sol12-5 - 2 = 3 x 2 ( x 4 + 1)-x 3 (4 x 3 ) ( x 4 + 1) 2...

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SOLUTIONS CHAPTER 12.5 MATH 132 WI01 6. Proof. C 0 ( I ) = (2 I 2 - 3) 0 · (3 I 2 - 4 I + 1) + (2 I 2 - 3) · (3 I 2 - 4 I + 1) 0 = = (4 I )(3 I 2 - 4 I + 1) + (2 I 2 - 3)(6 I - 4) ± 10. Proof. y 0 = (2 - 3 x + 4 x 2 ) 0 · (1 + 2 x - 3 x 2 ) + (2 - 3 x + 4 x 2 ) · (1 + 2 x - 3 x 2 ) 0 = = ( - 3 + 8 x )(1 + 2 x - 3 x 2 ) + (2 - 3 x + 4 x 2 )(2 - 6 x ) ± 24. Proof. f ( x ) = 5( x 2 - 2) 7 = 5 7 ( x 2 - 2) so then f 0 ( x ) = 5 7 ( x 2 - 2) 0 = 5 7 (2 x ) ± 28. Proof. y 0 = ( x 2 - 4 x + 2) 0 · ( x 2 + x + 1) - ( x 2 - 4 x + 2) · ( x 2 + x + 1) 0 ( x 2 + x + 1) 2 = = (2 x - 4)( x 2 + x + 1) - ( x 2 - 4 x + 2)(2 x + 1) ( x 2 + x + 1) 2 ± Date : 01/21/2000. 1
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2 MATH 132 WI01 36. Proof. y = x - 5 2 x = x 2 x - 5 2 x = 1 2 x - 5 2 1 x = = 1 2 x 1 2 - 5 2 x - 1 2 so then y 0 = 1 2 · 1 2 x 1 2 - 1 - 5 2 · ( - 1 2 ) x - 1 2 - 1 = = 1 4 x - 1 2 + 5 4 x - 3 2 ± 50. Proof. First compute the derivative of y : y 0 = ( x 3 ) 0 · ( x 4 + 1) - x 3 · ( x 4 + 1) 0 ( x 4 + 1)
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Unformatted text preview: 2 = 3 x 2 ( x 4 + 1)-x 3 (4 x 3 ) ( x 4 + 1) 2 Plug in x = 1; we get: slope of tangent line = 3 1 2 (1 4 + 1)-1 3 (4 1 3 ) (1 4 + 1) 2 = = 3 2-4 2 2 = 6-4 4 = 2 4 = 0 . 5 Based on point-slope formula we get, for the point (1 , 1 2 ) and slope 0 . 5: y-1 2 = 0 . 5 ( x-1)...
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sol12-5 - 2 = 3 x 2 ( x 4 + 1)-x 3 (4 x 3 ) ( x 4 + 1) 2...

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