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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. 3 2 2 ( ) 2 5 4 2 ( ) 6 10 4 0 0 . 0 0 ¡ f x x x x f x x x + , 2 2 6 10 4 0 2 3 5 2 0 2(3 1)( 2) 0 0 0 0 0 . 0 x x x x x x 1 or 2 3 0 x x Test ( ) ¡ f x at x = –2, x = 0, x = 3. 2 2 2 ( 2) 6( 2) 10( 2) 4 40 0 (0) 6(0) 10(0) 4 4 0 (3) 6(3) 10(3) 4 20 0 0 0 0 0 0 ! ¡ 0 0 0 ? ¡ 0 0 ! ¡ f f f ( ) ¡ f x is positive on 1 , 3 È Ø 0¢ 0 É Ù Ê Ú , so f ( x ) is increasing. ( ) ¡ f x is negative on 1 , 2 3 È Ø 0 É Ù Ê Ú , so f ( x ) is decreasing. ( ) ¡ f x is positive on (2, ), so f ( x ) is increasing. 10. 3 2 2 ( ) 4 9 12 7 ( ) 12 18 12 0 0 . 0 0 ¡ f x x x x f x x x + , 2 2 12 18 12 0 6 2 3 2 0 6(2 1)( 2) 0 0 0 0 0 . 0 x x x x x x 1 or 2 2 0 x x 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 0 0 ! ¡ 0 0 0 ? ¡ 0 0 ! ¡ f f f ( ) ¡ f x is positive on 1 , 2 È Ø 0¢ 0 É Ù Ê Ú , so f ( x ) is increasing. ( ) ¡ f x is negative on 1 , 2 2 È Ø 0 É Ù Ê Ú , so f ( x ) is decreasing. ( ) ¡ f x is positive (2, ), so f ( x ) is increasing.

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11. 2 2 1 ( ) 3 ( 3)(1) ( 1)(1) 3 ( ) ( 3) ( 3) . . . 0 . ¡ . . x f x x x x f x x x so 2 3 ( ) 0 ( 3) ¡ . f x x has no solution. ( ) ¡ f x does not exist if 2 ( 3) 0 . x x = –3. Test ( ) ¡ f x at x = –4 and x = 0. 2 2 3 ( 4) 3 ( 4 3) 3 3 1 (0) 9 3 (0 3) 0 ¡ 0 . ¡ . f f ( ) ¡ f x is positive on (– , –3) and (–3, ), so f ( x ) is increasing on those intervals.
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chapter 12 homework problems - Chapter 12 Applications of...

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