chapter 12 homework problems

# chapter 12 homework problems - Chapter 12: Applications of...

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Chapter 12: Applications of the Derivative Section 12.1 Derivatives and Graphs 1. The function is increasing on (1, ), and decreasing on (– , 1). The lowest point on the graph has coordinates (1, –4), so the local minimum of –4 occurs at x = 1. 2. The function is increasing on (– , 4) and decreasing on (4, ). The highest point on the graph has coordinates (4, 1), so the local maximum of 1 occurs at x = 4. 3. The function is increasing on (– , –2) and decreasing on (–2, ). The highest point on the graph has coordinates (–2, 3), so a local maximum of 3 occurs at x = –2. 4. The function is increasing on (3, ) and decreasing on (– , 3). The lowest point has coordinates (3, –4), so a local minimum of –4 occurs at x = 3. 5. The function is increasing on (– , –4) and (–2, ), and decreasing on (–4, –2). A relatively high point occurs at (–4, 3) and a relatively low point occurs at (–2, 1), so a local maximum of 3 occurs at x = –4 and a local minimum of 1 occurs at x = –2. 6. The function is increasing on (1, 5), and decreasing on (– , 1) and (5, ). A relatively low point occurs at (1, –6) and a relatively high point occurs at (5, 2), so a local minimum of –6 occurs at x = 1 and a local maximum of 2 occurs at x = 5. 7. The function is increasing on (–7, –4) and (–2, ) and decreasing on (– , –7) and (–4, –2). Relatively low points occur at (–7, –2) and (–2, –2), and relatively high point occurs at (–4, 3). Thus a local minimum of –2 occurs at x = –7 and x = –2, and a local maximum of 3 occurs at x = – 4. 8. The function is increasing on (–3, 0) and (3, ) and decreasing on (– , –3) and (0, 3). Relatively low points occur at (–3, 0) and (3, 0) and a relatively high point occurs at (0, 4). Thus a local minimum of 0 occurs at x = –3 and x = 3, and a local maximum of 4 occurs at x = 0. 9. 32 2 () 2 5 4 2 () 6 1 0 4 00 . ¡ fx x x x x x +, 2 2 61 0 4 0 23 5 2 0 2(3 1)( 2) 0 00 .0 xx 1 or 2 3 0 Test ( ) ¡ f x at x = –2, x = 0, x = 3. 2 2 2 (2 ) 6 ) 1 0 ) 4 4 0 0 (0) 6(0) 10(0) 4 4 0 (3) 6(3) 10(3) 4 20 0 0 0000 ! ¡ 0 ? ¡ ! ¡ f f f () ¡ f x is positive on 1 , 3 ÈØ 0¢ 0 ÉÙ ÊÚ , so f ( x ) is increasing. ¡ f x is negative on 1 ,2 3 0 , so f ( x ) is decreasing. ¡ f x is positive on (2, ), so f ( x ) is increasing. 10. 2 () 4 9 1 2 7 ( ) 12 18 12 00. 0 0 ¡ x x x x x 2 2 12 18 12 0 62 3 2 0 6(2 1)( 2) 0 1 or 2 2 0 Test ( ) ¡ f x at x = –1, x = 0, x = 3. 2 2 2 ( 1) 12( 1) 18( 1) 12 18 0 (0) 12(0) 18(0) 12 12 0 (3) 12(3) 18(3) 12 42 0 0 0 0 00 ! ¡ 0 0 0 ? ¡ 0 0 ! ¡ f f f ¡ f x is positive on 1 , 2 0¢ 0 , so f ( x ) is increasing. ¡ f x is negative on 1 2 0 , so f ( x ) is decreasing. ¡ f x is positive (2, ), so f ( x ) is increasing.

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11. 22 1 () 3 ( 3)(1) ( 1)(1) 3 (3 ) ) . . .0 . ¡ .. x fx x xx so 2 3 0 ) ¡ . x has no solution. ¡ f x does not exist if 2 )0 . x x = –3. Test ( ) ¡ f x at x = –4 and x = 0. 2 2 3 (4 ) 3 (4 3 ) 33 1 (0) 93 (0 3) 0 ¡ 0. ¡ . f f ¡ f x is positive on (– , –3) and (–3, ), so f ( x ) is increasing on those intervals. 12. +, 2 2 2 4 (2 ) 4 (1) 4 . 0 ¡ x x xx x x x x 2 2 4 0 0 x x when 2 10 x ( x + 2)( x – 2) = 0 2o r 2 0 ¡ f x does not exist if 2 0 0.
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## This note was uploaded on 04/24/2010 for the course MARK 3336 taught by Professor Cox during the Spring '10 term at University of Houston - Downtown.

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chapter 12 homework problems - Chapter 12: Applications of...

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