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Unformatted text preview: Lecture 9  Increasing and Decreasing Functions, Extrema, and the First Derivative Test 9.1 Increasing and Decreasing Functions One of our goals is to be able to solve max/min problems, especially economics related examples. We start with the following definitions: Definition 9.1 A function f is called increasing on an interval ( a,b ) if for any x 1 ,x 2 ( a,b ) , we have that x 1 < x 2 f ( x 1 ) < f ( x 2 ) A function f is called decreasing on an interval ( a,b ) if for any x 1 ,x 2 ( a,b ) , we have that x 1 < x 2 f ( x 1 ) > f ( x 2 ) Note that some books call these strictly increasing and strictly decreasing respectively. Examples: 1. f ( x ) = x 2 . f ( x ) is increasing on (0 , ) and decreasing ( , 0). 2. f ( x ) = x 3 . f ( x ) is increasing on ( , ). As you may have guessed, we can use the derivative to test for increasing/decreasing. Let f be differentiable on the interval ( a,b ). 1. If f ( x ) > 0 for all x in ( a,b ), then f is increasing on ( a,b ). 2. If f ( x ) < 0 for all x in ( a,b ), then f is decreasing on ( a,b ). 3. If f ( x ) = 0 for all x in ( a,b ), then f must be constant on ( a,b ). FACT: If f is a continuous function, then f ( x ) can only change signs at values of x where f ( x ) = 0 or f ( x ) doesnt exist. Examples: 1. Let f ( x ) = x 3 4 3 x . Find the intervals on which f ( x ) is increasing or decreasing. solution: We will use the test and fact above. The derivative is f ( x ) = 3 x 2 4 3 = 3 4 ( x 2 4) = 3 4 ( x 2)( x + 2) Because of the above, we split our real line up where f ( x ) = 0: f ( x ) % & % 2 2 On ( , 2), ( x 2) and ( x + 2) are both negative, so f ( x ) = 3 4 ( x 2)( x + 2) > 0, so f is increasing. On ( 2 , 2), ( x 2) is negative while ( x +2) is positive, so f ( x ) = 3 4 ( x 2)( x +2) < 0, so f is decreasing. On (2 , ), ( x 2) and ( x + 2) are both positive, so f ( x ) = 3 4 ( x 2)( x + 2) > 0, so f is increasing. To summarize, f ( x ) is increasing on ( , 2) and (2 , ) while f ( x ) is decreasing on ( 2 , 2). 2. Let g ( x ) = x + 32 x 2 . Find the intervals on which g ( x ) is increasing and decreasing. solution: We first need to find the derivative. g ( x ) = 1 64 x 3 Note that g ( x ) = 0 when x 3 = 64 x = 4. The derivative is also not defined when x = 0, so these are the places we divide our real line. g ( x ) % & % 4 To find out what sign the derivative is on ( , 0), we can substitute in a test point in that interval and see if the result is positive or negative. The point x = 1 is a convenient one to use: g ( 1) = 1 64 ( 1) 3 = 65 > Because of our fact,...
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This note was uploaded on 01/07/2010 for the course MAT mat1300 taught by Professor Pieterhofstra during the Fall '09 term at University of Ottawa.
 Fall '09
 PIETERHOFSTRA
 Derivative

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