Section	4.1
     Maximum	and	Minimum	Values

               V63.0121.006/016, Calculus	I



                      March	23...
Announcements




     Welcome	back	from	Spring	Break!
     Quiz	3: April	2, Sections	2.6–3.5




                        ...
Computation	of	Midterm	Letter	Grades
                   Q
                   . uizzes
                              .     ...
Computation	of	Midterm	Letter	Grades
                   Q
                   . uizzes
                              .     ...
Computation	of	Midterm	Letter	Grades
                            W
                            . ebAssign
                ...
Distribution	of	Midterm	Averages	and	Letter	Grades




      Median	=	81.35%	(curved	to	B)
      Average	=	74.39%	(curved	...
What	can	I do	to	improve	my	grade?
                   Q
                   . uizzes
                              .       ...
What	can	I do	to	improve	my	grade?
           f
           .uture	Quizzes          p
                                   . ...
What	can	I do	to	improve	my	grade?
                       M
                       . idterm
                              ...
Help!



 Free	resources:
        recitation
        TAs’	office	hours
        my	office	hours
        Math	Tutoring	Center
...
Outline

  Introduction

  The	Extreme	Value	Theorem

  Fermat’s	Theorem	(not	the	last	one)
     Tangent: Fermat’s	Last	Th...
Optimize




  .   .   .   .   .   .
Why	go	to	the	extremes?


     Rationally	speaking, it	is
     advantageous	to	find	the
     extreme	values	of	a
     funct...
Design




.
Image	credit: Jason	Tromm
                                    .
                            .   .       .   ....
Why	go	to	the	extremes?


     Rationally	speaking, it	is
     advantageous	to	find	the
     extreme	values	of	a
     funct...
Optics




                                               .
.
Image	credit: jacreative
                           .   .   ...
Why	go	to	the	extremes?


     Rationally	speaking, it	is
     advantageous	to	find	the
     extreme	values	of	a
     funct...
Outline

  Introduction

  The	Extreme	Value	Theorem

  Fermat’s	Theorem	(not	the	last	one)
     Tangent: Fermat’s	Last	Th...
Extreme	points	and	values

    Definition
    Let f have	domain D.




                                                    ...
Extreme	points	and	values

    Definition
    Let f have	domain D.
           The	function f has	an absolute
           max...
Extreme	points	and	values

    Definition
    Let f have	domain D.
           The	function f has	an absolute
           max...
Extreme	points	and	values

    Definition
    Let f have	domain D.
           The	function f has	an absolute
           max...
Theorem	(The	Extreme	Value	Theorem)
Let f be	a	function	which	is	continuous	on	the	closed	interval
[a, b]. Then f attains	...
Theorem	(The	Extreme	Value	Theorem)
Let f be	a	function	which	is	continuous	on	the	closed	interval
[a, b]. Then f attains	...
Theorem	(The	Extreme	Value	Theorem)
    Let f be	a	function	which	is	continuous	on	the	closed	interval
    [a, b]. Then f ...
No	proof	of	EVT forthcoming




      This	theorem	is	very	hard	to	prove	without	using	technical
      facts	about	continu...
Bad	Example	#1


  Example

 Consider	the	function
          {
           x      0≤x<1
  f(x ) =
           x − 2 1 ≤ x ≤ ...
Bad	Example	#1


  Example

 Consider	the	function                     .
          {
           x      0≤x<1
  f(x ) =    ...
Bad	Example	#1


  Example

 Consider	the	function                                     .
          {
           x      0≤x...
Bad	Example	#1


  Example

 Consider	the	function                                     .
          {
           x      0≤x...
Bad	Example	#2

  Example
  Consider	the	function f(x) = x restricted	to	the	interval [0, 1).




                        ...
Bad	Example	#2

  Example
  Consider	the	function f(x) = x restricted	to	the	interval [0, 1).


                          ...
Bad	Example	#2

  Example
  Consider	the	function f(x) = x restricted	to	the	interval [0, 1).


                          ...
Bad	Example	#2

  Example
  Consider	the	function f(x) = x restricted	to	the	interval [0, 1).


