MAT01B1: the Mean Value Theorem Dr Craig 21 August 2017
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Semester Test 1 Saturday 26 August D1 Lab 208 Starts at 08h30. Be seated by 08h15. Scope: Ch 7.1 7.5, 7.8, 4.1, 4.2 Also examinable: Proofs of Fermat s Theorem, Rolle s Theorem, Mean Value Theorem
Today Recap Rolle s Theorem Historical interlude Mean Value Theorem
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].
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.
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.
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)
Proof: Case 1: f(x) = k, a constant
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).
Proof: Case 2: f(x) > f(a) for some x (a, b)
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.
Proof: Case 3: f(x) < f(a) for some x (a, b)
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.
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.
Example: 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.
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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.
Some examples: 1 = 0 2 + 0 2 + 0 2 + 1 2 2 = 0 2 + 0 2 + 1 2 + 1 2 3 = 0 2 + 1 2 + 1 2 + 1 2 4 = 0 2 + 0 2 + 0 2 + 2 2... 13 = 0 2 + 0 2 + 2 2 + 3 2... 99 = 1 2 + 1 2 + 4 2 + 9 2...
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.
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)
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
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.
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
We must show that h(a) = h(b). 3. h(a) = 0 = h(b) [show details] 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
Example using MVT: Consider f(x) = x 3 x and a = 0, b = 2. c = 2 3
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]
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.
Example: Suppose that f(0) = 3 and that f (x) 5 for all x. What is the largest possible value of f(2)?
The Mean Value Theorem can be used to help us prove many useful results in differential calculus. Theorem If f (x) = 0 for all x in an interval (a, b) then f is constant on (a, b).
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 )
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).
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.
Example: Use calculus to prove that tan 1 x + cot 1 x = π 2