or negative on the interval [a,b] If f is not positive or negative on the interval [a,b], then f must be zero everywhere on the interval. That is, f = 0, and in particular, f is constant. The derivative of a constant function is zero, so f'(c) = 0 for all c between a and b. Thus, the conclusion of Rolle's theorem holds in this case. Case 2: f is positive somewhere on the interval [a,b] Suppose that f is positive somewhere on the interval [a,b]. A continuous function on a closed interval takes on a maximum value somewhere on the interval; suppose that f attains its max at x = c. Since we know f is positive somewhere on the interval, the max f(c) must be positive; since f(a) = 0 and f(b) = 0, c can't be either a or b. Therefore, a < c < b. In particular, f is differentiable at c. But at a point where a differentiable function attains a max, the derivative is zero. Thus, f'(c) = 0, and Rolle's theorem holds in this case as well.

5. Case 3: f is negative somewhere on the interval [a,b] This case is similar to Case 2. Suppose that f is negative somewhere on the interval [a,b]. A continuous function on a closed interval takes on a minimum value somewhere on the interval; suppose that f attains its min at x = c. Since I know f is negative somewhere on the interval, the min f(c) must be negative; since f(a) = 0 and f(b) = 0, c can't be either a or b. Therefore, a < c < b. In particular, f is differentiable at c. But at a point where a differentiable function attains a min, the derivative is zero. Thus, f'(c) = 0, and Rolle's theorem holds in Case 3. Since these three cases cover all the possibilities, the proof of Rolle's theorem is complete.

6. First example A semicircle of radius r. For an radius r > 0 consider the function Its graph is the upper semicircle centered at the origin. This function is continuous on the closed interval [r,r] and differentiable in the open interval (r,r), but not differentiable at the endpoints r and r. Since f(r) = f(r), Rolle's theorem applies, and indeed, there is a single point where the derivative of f is zero. A semicircle of radius r.

7. Second example If differentiability fails at a single inner point of the interval, the standard version of Rolle's theorem can fail. For some real a > 0 consider the absolute value function Then f(a) = f(a), but there is no c between a and a for which the derivative is zero. This is because that function, although continuous, is not differentiable at x = 0. Note that the derivative of f changes its sign at x = 0, but without attaining the value 0. The graph of the absolute value function.

8. Third example Consider the function f on [2,2] given by Its graph consists of two semicircles put together is a smooth way at the origin. The function is odd and satisfies f(2) = 0 = f(2). It has infinite slope at the origin and is therefore not considered differentiable at x = 0. However, the assertion of Rolle's theorem is still true, there is (even more than) one point in (2,2) where the derivative is zero. Function consisting of two semicirles joined smoothly at the origin

9. Sources Proofs from http://marauder.millersville.edu/ Examples from http://en.wikipedia.org www.maths.abdn.ac.uk