                          ...
Final	Bad	Example


  Example
                                 1
  Consider	the	function f(x) =     is	continuous	on	the	c...
Final	Bad	Example


  Example
                                 1
  Consider	the	function f(x) =     is	continuous	on	the	c...
Final	Bad	Example


  Example
                                 1
  Consider	the	function f(x) =     is	continuous	on	the	c...
Final	Bad	Example


  Example
                                 1
  Consider	the	function f(x) =     is	continuous	on	the	c...
Outline

  Introduction

  The	Extreme	Value	Theorem

  Fermat’s	Theorem	(not	the	last	one)
     Tangent: Fermat’s	Last	Th...
Local	extrema

  Definition
      A function f has	a local	maximum or relative	maximum at c
      if f(c) ≥ f(x) when x is	...
Local	extrema

  Definition
      A function f has	a local	maximum or relative	maximum at c
      if f(c) ≥ f(x) when x is	...
Local	extrema
      So	a	local	extremum	must	be inside the	domain	of f (not	on
      the	end).
      A global	extremum	tha...
Theorem	(Fermat’s	Theorem)
Suppose f has	a	local	extremum	at c and f is	differentiable	at c.
Then f′ (c) = 0.

           ...
Theorem	(Fermat’s	Theorem)
Suppose f has	a	local	extremum	at c and f is	differentiable	at c.
Then f′ (c) = 0.

           ...
Sketch	of	proof	of	Fermat’s	Theorem

   Suppose	that f has	a	local	maximum	at c.




                                     ...
Sketch	of	proof	of	Fermat’s	Theorem

   Suppose	that f has	a	local	maximum	at c.
       If x is	slightly	greater	than c, f...
Sketch	of	proof	of	Fermat’s	Theorem

   Suppose	that f has	a	local	maximum	at c.
       If x is	slightly	greater	than c, f...
Sketch	of	proof	of	Fermat’s	Theorem

   Suppose	that f has	a	local	maximum	at c.
       If x is	slightly	greater	than c, f...
Sketch	of	proof	of	Fermat’s	Theorem

   Suppose	that f has	a	local	maximum	at c.
       If x is	slightly	greater	than c, f...
Sketch	of	proof	of	Fermat’s	Theorem

   Suppose	that f has	a	local	maximum	at c.
       If x is	slightly	greater	than c, f...
Meet	the	Mathematician: Pierre	de	Fermat




     1601–1665
     Lawyer	and	number
     theorist
     Proved	many	theorems...
Tangent: Fermat’s	Last	Theorem

     Plenty	of	solutions	to
     x2 + y2 = z2 among
     positive	whole	numbers
     (e.g....
Tangent: Fermat’s	Last	Theorem

     Plenty	of	solutions	to
     x2 + y2 = z2 among
     positive	whole	numbers
     (e.g....
Tangent: Fermat’s	Last	Theorem

     Plenty	of	solutions	to
     x2 + y2 = z2 among
     positive	whole	numbers
     (e.g....
Tangent: Fermat’s	Last	Theorem

     Plenty	of	solutions	to
     x2 + y2 = z2 among
     positive	whole	numbers
     (e.g....
Outline

  Introduction

  The	Extreme	Value	Theorem

  Fermat’s	Theorem	(not	the	last	one)
     Tangent: Fermat’s	Last	Th...
Flowchart	for	placing	extrema
Thanks	to	Fermat
    Suppose f is	a	continuous	function	on	the	closed, bounded
    interval ...
The	Closed	Interval	Method



   This	means	to	find	the	maximum	value	of f on [a, b], we	need	to:
       Evaluate f at	the ...
Outline

  Introduction

  The	Extreme	Value	Theorem

  Fermat’s	Theorem	(not	the	last	one)
     Tangent: Fermat’s	Last	Th...
Example
Find	the	extreme	values	of f(x) = 2x − 5 on [−1, 2].




                                            .    .     . ...
Example
Find	the	extreme	values	of f(x) = 2x − 5 on [−1, 2].

Solution
Since f′ (x) = 2, which	is	never	zero, we	have	no	c...
Example
Find	the	extreme	values	of f(x) = 2x − 5 on [−1, 2].

Solution
Since f′ (x) = 2, which	is	never	zero, we	have	no	c...
Example
Find	the	extreme	values	of f(x) = x2 − 1 on [−1, 2].




                                            .    .     . ...
Example
Find	the	extreme	values	of f(x) = x2 − 1 on [−1, 2].

Solution
We	have f′ (x) = 2x, which	is	zero	when x = 0.




