MAT01B1: the Mean Value Theorem

Transcription

1 MAT01B1: the Mean Value Theorem Dr Craig 15 August 2018

2 My details: Consulting hours: Monday 14h40 15h25 Thursday 11h20 12h55 Friday (this week): 11h20 12h30 Office C-Ring (Or, just google Andrew Craig maths.)

3 Assignments and Class Tests Assignment 1 and Class Test 1 are available from the collection facility. Check the memo and your scripts to learn from these assessments. e-quiz 1 Open until 23h59 Friday 17 August. Three attempts: 60min time limit. Saturday class D-LES 101 from 09h00-12h00

4 Semester Test 1 Saturday 25 August D1 Lab 308 Starts at 09h00. Be seated by 08h45. Scope: Ch , 7.8, 4.1, 4.2, 4.3 Also examinable: Proofs of Fermat s Theorem, Rolle s Theorem, Mean Value Theorem

5 Today Recap Rolle s Theorem Historical interlude Mean Value Theorem

6 From last time: Extreme Value Theorem If f is continuous on a closed interval [a, b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some numbers c and d in [a, b].

7 Also from last time: Fermat s Theorem: If f has a local maximum or minimum at c, and if f (c) exists, then f (c) = 0.

8 Rolle s Theorem Let f be a function that satisfies the following three hypotheses: 1. f is continuous on the closed interval [a, b] 2. f is differentiable on the open interval (a, b) 3. f(a) = f(b) Then there is a number c (a, b) such that f (c) = 0.

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10 Proof: We proceed using proof by cases. Case 1: f(x) = k, a constant Case 2: f(x) > f(a) for some x (a, b) Case 3: f(x) < f(a) for some x (a, b)

11 Proof: Case 1: f(x) = k, a constant

12 Proof: Case 1: f(x) = k, a constant In this case we have f (x) = 0 so the number c can be taken as any number in (a, b).

13 Proof: Case 2: f(x) > f(a) for some x (a, b)

14 Proof: Case 2: f(x) > f(a) for some x (a, b) By EVT (applicable by hypothesis 1), f has an absolute maximum value somewhere in [a, b]. Since f(a) = f(b) and since f(x) > f(a) for some x (a, b), it must attain this value at a number c in the open interval (a, b). Then f has a local maximum at c and, by hypothesis 2, f is differentiable at c. Therefore, by Fermat s Theorem, f (c) = 0.

15 Proof: Case 3: f(x) < f(a) for some x (a, b)

16 Proof: Case 3: f(x) < f(a) for some x (a, b) By EVT f has an absolute minimum value in [a, b]. Since f(a) = f(b) and f(x) < f(a) for some x (a, b), it must attain this minimum value at some point c (a, b). Thus f(c) is a local minimum and, since f is differentiable on (a, b), Fermat s Theorem again tells us that f (c) = 0.

17 Rolle s Theorem Let f be a function that satisfies the following three hypotheses: 1. f is continuous on the closed interval [a, b] 2. f is differentiable on the open interval (a, b) 3. f(a) = f(b) Then there is a number c (a, b) such that f (c) = 0.

18 Rolle s Theorem applied to motion Consider s = f(t), the position function of an object. If the object is in the same place at two different moments t = a and t = b, then f(a) = f(b). Rolle s Theorem says that there exists a moment c between a and b when we must have f (c) = 0, i.e. the velocity is 0.

19 Example (applying Rolle s Theorem): Prove that the equation x 3 + x 1 = 0 has exactly one root. Solution: we use the Intermediate Value Theorem to show that it has at least one root. Then assume that it has two roots (i.e. assume f(a) = 0 = f(b) with a b) and use Rolle s Theorem to reach a contradiction.

20 SHORT BREAK

21 Joseph Lagrange is credited as being the first mathematician to formulate the Mean Value Theorem. He was also active in another area of mathematics: number theory. Lagrange s Theorem: Every positive integer can be written as the sum of four perfect squares.

22 Some examples: 1 = = = = = =

23 Lagrange s Theorem: Every positive integer can be written as the sum of four perfect squares. Historians believe that people were aware of this property of numbers around 250AD. However, it wasn t until 1770 (more than 1,500 years later) that Joseph Louis Lagrange was able to prove it.

24 Mean Value Theorem Let f be a function that satisfies: 1. f is continuous on [a, b] 2. f is differentiable on (a, b). Then there exists c (a, b) such that f (c) = f(b) f(a) b a or, equivalently f(b) f(a) = f (c)(b a)

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27 Proof: We will apply Rolle s Theorem to a new function h. This function h is constructed as the difference between f and the function whose graph is the secant line AB.

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29 Proof: We will apply Rolle s Theorem to a new function h. This function h is constructed as the difference between f and the function whose graph is the secant line AB. The equation of the secant line can be written as: or y f(a) = y = f(a) + f(b) f(a) (x a) b a f(b) f(a) (x a) b a

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31 The equation of the secant AB is given by y = f(a) + f(b) f(a) (x a) b a So, the equation of the difference function will be given by: h(x) = f(x) f(a) f(b) f(a) (x a) b a Now we want to verify that h(x) satisfies the hypotheses of Rolle s Theorem.

32 h(x) = f(x) f(a) f(b) f(a) (x a) b a 1. h is continuous on [a, b] because it is the sum of f and a first-degree polynomial, both of which are continuous. 2. We can compute h (x) h (x) = f (x) f(b) f(a). b a

33 We must show that h(a) = h(b). 3. h(a) = 0 = h(b) Now since h satisfies the hypotheses of Rolle s Theorem, there exists c (a, b) such that h (c) = 0. Therefore Hence 0 = h (c) = f (c) f (c) = f(b) f(a). b a f(b) f(a). b a

34 Example using MVT: Consider f(x) = x 3 x and a = 0, b = 2. c = 2 3

35 Example: Verify that the function satisfies the hypotheses of the Mean Value Theorem and then find the number guaranteed to exist by the theorem. f(x) = 1 x [1, 3]

36 Speed and distance An object moving in a straight line with position function s = f(t), then the average velocity between t = a and t = b is f(b) f(a) b a The velocity at t = c is f (c). E.g.: if you drive from JHB to Durban and your average velocity is 100km/h, then there must have been a moment when your instantaneous velocity was 100km/h.

37 Example: Suppose that f(0) = 3 and that f (x) 5 for all x. What is the largest possible value of f(2)?

38 The Mean Value Theorem can be used to help us prove many useful results in differential calculus. Below are the two such results (the proofs of which are not examinable). Theorem If f (x) = 0 for all x in an interval (a, b) then f is constant on (a, b).

39 Theorem If f (x) = 0 for all x in an interval (a, b) then f is constant on (a, b). Proof: Let x 1, x 2 (a, b) with x 1 < x 2. Since f is differentiable on (a, b) it must be differentiable on (x 1, x 2 ) and continuous on [x 1, x 2 ]. Apply MVT to f on [x 1, x 2 ] we get c (x 1, x 2 ) such that f(x 2 ) f(x 1 ) = f (c)(x 2 x 1 )

40 Proof continued: Since f (x) = 0 for all x, we have that f (c) = 0 and so f(x 1 ) = f(x 2 ). Since x 1 and x 2 were arbitrary, we have that f has the same value at any two numbers in (a, b). That is, f is constant on (a, b).

41 Corollary of previous theorem If f (x) = g (x) for all x in an interval (a, b), then f g is constant on (a, b). That is, f(x) = g(x) + c where c is a constant.