chapter 6 homework

# chapter 6 homework - Chapter 6 APPLICATIONS OF THE...

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Unformatted text preview: Chapter 6 APPLICATIONS OF THE DERIVATIVE 6.1 Absolute Extrema 1. As shown on the graph, the absolute maximum oc- curs at x 3 ; there is no absolute minimum. (There is no functional value that is less than all others.) 2. As shown on the graph, the absolute minimum oc- curs at x 1 ; there is no absolute maximum. (There is no functional value that is greater than all oth- ers.) 3. As shown on the graph, there are no absolute ex- trema. 4. As shown on the graph, there are no absolute ex- trema. 5. As shown on the graph, the absolute minimum occurs at x 1 ; there is no absolute maximum. 6. As shown on the graph, the absolute maximum occurs at x 1 ; there is no absolute minimum. 7. As shown on the graph, the absolute maximum occurs at x 1; the absolute minimum occurs at x 2 : 8. As shown on the graph, the absolute maximum occurs at x 2 ; the absolute minimum occurs at x 1 : 10. f ( x ) = x 3 ¡ 3 x 2 ¡ 24 x + 5; [ ¡ 3 ; 6] Find critical numbers: f ( x ) = 3 x 2 ¡ 6 x ¡ 24 = 0 3( x 2 ¡ 2 x ¡ 8) = 0 3( x + 2)( x ¡ 4) = 0 x = ¡ 2 or x = 4 x f ( x ) ¡ 3 23 ¡ 2 33 Absolute maximum 4 ¡ 75 Absolute minimum 6 ¡ 31 11. f ( x ) = x 3 ¡ 6 x 2 + 9 x ¡ 8; [0 ; 5] Find critical numbers: f ( x ) =3 x 2 ¡ 12 x + 9 = 0 x 2 ¡ 4 x + 3 = 0 ( x ¡ 3)( x ¡ 1) = 0 x = 1 or x = 3 x f ( x ) ¡ 8 Absolute minimum 1 ¡ 4 3 ¡ 8 Absolute minimum 5 12 Absolute maximum 12. f ( x ) = 1 3 x 3 ¡ 1 2 x 2 ¡ 6 x + 3; [ ¡ 4 ; 4] Find critical numbers: f ( x ) = x 2 ¡ x ¡ 6 = 0 ( x + 2)( x ¡ 3) = 0 x = ¡ 2 or x = 3 x f ( x ) ¡ 4 ¡ 7 3 ¼ ¡ 2 : 3 ¡ 2 31 3 ¼ 10 : 3 Absolute maximum 3 ¡ 21 2 ¼ ¡ 10 : 5 Absolute minimum 4 ¡ 23 3 ¼ ¡ 7 : 7 13. f ( x ) = 1 3 x 3 + 3 2 x 2 ¡ 4 x + 1; [ ¡ 5 ; 2] Find critical numbers: f ( x ) = x 2 + 3 x ¡ 4 = 0 ( x + 4)( x ¡ 1) = 0 x = ¡ 4 or x = 1 x f ( x ) ¡ 4 59 3 ¼ 19 : 67 Absolute maximum 1 ¡ 7 6 ¼¡ 1 : 17 Absolute minimum ¡ 5 101 6 ¼ 16 : 83 2 5 3 ¼ 1 : 67 376 Section 6.1 Absolute Extrema 377 14. f ( x ) = x 4 ¡ 32 x 2 ¡ 7; [ ¡ 5 ; 6] f ( x ) = 4 x 3 ¡ 64 x = 0 4 x ( x 2 ¡ 16) = 0 4 x ( x ¡ 4)( x + 4) = 0 x = 0 or x = 4 or x = ¡ 4 x f ( x ) ¡ 5 ¡ 182 ¡ 4 ¡ 263 Absolute minimum ¡ 7 4 ¡ 263 Absolute minimum 6 137 Absolute maximum 15. f ( x ) = x 4 ¡ 18 x 2 + 1; [ ¡ 4 ; 4] f ( x ) = 4 x 3 ¡ 36 x = 0 4 x ( x 2 ¡ 9) = 0 4 x ( x + 3)( x ¡ 3) = 0 x = 0 or x = ¡ 3 or x = 3 x f ( x ) ¡ 4 ¡ 31 ¡ 3 ¡ 80 Absolute minimum 1 Absolute maximum 3 ¡ 80 Absolute minimum 4 ¡ 31 16. f ( x ) = 8 + x 8 ¡ x ; [4 ; 6] f ( x ) = (8 ¡ x )(1) ¡ (8 + x )( ¡ 1) (8 ¡ x ) 2 = 16 (8 ¡ x ) 2 f ( x ) is never zero. Although f ( x ) fails to exist if x = 8 ; 8 is not in the given interval. x f ( x ) 4 3 Absolute minimum 6 7 Absolute maximum 17. f ( x ) = 1 ¡ x 3 + x ; [0 ; 3] f ( x ) = ¡ 4 (3+ x ) 2 No critical numbers x f ( x ) 1 3 Absolute maximum 3 ¡ 1 3 Absolute minimum 18. f ( x ) = x x 2 + 2 ; [0 ; 4] f ( x ) = ( x 2 + 2)1 ¡ x (2 x ) ( x 2 + 2) 2 = ¡ x 2 + 2 ( x 2 + 2) 2 = 0 ¡ x 2 + 2 = 0 x 2 = 2 x = p...
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chapter 6 homework - Chapter 6 APPLICATIONS OF THE...

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