...
Example
Find	the	extreme	values	of f(x) = x2 − 1 on [−1, 2].

Solution
We	have f′ (x) = 2x, which	is	zero	when x = 0. So	o...
Example
Find	the	extreme	values	of f(x) = x2 − 1 on [−1, 2].

Solution
We	have f′ (x) = 2x, which	is	zero	when x = 0. So	o...
Example
Find	the	extreme	values	of f(x) = x2 − 1 on [−1, 2].

Solution
We	have f′ (x) = 2x, which	is	zero	when x = 0. So	o...
Example
Find	the	extreme	values	of f(x) = x2 − 1 on [−1, 2].

Solution
We	have f′ (x) = 2x, which	is	zero	when x = 0. So	o...
Example
Find	the	extreme	values	of f(x) = x2 − 1 on [−1, 2].

Solution
We	have f′ (x) = 2x, which	is	zero	when x = 0. So	o...
Example
Find	the	extreme	values	of f(x) = x2 − 1 on [−1, 2].

Solution
We	have f′ (x) = 2x, which	is	zero	when x = 0. So	o...
Example
Find	the	extreme	values	of f(x) = 2x3 − 3x2 + 1 on [−1, 2].




                                            .   . ...
Example
Find	the	extreme	values	of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we	have...
Example
Find	the	extreme	values	of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we	have...
Example
Find	the	extreme	values	of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we	have...
Example
Find	the	extreme	values	of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we	have...
Example
Find	the	extreme	values	of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we	have...
Example
Find	the	extreme	values	of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we	have...
Example
Find	the	extreme	values	of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we	have...
Example
Find	the	extreme	values	of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we	have...
Example
Find	the	extreme	values	of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we	have...
Example
Find	the	extreme	values	of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we	have...
Example
Find	the	extreme	values	of f(x) = x2/3 (x + 2) on [−1, 2].




                                             .    ....
Example
Find	the	extreme	values	of f(x) = x2/3 (x + 2) on [−1, 2].

Solution
Write f(x) = x5/3 + 2x2/3 , then

           ...
Example
Find	the	extreme	values	of f(x) = x2/3 (x + 2) on [−1, 2].

Solution
Write f(x) = x5/3 + 2x2/3 , then

           ...
Example
Find	the	extreme	values	of f(x) = x2/3 (x + 2) on [−1, 2].

Solution
Write f(x) = x5/3 + 2x2/3 , then

           ...
Example
Find	the	extreme	values	of f(x) = x2/3 (x + 2) on [−1, 2].

Solution
Write f(x) = x5/3 + 2x2/3 , then

           ...
Example
Find	the	extreme	values	of f(x) = x2/3 (x + 2) on [−1, 2].

Solution
Write f(x) = x5/3 + 2x2/3 , then

           ...
Example
Find	the	extreme	values	of f(x) = x2/3 (x + 2) on [−1, 2].

Solution
Write f(x) = x5/3 + 2x2/3 , then

           ...
Example
Find	the	extreme	values	of f(x) = x2/3 (x + 2) on [−1, 2].

Solution
Write f(x) = x5/3 + 2x2/3 , then

           ...
Example
Find	the	extreme	values	of f(x) = x2/3 (x + 2) on [−1, 2].

Solution
Write f(x) = x5/3 + 2x2/3 , then

           ...
Example
Find	the	extreme	values	of f(x) = x2/3 (x + 2) on [−1, 2].

Solution
Write f(x) = x5/3 + 2x2/3 , then

           ...
Example                             √
Find	the	extreme	values	of f(x) =       4 − x2 on [−2, 1].




                     ...
Example                             √
Find	the	extreme	values	of f(x) =       4 − x2 on [−2, 1].

Solution
               ...
Example                             √
Find	the	extreme	values	of f(x) =       4 − x2 on [−2, 1].

Solution
               ...
Example                             √
Find	the	extreme	values	of f(x) =       4 − x2 on [−2, 1].

Solution
               ...
Example                             √
Find	the	extreme	values	of f(x) =       4 − x2 on [−2, 1].

Solution
               ...
Example                             √
Find	the	extreme	values	of f(x) =       4 − x2 on [−2, 1].

Solution
               ...
Example                             √
Find	the	extreme	values	of f(x) =       4 − x2 on [−2, 1].

Solution
               ...
Example                             √
Find	the	extreme	values	of f(x) =       4 − x2 on [−2, 1].

