Math 131. Rolle s and Mean Value Theorems Larson Section 3.2

Transcription

1 Math 3. Rolle s and Mean Value Theorems Larson Section 3. Many mathematicians refer to the Mean Value theorem as one of the if not the most important theorems in mathematics. Rolle s Theorem. Suppose f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). If f(a) = f(b), then there is some number c between a and b such that 0. Proof. If f is constant, there is nothing to do, because 0 for each a < c < b. If f is not constant, then f has a maximum or minimum on the interval [a, b] at some number c between a and b. Then c is a critical number of f, and f is differentiable at c and so 0. An illustration of Rolle s theorem is as follows Example. Find the two x-intercepts of the function f and show that f (x) = 0 at some point between the two x-intercepts. f(x) = x 4x 5 Solution: First, factor f(x) = x 4x 5 = (x + )(x 5) and so the x-intercepts are: (, 0) and (5, 0) Now f (x) = x 4, and setting f (x) = 0, we find x =.0 which is between and 5. Example. Determine whether Rolle s theorem can be applied to the function f(x) = x 4x 5 x 8 on the interval [, 5]. (Select all that apply)

2 (a) Yes, Rolle s theorem can be applied. (b) No, because f is not continuous on the closed interval [, 5]. (c) No, because f is not differentiable on the open interval (, 5). (d) No, because f( ) f(5). If Rolle s theorem can be applied, find all c in the interval (, 5) such that 0. Solution: Only (a) applies because f is continuous and differentiable when x 8, so f is continuous on the closed interval [, 5] and f is differentiable on the open interval (, 5). Moreover, f( ) = 0 = f(5) and so Rolle s theorem applies. To find c, we first find f (x) and set it to 0: f (x) = (x 4)(x 8) (x 4x 5) = x 6x + 37 (x 8) (x 8) This implies we find c so that c 6c + 37 = 0 and < c < 5. The quadratic formula implies c = 6 ± 6 4()(37) = 6 ± 08 the only value of c between and 5 is c = A graph of f on the interval [, 5] along with the horizontal tangent line at (c, f(c)) for c as just found, is as follows: y x Mean Value Theorem. Suppose f is continuous on [a, b] and differentiable on (a, b). Then there is a number c between a and b so that

3 This says at the point c, the slope of the tangent line is the same as the slope of the secant line through (a, f(a)) and (b, f(b)). A graph illustrating this is a follows. Proof. Let h(x) = f(x) (x a) Then h is continuous on [a, b] and differentiable on (a, b) and h(a) = f(a) and h(b) = f(a) and so Rolle s theorem implies there is a number c between a and b with h (c) = 0. This means f (c) = 0 and so as desired. Example 3. Let f(x) = /x. Verify that the mean value theorem can be applied to f on the interval [0.5, 3] and then find a number c between 0.5 and 3 such that f(3) f(0.5) Solution: The function f is continuous and differentiable if x 0, therefore it is continuous on the interval [0.5, 3] and differentiable on the interval (0.5, 3) and so there is a number c between 0.5 and 3 so that f(3) f(0.5) = /3 5/ = 5/3 5/ = 3 Now c and so c = 3 which means c = 3/ since we need c between 0.5 and 3.

4 3 y 3 x Example 4. Consider the function f(x) = x. Then the Mean Value theorem (can or cannot) be applied to f on the interval [ 4, ] because: f is on [ 4, ]; and f is on ( 4, ). Compute: f( 4) = ; and f() =. If the Mean Value theorem can be applied, it ensures there is at least one number c between 4 and such that and all possible such numbers c between 4 and are: c = Solution: Consider the function f(x) = x. Then the Mean Value theorem can be applied to f on the interval [ 4, ] because: f is continuous on [ 4, ] and f is differentiable on ( 4, ). Now, f( 4) = 5 and f() = 0. Thus, the Mean Value theorem ensures there is at least one number c between 4 and such that f() f( 4) ( 4) To find c we compute f (c) = and so c and thus c = 9 4. = = 5 c = 5 or 5 = c or c = 5 4

5 Some further examples related to assignment questions are as follows. Example 5. Determine whether Rolle s theorem can be applied to the function (Select all that apply) f(x) = (x 3) on the interval [, 5]. (a) Yes, Rolle s theorem can be applied. (b) No, because f is not continuous on the closed interval [, 5]. (c) No, because f is not differentiable on the open interval (, 5). (d) No, because f() f(5). If Rolle s theorem can be applied, find all c in the interval (, 5) such that 0. Solution: (a) is not true (see (b) and (c)). (b) applies because f is not continuous at x = 3. (c) applies because f is not differentiable at x = 3. (d) f() = = f(5), so (d) does not apply. The last question concerning finding c is not applicable since Rolle s theorem does not apply. Example 6. Determine whether Rolle s theorem can be applied to the function (Select all that apply) f(x) = 4 x 7 on the interval [3, ]. (a) Yes, Rolle s theorem can be applied. (b) No, because f is not continuous on the closed interval [3, ]. (c) No, because f is not differentiable on the open interval (3, ). (d) No, because f(3) f(). If Rolle s theorem can be applied, find all c in the interval (3, ) such that 0.

6 Solution: (a) is not true see (c). (b) does not apply because f is continuous everywhere, and thus on [3, ]. (c) applies because f is not differentiable at x = 7. (d) f(3) = 0 = f(), so (d) does not apply. The last question concerning finding c is not applicable since Rolle s theorem does not apply because of (b). The graph below shows f(x) = 4 x 7 on the interval [3, ], and the sharp point above x = 7 shows f is not differentiable at x = 7 because the graph has no tangent line there. y x Example 7. A rock thrown off of a cliff over the sea has its height in feet above the sea given by the function s(t) = 6t + t + 480, where t is the time in seconds after the ball was released. (a) Find the average velocity of the rock during the first 6 seconds. (b) Use the Mean Value theorem to verify that at some time during the first 6 seconds of the flight of the rock, the instantaneous velocity equals the average velocity. (c) Find the time requested in (b). Solution: (a) The average velocity is given by s(6) s(0) 6 0 = = 6 feet per second. (b) The function s(t) is continuous on [0, 6] and differentiable on (0, 6). The Mean Value

7 theorem ensures that at some point c between 0 and 6 will have s (t) = s(6) s(0) 6 0 = 6 (c) The velocity is given by v(t) = s (t) = 3t +. Then v(c) = 6 implies 3c + = 6 and so c = 6 3 = 3.0 That is, after exactly 3.0 seconds into its flight, the instantaneous velocity of the rock will be 6 feet per second.