Todays Agenda Upcoming Homework What to study for Exam 2 Section - PowerPoint PPT Presentation

Today’s Agenda • Upcoming Homework • What to study for Exam 2 • Section 4.1: Maximum and Minimum Values Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Monday, 26 October 2015 1 / 10

Upcoming Homework • Study for Exam 2 on 10/28!!! • WeBWorK HW #17: Section 4.1, due 10/30/2015 • Written HW I: Section 3.7, #2,4,6,10,12,28,38. Section 4.1, #28,32,36,46,48,50. Due 11/2/2015. • WeBWorK HW #18: Section 4.2, due 11/4/2015 • Written HW J: Section 4.2, #10,12,16,24. Due 11/6/2015. Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Monday, 26 October 2015 2 / 10

What to study for Exam 2 1 Product Rule, Chain Rule, Quotient Rule 2 Implicit differentiation, slope of the tangent line at a given point 3 Related rates problems 4 Differentials and how to evaluate them 5 L’Hospital’s Rule 6 Inverse Function Theorem 7 Linear approximation centered at a point a Suggested practice problems: 1 Pages 142-143, #43-46,52-60,64,67-68,72-75. 2 Page 199, True/False #1-16. 3 Pages 200-201, #3-4,49-50,61-77. 4 Also review derivatives from the Mastery Test review sheet, and look at the exam review problems and old exams at: https://math.asu. edu/first-year-math/mat-265-calculus-engineers-i Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Monday, 26 October 2015 3 / 10

Section 4.1 Definition 4.1.1 Let c be a number in the domain D of a function f . Then f ( c ) is the 1 absolute maximum value of f on D if f ( c ) ≥ f ( x ) for all x ∈ D . 2 absolute minimum value of f on D if f ( c ) ≤ f ( x ) for all x ∈ D . The absolute maximum and minimum values of f are called extreme values of f . Definition 4.1.2 Let c be a number in the domain D of a function f . Then f ( c ) is a 1 local maximum value of f if f ( c ) ≥ f ( x ) whenever x is sufficiently near to c . 2 local minimum value of f if f ( c ) ≤ f ( x ) whenever x is sufficiently near to c . Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Monday, 26 October 2015 4 / 10

Section 4.1 The Extreme Value Theorem If f is continuous on a closed interval [ a , b ], then f attains an absolute minimum value f ( c ) and an absolute maximum value f ( d ) at some numbers c and d in the interval [ a , b ]. Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Monday, 26 October 2015 5 / 10

Section 4.1 Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Monday, 26 October 2015 6 / 10

Section 4.1 Fermat’s Theorem If f has a local maximum or minimum at c , and if f ′ ( c ) exists, then f ′ ( c ) = 0. Long Side Note: This is NOT the same as Fermat’s Last Theorem, which you may have heard of. The statement of Fermat’s Last Theorem is simple: if n > 2, there are no integer solutions to the equation a n + b n = c n . Around 1637, Fermat wrote a note in his copy of Diophantus’ Arithmetica : ”I have discovered a truly marvellous proof of this, which this margin is too narrow to contain.” In fact, the proof of Fermat’s Last Theorem incredibly difficult; Andrew Wiles famously proved it in a series of lectures and peer-reviewed papers in the mid-1990’s. It was no small task. He essentially locked himself away in his office for six years, only telling his wife about his work. After presenting his results, an error was found and he spent another two years fixing it. In other words, it took three and a half centuries of modern mathematics followed by the better part of a decade’s worth of work from one of the world’s most brilliant human beings to solve Fermat’s Last Theorem. If you’re curious about the story, there’s an absolutely fantastic book about it called Fermat’s Enigma by Simon Singh. Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Monday, 26 October 2015 7 / 10

Section 4.1 Back to the theorem of Fermat that we are concerned with in our class (the proof of which is much shorter and much easier than the one just discussed)... If the function f has a local maximum at c , then for small (positive or negative) values of the number h , f ( c ) ≥ f ( c + h ). If h > 0, then f ( c + h ) − f ( c ) ≤ 0, so h f ( c + h ) − f ( c ) f ′ ( c ) = lim ≤ 0 . h h → 0 + If h < 0, then f ( c + h ) − f ( c ) ≥ 0, so h f ( c + h ) − f ( c ) f ′ ( c ) = lim ≥ 0 . h h → 0 − The only way for the statement 0 ≤ f ′ ( c ) ≤ 0 to be true is if f ′ ( c ) = 0. Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Monday, 26 October 2015 8 / 10

Section 4.1 The converse of Fermat’s Theorem is not necessarily true. For example, if f ( x ) = x 3 , then f ′ (0) = 3 · 0 2 = 0, but 0 is not a local maximum or minimum of f ( x ). However, finding a derivative and setting it equal to 0 is a good place to start looking for local maxima or local minima. Definition 4.1.3 A critical number of a function f is a number c in the domain D of f such that either f ′ ( c ) = 0 or f ′ ( c ) does not exist. Note that if f has a local maximum or minimum at c , then c is a critical number of f . Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Monday, 26 October 2015 9 / 10

Section 4.1 Example Problems Find the absolute maximum and absolute minimum values of f on the given interval: 1 f ( x ) = 12 + 4 x − x 2 , [0 , 5] 2 f ( x ) = 2 x 3 − 3 x 2 − 12 x + 1, [ − 2 , 3] 3 f ( x ) = 3 x 4 − 4 x 3 − 12 x 2 + 1, [ − 2 , 3] √ 4 f ( t ) = t 4 − t 2 , [ − 1 , 2] 5 f ( t ) = 2 cos( t ) + sin(2 t ), [0 , π/ 2] 6 f ( x ) = xe − x 2 / 8 , [ − 1 , 4] 7 f ( x ) = ln( x 2 + x + 1), [ − 1 , 1] Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Monday, 26 October 2015 10 / 10