Solution
               ...
Outline

  Introduction

  The	Extreme	Value	Theorem

  Fermat’s	Theorem	(not	the	last	one)
     Tangent: Fermat’s	Last	Th...
Challenge: Cubic	functions




   Example
   How	many	critical	points	can	a	cubic	function

                      f(x) = a...
Solution
If f′ (x) = 0, we	have

                         3ax2 + 2bx + c = 0,

and	so
                         √          ...
Solution
If f′ (x) = 0, we	have

                         3ax2 + 2bx + c = 0,

and	so
                         √          ...
Solution
If f′ (x) = 0, we	have

                         3ax2 + 2bx + c = 0,

and	so
                         √          ...
Solution
If f′ (x) = 0, we	have

                         3ax2 + 2bx + c = 0,

and	so
                         √          ...
What	have	we	learned	today?




      The	Extreme	Value	Theorem: a	continuous	function	on	a
      closed	interval	must	ach...
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Lesson 16: Maximum and Minimum Values

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Lesson 16: Maximum and Minimum Values

  1. 1. Section 4.1 Maximum and Minimum Values V63.0121.006/016, Calculus I March 23, 2010 Announcements Welcome back from Spring Break! Quiz 3: April 2, Sections 2.6–3.5 . . . . . .
  2. 2. Announcements Welcome back from Spring Break! Quiz 3: April 2, Sections 2.6–3.5 . . . . . .
  3. 3. Computation of Midterm Letter Grades Q . uizzes . W . ebAssign . 1 . 5% 1 . 0% . . W H M . idterm . 1 . 0% 2 . 5% . 4 . 0% . F . inal . . . . . .
  4. 4. Computation of Midterm Letter Grades Q . uizzes . W . ebAssign . 1 . 5% 1 . 0% . . W H M . idterm . 1 . 0% 2 . 5% . 4 . 0% . F . inal (to be determined) . . . . . .
  5. 5. Computation of Midterm Letter Grades W . ebAssign . 1 . 7% H . W . Q . uizzes 1 . 7% . 2 . 5% . 4 . 1% . M . idterm . . . . . .
  6. 6. Distribution of Midterm Averages and Letter Grades Median = 81.35% (curved to B) Average = 74.39% (curved to B-) Standard Deviation = 21% . . . . . .
  7. 7. What can I do to improve my grade? Q . uizzes . W . ebAssign . 1 . 5% 1 . 0% . . W H M . idterm . 1 . 0% 2 . 5% . 4 . 0% . F . inal . . . . . .
  8. 8. What can I do to improve my grade? f .uture Quizzes p . ast Quizzes . . f .uture WA . p . ast WA 6 .% . 9 .% 5 .% 5 .% f .uture HW . 5 .% M . idterm . p . ast HW . 2 . 5% 5 .% . 4 . 0% . F . inal . . . . . .
  9. 9. What can I do to improve my grade? M . idterm . p . ast Quizzes . 2 . 5% 6 .% p . ast WA f .uture HW . . 5 .% 5 .% p . ast HW f .uture WA . . 5 .% 5 .% . . .% 9 f .uture Quizzes 4 . 0% . F . inal 59% of your grade is still in play! . . . . . .
  10. 10. Help! Free resources: recitation TAs’ office hours my office hours Math Tutoring Center (CIWW 524) College Learning Center . . . . . .
  11. 11. Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples Challenge: Cubic functions . . . . . .
  12. 12. Optimize . . . . . .
  13. 13. Why go to the extremes? Rationally speaking, it is advantageous to find the extreme values of a function (maximize profit, minimize costs, etc.) Pierre-Louis Maupertuis (1698–1759) . . . . . .
  14. 14. Design . Image credit: Jason Tromm . . . . . . .
  15. 15. Why go to the extremes? Rationally speaking, it is advantageous to find the extreme values of a function (maximize profit, minimize costs, etc.) Many laws of science are derived from minimizing principles. Pierre-Louis Maupertuis (1698–1759) . . . . . .
  16. 16. Optics . . Image credit: jacreative . . . . . .
  17. 17. Why go to the extremes? Rationally speaking, it is advantageous to find the extreme values of a function (maximize profit, minimize costs, etc.) Many laws of science are derived from minimizing principles. Maupertuis’ principle: “Action is minimized through the wisdom of God.” Pierre-Louis Maupertuis (1698–1759) . . . . . .
  18. 18. Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples Challenge: Cubic functions . . . . . .
  19. 19. Extreme points and values Definition Let f have domain D. . . Image credit: Patrick Q . . . . . .
  20. 20. Extreme points and values Definition Let f have domain D. The function f has an absolute maximum (or global maximum) (respectively, absolute minimum) at c if f(c) ≥ f(x) (respectively, f(c) ≤ f(x)) for all x in D . . Image credit: Patrick Q . . . . . .
