MAT01B1: Maximum and Minimum Values

Transcription

1 MAT01B1: Maximum and Minimum Values Dr Craig 14 August 2018

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

3 Assignments and Class Tests Collect Assignment 1 and Class Test 1 from the collection facility. Check the memo and your scripts to learn from these assessments.

4 e-quiz 1 Covers Chapter 7. Currently live. 60min time limit per attempt. Three attempts. Saturday classes This week: 09h00 12h00 in D-LES 101.

5 Today Pop Quiz Maxima, Minima and Extreme Values Fermat s Theorem (with proof) Critical numbers

6 Pop Quiz: write down the following sec x dx formula for integration by parts squared identity with cot θ formula for sin A sin B cos 2 x in terms of cos 2x trig substitution to help solve an integral containing 3 + x 2 form of the partial fraction decomposition 4x + 2 for (2x + 7) 3

7 Maximum and Minimum Values Some examples of where we might want to be able to calculate maximum and minimum values: What shape will minimize the manufacturing cost of a tin can? What is the maximum acceleration of a space shuttle during take-off? At what angle should blood vessels branch so as to minimize the energy expended by the heart in pumping blood?

8 Definition: Let c be a number in the domain D of a function f. Then f(c) is the absolute maximum value of f on D if f(c) f(x) for all x D; absolute minimum value of f on D if f(c) f(x) for all x D. The absolute maximum/minimum is often called the global maximum/minimum.

9 f(a) is the abs. min., f(d) is the abs. max.

10 Definition: the number f(c) is a local maximum value of f if f(c) f(x) when x is near c. local minimum value of f if f(c) f(x) when x is near c. x near c = when x is in some open interval containing c, i.e. x (c ε, c + ε). [Recall that ε is a small positive number.]

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12 Important: When we refer to a R as being a local or absolute maximum or minimum we are referring to a as a y-value.

13 Examples of maxima and minima: f(x) = cos x f(x) = x 2 f(x) = x 3

14 Examples of maxima and minima:

15 An important point about end-points The end point of a closed interval cannot be used as the x-value for a local minimum or local maximum. In the picture on the previous slide the function is defined on [ 1, 4]. Therefore neither 37 (f( 1) = 37) nor 32 (f(4) = 32) are local maxima.

16 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]. We do not cover the proof of this theorem but we will use it throughout this chapter.

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18 What if the hypothesis does not hold?

19 Fermat s Theorem: If f has a local maximum or minimum at c, and if f (c) exists, then f (c) = 0. Proof: Suppose that f has a local maximum at c and f (c) exists. By the definition of a local maximum we have that f(c) f(x) for any x sufficiently close to c. Thus, for h positive or negative and sufficiently close to 0, we have f(c) f(c + h).

20 Proof of Fermat s Theorem continued From f(c) f(c + h) we get that f(c + h) f(c) 0. For h > 0 and sufficiently small, we get f(c + h) f(c) 0. h Now take limits to get f(c + h) f(c) lim h 0 + h lim h = 0.

21 We took limits to get f(c + h) f(c) lim h 0 + h Since f (c) exists, we have lim h = 0. f f(c + h) f(c) (c) = lim h 0 h f(c + h) f(c) = lim. h 0 + h Hence f (c) 0.

22 If h < 0 but sufficiently close to 0, then when we divide both sides of the inequality f(c + h) f(c) 0 f(c + h) f(c) by h, we get 0. h Taking lim h 0 we have f f(c + h) f(c) (c) = lim h 0 h f(c + h) f(c) = lim 0. h 0 h

23 We have shown that f (c) 0 and f (c) 0 and hence f (c) = 0. We assumed at the beginning that f had a local maximum at c. If we had assumed that f had a local minimum at c then we can use a similar approach to prove that we will have f (c) = 0. (Exercise 79 asks you to go through the details of the proof of the case that f has a local minimum at c.)

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

25 Definition: a critical number of a function f is a number c in the domain of f such that either f (c) = 0 or f (c) does not exist. Example: Find the critical numbers of f(x) = x 3/5 (4 x) Solution: x = 0, x = 3/2.

26 A different version of Fermat s Theorem If f has a local maximum or minimum at c, then c is a critical number of f. How do we get this? (p q) r s [ ((p q) r) ] s [ (p q) r ] s (p q) ( r s) (p q) ( r s)

27 Example: Prove that the function f(x) = x x 51 + x + 1 has neither a local maximum nor a local minimum. Hint: Use proof by contradiction. Assume and that f(x) does have a local maximum and apply Fermat s Theorem. Then reach a contradiction by showing that f (c) 0 for all c R.

28 Closed Interval Method: to find the absolute maximum and absolute minimum values of a continuous function f on a closed interval [a, b]: 1. Calculate f(c) for every critical number c (a, b). 2. Calculate f(a) and f(b), i.e. find the value of f at the end points. 3. The largest value from Steps 1 & 2 is the absolute maximum. The smallest value from Steps 1 & 2 is the absolute min.

29 Example: Find the absolute maximum and minimum values of the function where 1 2 f(x) = x 3 3x x 4. Solution: critical numbers x = 0, x = 2. f(0) = 1, f(2) = 3 f( 1 2 ) = 1 8, f(4) = 17 Abs. min = 3 and abs. max. = 17

30 Example: Use calculus to find the absolute minimum and maximum values of on the interval [0, 2π]. f(x) = x 2 sin x Solution: critical numbers x = π 3, x = 5π 3 f(0) = 0, f(2π) = 2π f( π 3 ) = π/3 3, f( 5π 3 ) = 5π/3 + 3 Abs. min. = π/3 3 Abs. max.= 5π/3 + 3.

31 Tomorrow s lecture on Ch 4.2: Use Fermat s Theorem to prove Rolle s Theorem Use Rolle s Theorem to prove the Mean Value Theorem Applications of Rolle s Theorem and the Mean Value Theorem