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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.

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Maximum and Minimum Values ( Section 3.1)

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  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

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