  21. 21. Extreme points and values Definition Let f have domain D. The function f has an absolute maximum (or global maximum) (respectively, absolute minimum) at c if f(c) ≥ f(x) (respectively, f(c) ≤ f(x)) for all x in D The number f(c) is called the maximum value (respectively, minimum value) of f on D. . . Image credit: Patrick Q . . . . . .
  22. 22. Extreme points and values Definition Let f have domain D. The function f has an absolute maximum (or global maximum) (respectively, absolute minimum) at c if f(c) ≥ f(x) (respectively, f(c) ≤ f(x)) for all x in D The number f(c) is called the maximum value (respectively, minimum value) of f on D. An extremum is either a maximum or a minimum. An extreme value is . either a maximum value or minimum value. . Image credit: Patrick Q . . . . . .
  23. 23. Theorem (The Extreme Value Theorem) Let f be a function which is continuous on the closed interval [a, b]. Then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at numbers c and d in [a, b]. . . . . . .
  24. 24. Theorem (The Extreme Value Theorem) Let f be a function which is continuous on the closed interval [a, b]. Then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at numbers c and d in [a, b]. . . . . a . b . . . . . . .
  25. 25. Theorem (The Extreme Value Theorem) Let f be a function which is continuous on the closed interval [a, b]. Then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at numbers c and d in [a, b]. . maximum .(c) f . value . . minimum .(d) f . value . . .. a . d c b . minimum maximum . . . . . .
  26. 26. No proof of EVT forthcoming This theorem is very hard to prove without using technical facts about continuous functions and closed intervals. But we can show the importance of each of the hypotheses. . . . . . .
  27. 27. Bad Example #1 Example Consider the function { x 0≤x<1 f(x ) = x − 2 1 ≤ x ≤ 2. . . . . . .
  28. 28. Bad Example #1 Example Consider the function . { x 0≤x<1 f(x ) = . . | . x − 2 1 ≤ x ≤ 2. 1 . . . . . . . .
  29. 29. Bad Example #1 Example Consider the function . { x 0≤x<1 f(x ) = . . | . x − 2 1 ≤ x ≤ 2. 1 . . Then although values of f(x) get arbitrarily close to 1 and never bigger than 1, 1 is not the maximum value of f on [0, 1] because it is never achieved. . . . . . .
  30. 30. Bad Example #1 Example Consider the function . { x 0≤x<1 f(x ) = . . | . x − 2 1 ≤ x ≤ 2. 1 . . Then although values of f(x) get arbitrarily close to 1 and never bigger than 1, 1 is not the maximum value of f on [0, 1] because it is never achieved. This does not violate EVT because f is not continuous. . . . . . .
  31. 31. Bad Example #2 Example Consider the function f(x) = x restricted to the interval [0, 1). . . . . . .
  32. 32. Bad Example #2 Example Consider the function f(x) = x restricted to the interval [0, 1). . . . | 1 . . . . . . .
  33. 33. Bad Example #2 Example Consider the function f(x) = x restricted to the interval [0, 1). . . . | 1 . There is still no maximum value (values get arbitrarily close to 1 but do not achieve it). . . . . . .
  34. 34. Bad Example #2 Example Consider the function f(x) = x restricted to the interval [0, 1). . . . | 1 . There is still no maximum value (values get arbitrarily close to 1 but do not achieve it). This does not violate EVT because the domain is not closed. . . . . . .
  35. 35. Final Bad Example Example 1 Consider the function f(x) = is continuous on the closed x interval [1, ∞). . . . . . .
  36. 36. Final Bad Example Example 1 Consider the function f(x) = is continuous on the closed x interval [1, ∞). . . . 1 . . . . . . .
  37. 37. Final Bad Example Example 1 Consider the function f(x) = is continuous on the closed x interval [1, ∞). . . . 1 . There is no minimum value (values get arbitrarily close to 0 but do not achieve it). . . . . . .
  38. 38. Final Bad Example Example 1 Consider the function f(x) = is continuous on the closed x interval [1, ∞). . . . 1 . There is no minimum value (values get arbitrarily close to 0 but do not achieve it). This does not violate EVT because the domain is not bounded. . . . . . .
  39. 39. Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples Challenge: Cubic functions . . . . . .
  40. 40. Local extrema Definition A function f has a local maximum or relative maximum at c if f(c) ≥ f(x) when x is near c. This means that f(c) ≥ f(x) for all x in some open interval containing c. Similarly, f has a local minimum at c if f(c) ≤ f(x) when x is near c. . . . . . .
