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 2 : 3 2 31 3 10 : 3 Absolute maximum 3 21 2 10 : 5 Absolute minimum 4 23 3 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 19 : 67 Absolute maximum 1 7 6 1 : 17 Absolute minimum 5 101 6 16 : 83 2 5 3 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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