This theorem guarantees solutions to many problems you will encounter. exists, then f ( c)
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1 Maimum and Minimum Values Etreme Value Theorem If f () is continuous on the closed interval [a, b], then f () achieves both a global (absolute) maimum and global minimum at some numbers c and d in [a, b]. This theorem guarantees solutions to many problems you will encounter. Fermat s Theorem If f () has a local maimum or minimum value at c, and if f (c) eists, then f ( c) = 0 MATH 180 Lecture 17 1 of 15 Ronald Brent 018 All rights reserved.
2 Recall: A critical number (point) of a function f is a number c in the domain of f such that either f ( c) = 0 or f (c) does not eist. Eample: Find all critical numbers for the function f ( ) = +. Compute f ( ) = + 1. So the only critical number is = Eample: Find all critical points for the function f ( ) = 18. f ( ) = = 0, dividing by, we need to solve 5 6 = 0. Factoring is easiest so rewrite this as ( + 1)( 6) = 0. Critical points are = 1, and = 6. MATH 180 Lecture 17 of 15 Ronald Brent 018 All rights reserved.
3 Eample: Find all critical numbers for the function Compute f ( ) = 4 9 = (4 9). 9 So the critical numbers are = 0,. 4 f ( ) 4 =. Eample: Find all critical points for the function f ( ) 54 = +. f 54 ( ) =, so = 0 is a critical point ( f () undefined) Also = 0 = = 7 = So, the critical points are = 0,. MATH 180 Lecture 17 of 15 Ronald Brent 018 All rights reserved.
4 The Closed Interval Method To find the absolute maimum and minimum values of a continuous function f on a closed interval [a, b]: 1) Find the values of f at the critical numbers of f in (a, b) ) Find the values of f at the endpoints of the interval. ) The largest value from steps 1 and is the absolute maimum value, and the smallest of these values is the absolute minimum value. MATH 180 Lecture 17 4 of 15 Ronald Brent 018 All rights reserved.
5 Eample: Find the absolute maimum and minimum values for the function f ( ) = + 4 on [ 1, 5]. For critical numbers = 0. Solving the equation we get =, and = 4. We consider only =, since = 4 is not in the interval [ 1, 5]. Now check the values: f ( 1) = 6, f ( ) = 8, and f ( 5) = 80 and so the absolute maimum is 80, and the absolute minimum is -8 y f ( ) = MATH 180 Lecture 17 5 of 15 Ronald Brent 018 All rights reserved.
6 Eample: Find the absolute maimum and minimum values for the function f ( ) = on [ 1, 5]. 1 First note that f is not continuous in this interval. It may or may not have a maimum or minimum value. Critical numbers are = 0, 1, and. However, in this case, the function has a vertical asymptote at = 1. So there is no maimum or minimum values. MATH 180 Lecture 17 6 of 15 Ronald Brent 018 All rights reserved.
7 Eample: Find the absolute maimum and minimum values 4 for the function f ( ) = 8 on [ 0,]. y f ( ) = 4 16 = 4( 4) = 4( + )( ) = 0 gives = 0, ±. We consider = 0,, and. f ( 0) = 0 f ( ) = 16, and f ( ) = 9. So the maimum value is 9 and minimum is MATH 180 Lecture 17 7 of 15 Ronald Brent 018 All rights reserved.
8 Eample: Find the absolute maimum and minimum values for the function f ( ) = on [ 1, ]. f ( ) = 6 = ( ) = 0 gives = 0, as critical points. We consider = 1, 0,, and. f ( 1) = 4 f ( 0) = 0 f ( ) = 4, and f ( ) = 0. So the maimum is 0 and minimum is 4. y MATH 180 Lecture 17 8 of 15 Ronald Brent 018 All rights reserved.
9 Eample: Find the absolute maimum and minimum values for the function f ( ) = + sin on [ 0, π ]. f ( ) = 1+ cos, and so critical numbers are 1 determined by 1 + cos = 0, or cos =. This π occurs for = in the given interval. Now checking all the values: π π π π π f = + sin = + = + Checking the endpoints f ( 0) = 0 + sin(0) = 0, and f ( π ) = π + sin( π ) = π so the absolute min is 0, and ma is π +..8 MATH 180 Lecture 17 9 of 15 Ronald Brent 018 All rights reserved.
10 y MATH 180 Lecture of 15 Ronald Brent 018 All rights reserved.
11 Eample: Find the absolute maimum and minimum values 7 for the function f ( ) = + on [ 1,9 ]. 7 f ( ) = gives = ± 6. We consider =1,6,9. f ( 1) = 74 f ( 6) = 4, and f ( 9) = 6. So the maimum is 74 and minimum is 4. y MATH 180 Lecture of 15 Ronald Brent 018 All rights reserved.
12 Eample: Find the absolute maimum and minimum values for the function f ( ) = on [ 1, ]. + 1 f ( ) = = 0 gives = 0 ( + 1) 1-1, and. f ( 1) = f ( 0) = 0 and f maimum is 4/5 and minimum is 0.. We consider = 0, ( ) = 4. So the 5 y MATH 180 Lecture 17 1 of 15 Ronald Brent 018 All rights reserved.
13 Eample: Find the absolute maimum and minimum values 4 for the function f ( ) = on [ 0,]. 1 f ( ) = 1 4 so that critical points are = ±. We 1 consider = 0 and 1/. f ( 0) = 0 and f ( 1/ ) = and f ( ) = so is the maimum value and is the minimum value. y MATH 180 Lecture 17 1 of 15 Ronald Brent 018 All rights reserved.
14 Eample: Find the absolute maimum and minimum values for the function f ( ) = on [ 0,4]. f ( ) = 1 so that critical points are = 0, 4. f ( 0) = 10 and f ( 4) = so 10 is the maimum value and is the minimum value. y MATH 180 Lecture of 15 Ronald Brent 018 All rights reserved.
15 Eample: Find the absolute maimum and minimum values 4 / 4 for the function f ( ) = on [ 0,]. 1 f ( ) = 1 so that only critical point is = 0. We consider = 0 and. f ( 0) = 0 and f ( ) = and So is the minimum value and is the maimum value MATH 180 Lecture of 15 Ronald Brent 018 All rights reserved.
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