  41. 41. Local extrema Definition A function f has a local maximum or relative maximum at c if f(c) ≥ f(x) when x is near c. This means that f(c) ≥ f(x) for all x in some open interval containing c. Similarly, f has a local minimum at c if f(c) ≤ f(x) when x is near c. . . . . .... | a local . . | local b . maximum minimum . . . . . .
  42. 42. Local extrema So a local extremum must be inside the domain of f (not on the end). A global extremum that is inside the domain is a local extremum. . . . . .... . a local . local and global . . | | b global maximum min max . . . . . .
  43. 43. Theorem (Fermat’s Theorem) Suppose f has a local extremum at c and f is differentiable at c. Then f′ (c) = 0. . . . . .... | a local . . | local b . maximum minimum . . . . . .
  44. 44. Theorem (Fermat’s Theorem) Suppose f has a local extremum at c and f is differentiable at c. Then f′ (c) = 0. . . . . .... | a local . . | local b . maximum minimum . . . . . .
  45. 45. Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. . . . . . .
  46. 46. Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. If x is slightly greater than c, f(x) ≤ f(c). This means f(x) − f(c) ≤0 x−c . . . . . .
  47. 47. Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. If x is slightly greater than c, f(x) ≤ f(c). This means f(x) − f(c) f(x) − f(c) ≤ 0 =⇒ lim ≤0 x−c x →c + x−c . . . . . .
  48. 48. Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. If x is slightly greater than c, f(x) ≤ f(c). This means f(x) − f(c) f(x) − f(c) ≤ 0 =⇒ lim ≤0 x−c x →c + x−c The same will be true on the other end: if x is slightly less than c, f(x) ≤ f(c). This means f(x) − f(c) ≥0 x−c . . . . . .
  49. 49. Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. If x is slightly greater than c, f(x) ≤ f(c). This means f(x) − f(c) f(x) − f(c) ≤ 0 =⇒ lim ≤0 x−c x →c + x−c The same will be true on the other end: if x is slightly less than c, f(x) ≤ f(c). This means f(x) − f(c) f(x) − f(c) ≥ 0 =⇒ lim ≥0 x−c x →c − x−c . . . . . .
  50. 50. Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. If x is slightly greater than c, f(x) ≤ f(c). This means f(x) − f(c) f(x) − f(c) ≤ 0 =⇒ lim ≤0 x−c x →c + x−c The same will be true on the other end: if x is slightly less than c, f(x) ≤ f(c). This means f(x) − f(c) f(x) − f(c) ≥ 0 =⇒ lim ≥0 x−c x →c − x−c f(x) − f(c) Since the limit f′ (c) = lim exists, it must be 0. x →c x−c . . . . . .
  51. 51. Meet the Mathematician: Pierre de Fermat 1601–1665 Lawyer and number theorist Proved many theorems, didn’t quite prove his last one . . . . . .
  52. 52. Tangent: Fermat’s Last Theorem Plenty of solutions to x2 + y2 = z2 among positive whole numbers (e.g., x = 3, y = 4, z = 5) . . . . . .
  53. 53. Tangent: Fermat’s Last Theorem Plenty of solutions to x2 + y2 = z2 among positive whole numbers (e.g., x = 3, y = 4, z = 5) No solutions to x3 + y3 = z3 among positive whole numbers . . . . . .
  54. 54. Tangent: Fermat’s Last Theorem Plenty of solutions to x2 + y2 = z2 among positive whole numbers (e.g., x = 3, y = 4, z = 5) No solutions to x3 + y3 = z3 among positive whole numbers Fermat claimed no solutions to xn + yn = zn but didn’t write down his proof . . . . . .
  55. 55. Tangent: Fermat’s Last Theorem Plenty of solutions to x2 + y2 = z2 among positive whole numbers (e.g., x = 3, y = 4, z = 5) No solutions to x3 + y3 = z3 among positive whole numbers Fermat claimed no solutions to xn + yn = zn but didn’t write down his proof Not solved until 1998! (Taylor–Wiles) . . . . . .
  56. 56. Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples Challenge: Cubic functions . . . . . .
  57. 57. Flowchart for placing extrema Thanks to Fermat Suppose f is a continuous function on the closed, bounded interval [a, b], and c is a global maximum point. . . . c is a start local max . . . Is c an Is f diff’ble f is not n .o n .o endpoint? at c? diff at c y . es y . es . c = a or . f′ (c) = 0 c = b . . . . . .
  58. 58. The Closed Interval Method This means to find the maximum value of f on [a, b], we need to: Evaluate f at the endpoints a and b Evaluate f at the critical points or critical numbers x where either f′ (x) = 0 or f is not differentiable at x. The points with the largest function value are the global maximum points The points with the smallest or most negative function value are the global minimum points. . . . . . .
  59. 59. Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples Challenge: Cubic functions . . . . . .
