1 / 24

Maximum and Minimum Values ( Section 3.1)

Maximum and Minimum Values ( Section 3.1) . Alex Karassev. Absolute maximum values. A function f has an absolute maximum value on a set S at a point c in S if f(c) ≥ f(x) for all x in S. y. y = f(x). f(c). x. S. c. Absolute minimum values.

paley
Télécharger la présentation

Maximum and Minimum Values ( Section 3.1)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Maximum and Minimum Values(Section 3.1) Alex Karassev

  2. Absolute maximum values • A function f has an absolute maximum value on a set S at a point c in S if f(c) ≥ f(x) for all x in S y y = f(x) f(c) x S c

  3. Absolute minimum values • A function f has an absolute minimum value on a set S at a point c in S if f(c) ≤ f(x) for all x in S y y = f(x) x f(c) S c

  4. Example: f(x) = x2 • S = (-∞, ∞) • No absolute maximum • Absolute minimum:f(0) = 0 at c = 0 y x 0

  5. Example: f(x) = x2 • S = [0,1] • Absolute maximumf(1) = 1 at c = 1 • Absolute minimum:f(0) = 0 at c = 0 y x 0 1

  6. Example: f(x) = x2 • S = (0,1] • Absolute maximumf(1) = 1 at c = 1 • No absoluteminimum,although function isbounded from below:0 < x2 for allx in (0,1] ! y x 0 1

  7. Local maximum values • A function f has a local maximum value at a point c if f(c) ≥ f(x) for all x near c (i.e. for all x in some open interval containing c) y y = f(x) x c

  8. Local minimum values • A function f has a local minimum value at a point cif f(c) ≤ f(x) for all x near c(i.e. for all x in some open interval containing c) y y = f(x) x c

  9. Example: y = sin x f(x) = sin xhas local (and absolute) maximumat all points of the form π/2 + 2πk,and local (and absolute) minimumat all points of the form -π/2 + 2πk,where k is an integer 1 - π/2 π/2 -1

  10. Applications • Curve sketching • Optimization problems (with constraints),for example: • Finding parameters to minimizemanufacturing costs • Investing to maximize profit (constraint: amount of money to invest is limited) • Finding route to minimize the distance • Finding dimensions of containers to maximize volumes (constraint: amount of material to be used is limited)

  11. Extreme Value Theorem If f is continuous on a closed interval [a,b], then f attainsabsolute maximum value f(cMAX) andabsolute minimum value f(cMIN)at some numbers cMAX andcMIN in [a,b]

  12. Extreme Value Theorem - Examples y y y = f(x) y = f(x) x x a a b cMIN cMAX cMIN cMAX= b Both absolute max and absolute min are attained in the open interval (a,b) at the points of local max and min Absolute maximum is attained at the right end point: cMAX = b

  13. Continuity is important y x -1 1 0 No absolute maximum or minimumon [-1,1]

  14. Closed interval is important • f(x) = x2, S = (0,1] • No absoluteminimum in (0,1] y x 0 1

  15. How to find max and min values? • Absolute maximum or minimum values of a function, continuous on a closed interval are attained either at the points which are simultaneously the points of local maximum or minimum, or at the endpoints • Thus, we need to know how to find points of local maximums and minimums

  16. Fermat's Theorem • If f has a local maximum or minimum at c and f′(c) exists, then f′(c) = 0 y horizontal tangent line at the point of local max (or min) y = f(x) x c

  17. Converse of Fermat's theoremdoes not hold! • If f ′(c) = 0 it does not mean that c is a point of local maximum or minimum • Example: f(x) = x3, f ′(0) = 0, but 0 is not a point of local max or min • Nevertheless, points c wheref ′(c) = 0 are "suspicious" points(for local max or min) y x

  18. Problem: f′ not always exists • f(x) = |x| • It has local (and absolute) minimum at 0 • However, f′ (0) does not exists! y x

  19. Critical numbers • Two kinds of "suspicious" points(for local max or min): • f′(c) = 0 • f′(c) does not exists

  20. Critical numbers – definition • A number c is called a critical number of function f if the following conditions are satisfied: • c is in the domain of f • f′(c) = 0 or f′(c) does not exist

  21. Closed Interval Method • The method to find absolute maximum or minimum of a continuous function, defined on a closed interval [a,b] • Based on the fact that absolute maximum or minimum • either is attained at some point inside the open interval (a,b) (then this point is also a point of local maximum or minimum and hence isa critical number) • or is attained at one of the endpoints

  22. Closed Interval Method • To find absolute maximum and minimumof a function f, continuous on [a,b]: • Find critical numbers inside (a,b) • Find derivative f′ (x) • Solve equation f′ (x)=0 for x and choose solutions which are inside (a,b) • Find numbers in (a,b) where f′ (x) d.n.e. • Suppose that c1, c2, …, ckare all critical numbers in (a,b) • The largest of f(a), f(c1), f(c2), …, f(ck), f(b) is theabsolute maximum of f on [a,b] • The smallest of these numbers is theabsolute minimum of f on [a,b]

  23. Example • Find the absolute maximum and minimum values of f(x) = x/(x2+1) on the interval [0,2]

  24. Find the absolute maximum and minimum values of f(x) = x/(x2+1) on the interval [0,2] Solution • Find f′(x): • Critical numbers: f′(x) = 0 ⇔ 1– x2 = 0 • So x = 1 or x = – 1 • However, only 1 is inside [0,2] • Now we need to compare f(0), f(1), and f(2): • f(0) = 0, f(1) = 1/2, f(2)= 2/5 • Therefore 0 is absolute minimum and 1/2 is absolute maximum

More Related