Chapter12.1 - 2 <-= +-= +-= f , so f is...

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Chapter 12.1 Page 647 Definition: If 0 ) ( ' = c f , for some number c , then c is a critical value. Example 2. (A) 3 1 ) ( x x f + = 2 3 3 )' ( )' 1 ( ) ( ' x x x f = + = ; the solution to 0 ) ( ' = x f is the critical point. So solve 0 0 3 2 = = x x f . Page 651 Theorem 2. If ) ( x f is continuous at c and the point )) ( , ( c f c is an extremum, then 0 ) ( ' = c f . Page 653 Example 7. 1 9 6 ) ( 2 3 + + - = x x x x f (A) Find the critical value(s) of f ; 9 12 3 )' 1 ( )' 9 ( )' 6 ( )' ( ) ( ' 2 2 3 + - = + + - = x x x x x x f Solve 3 , 1 0 ) 3 )( 1 ( 3 0 ) 3 4 ( 3 0 9 12 3 2 2 = = = - - = + - = + - x x x x x x x x f f f . Remark: (i) If , 0 ) ( ' x f x in ( a , b ), then f is increasing over ( a , b ). (ii) If , 0 ) ( ' x f x in ( a , b ), then f is decreasing over ( a , b ). 1
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(B) Find the local maxima and minima of f ; 1 is in ] 1 , 0 [ , 0 9 9 ) 0 ( 12 ) 0 ( 3 ) 0 ( ' 2 = + - = f and 0 ) 1 ( ' = f , so f is increasing over (0, 1); {from the left hand side of 1}. 1 is in [1, 2], 0 ) 1 ( ' = f and 0 3 9 24 12 9 ) 2 ( 12 ) 2 ( 3 ) 2 ( '
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Unformatted text preview: 2 <-= +-= +-= f , so f is decreasing over (1, 2); {from the right hand side of 1}. So f has a local maxima at x = 1. 3 is in [2, 3], 3 9 24 12 9 ) 2 ( 12 ) 2 ( 3 ) 2 ( ' 2 <-= +-= +-= f and ) 3 ( ' = f , so f is decreasing over (2, 3); {from the left hand side of 3}. 3 is in [3, 4], ) 3 ( ' = f and 9 9 48 48 9 ) 4 ( 12 ) 4 ( 3 ) 4 ( ' 2 = +-= +-= f , so f is increasing over (3, 4); {from the right hand side of 3}. So f has a local minima at x = 3. 2...
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This note was uploaded on 04/09/2008 for the course MAT 220 taught by Professor For during the Spring '08 term at Community College of Allegheny County.

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Chapter12.1 - 2 <-= +-= +-= f , so f is...

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