  60. 60. Example Find the extreme values of f(x) = 2x − 5 on [−1, 2]. . . . . . .
  61. 61. Example Find the extreme values of f(x) = 2x − 5 on [−1, 2]. Solution Since f′ (x) = 2, which is never zero, we have no critical points and we need only investigate the endpoints: f(−1) = 2(−1) − 5 = −7 f(2) = 2(2) − 5 = −1 . . . . . .
  62. 62. Example Find the extreme values of f(x) = 2x − 5 on [−1, 2]. Solution Since f′ (x) = 2, which is never zero, we have no critical points and we need only investigate the endpoints: f(−1) = 2(−1) − 5 = −7 f(2) = 2(2) − 5 = −1 So The absolute minimum (point) is at −1; the minimum value is −7. The absolute maximum (point) is at 2; the maximum value is −1. . . . . . .
  63. 63. Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. . . . . . .
  64. 64. Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solution We have f′ (x) = 2x, which is zero when x = 0. . . . . . .
  65. 65. Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solution We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = f(0) = f(2) = . . . . . .
  66. 66. Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solution We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = 0 f(0) = f(2) = . . . . . .
  67. 67. Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solution We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = 0 f(0) = − 1 f(2) = . . . . . .
  68. 68. Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solution We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = 0 f(0) = − 1 f(2) = 3 . . . . . .
  69. 69. Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solution We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = 0 f(0) = − 1 (absolute min) f(2) = 3 . . . . . .
  70. 70. Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solution We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = 0 f(0) = − 1 (absolute min) f(2) = 3 (absolute max) . . . . . .
  71. 71. Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. . . . . . .
  72. 72. Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. . . . . . .
  73. 73. Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = f(0) = f(1) = f(2) = . . . . . .
  74. 74. Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 f(0) = f(1) = f(2) = . . . . . .
  75. 75. Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 f(0) = 1 f(1) = f(2) = . . . . . .
  76. 76. Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 f(0) = 1 f(1) = 0 f(2) = . . . . . .
  77. 77. Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 f(0) = 1 f(1) = 0 f(2) = 5 . . . . . .
  78. 78. Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 (global min) f(0) = 1 f(1) = 0 f(2) = 5 . . . . . .
  79. 79. Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 (global min) f(0) = 1 f(1) = 0 f(2) = 5 (global max) . . . . . .
  80. 80. Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 (global min) f(0) = 1 (local max) f(1) = 0 f(2) = 5 (global max) . . . . . .
  81. 81. Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 (global min) f(0) = 1 (local max) f(1) = 0 (local min) f(2) = 5 (global max) . . . . . .
  82. 82. Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. . . . . . .
  83. 83. Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x + x = x (5x + 4) 3 3 3 Thus f′ (−4/5) = 0 and f is not differentiable at 0. . . . . . .
  84. 84. Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x + x = x (5x + 4) 3 3 3 Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check are: f(−1) = f(−4/5) = f(0) = f(2) = . . . . . .
  85. 85. Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x + x = x (5x + 4) 3 3 3 Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check are: f(−1) = 1 f(−4/5) = f(0) = f(2) = . . . . . .
  86. 86. Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x + x = x (5x + 4) 3 3 3 Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check are: f(−1) = 1 f(−4/5) = 1.0341 f(0) = f(2) = . . . . . .
  87. 87. Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x + x = x (5x + 4) 3 3 3 Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check are: f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 f(2) = . . . . . .
  88. 88. Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x + x = x (5x + 4) 3 3 3 Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check are: f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 f(2) = 6.3496 . . . . . .
  89. 89. Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x + x = x (5x + 4) 3 3 3 Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check are: f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 (absolute min) f(2) = 6.3496 . . . . . .
  90. 90. Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x + x = x (5x + 4) 3 3 3 Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check are: f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 (absolute min) f(2) = 6.3496 (absolute max) . . . . . .
  91. 91. Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x + x = x (5x + 4) 3 3 3 Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check are: f(−1) = 1 f(−4/5) = 1.0341 (relative max) f(0) = 0 (absolute min) f(2) = 6.3496 (absolute max) . . . . . .
  92. 92. Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. . . . . . .
  93. 93. Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solution x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differentiable at ±2 as well.) . . . . . .
  94. 94. Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solution x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differentiable at ±2 as well.) So our points to check are: f(−2) = f(0) = f(1) = . . . . . .
  95. 95. Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solution x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differentiable at ±2 as well.) So our points to check are: f(−2) = 0 f(0) = f(1) = . . . . . .
  96. 96. Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solution x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differentiable at ±2 as well.) So our points to check are: f(−2) = 0 f(0) = 2 f(1) = . . . . . .
  97. 97. Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solution x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differentiable at ±2 as well.) So our points to check are: f(−2) = 0 f(0) = 2 √ f(1) = 3 . . . . . .
  98. 98. Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solution x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differentiable at ±2 as well.) So our points to check are: f(−2) = 0 (absolute min) f(0) = 2 √ f(1) = 3 . . . . . .
  99. 99. Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solution x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differentiable at ±2 as well.) So our points to check are: f(−2) = 0 (absolute min) f(0) = 2 (absolute max) √ f(1) = 3 . . . . . .
  100. 100. Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples Challenge: Cubic functions . . . . . .
  101. 101. Challenge: Cubic functions Example How many critical points can a cubic function f(x) = ax3 + bx2 + cx + d have? . . . . . .
  102. 102. Solution If f′ (x) = 0, we have 3ax2 + 2bx + c = 0, and so √ √ −2b ± 4b2 − 12ac −b ± b2 − 3ac x= = , 6a 3a and so we have three possibilities: . . . . . .
  103. 103. Solution If f′ (x) = 0, we have 3ax2 + 2bx + c = 0, and so √ √ −2b ± 4b2 − 12ac −b ± b2 − 3ac x= = , 6a 3a and so we have three possibilities: b2 − 3ac > 0, in which case there are two distinct critical points. An example would be f(x) = x3 + x2 , where a = 1, b = 1, and c = 0. . . . . . .
  104. 104. Solution If f′ (x) = 0, we have 3ax2 + 2bx + c = 0, and so √ √ −2b ± 4b2 − 12ac −b ± b2 − 3ac x= = , 6a 3a and so we have three possibilities: b2 − 3ac > 0, in which case there are two distinct critical points. An example would be f(x) = x3 + x2 , where a = 1, b = 1, and c = 0. b2 − 3ac < 0, in which case there are no real roots to the quadratic, hence no critical points. An example would be f(x) = x3 + x2 + x, where a = b = c = 1. . . . . . .
  105. 105. Solution If f′ (x) = 0, we have 3ax2 + 2bx + c = 0, and so √ √ −2b ± 4b2 − 12ac −b ± b2 − 3ac x= = , 6a 3a and so we have three possibilities: b2 − 3ac > 0, in which case there are two distinct critical points. An example would be f(x) = x3 + x2 , where a = 1, b = 1, and c = 0. b2 − 3ac < 0, in which case there are no real roots to the quadratic, hence no critical points. An example would be f(x) = x3 + x2 + x, where a = b = c = 1. b2 − 3ac = 0, in which case there is a single critical point. Example: x3 , where a = 1 and b = c = 0. . . . . . .
  106. 106. What have we learned today? The Extreme Value Theorem: a continuous function on a closed interval must achieve its max and min Fermat’s Theorem: local extrema are critical points The Closed Interval Method: an algorithm for finding global extrema Show your work unless you want to end up like Fermat! . . . . . .

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