sol12-2

sol12-2 - and so y = 7 2 x 5 2 1 2 x 1 2 = = 7 2 x 5 2 1 2...

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SOLUTIONS CHAPTER 12.2 MATH 132 WI01 2. Proof. f ( x ) is a constant! it does not contain any x in its formula, hence f 0 ( x ) = 0 ± 8. Proof. y 0 = (4 · x 4 ) 0 = 4( x 4 ) 0 = 4 · 4 x 3 = 16 x 3 ± 28. Proof. f ( x ) = 5( x 4 - 3) 2 = 5 2 ( x 4 - 3) and so f 0 ( x ) = ( 5 2 ( x 4 - 3)) 0 = 5 2 ( x 4 - 3) 0 = 5 2 (4 x 3 ) = 10 x 3 ± 48. Proof. y = 1 4 x 5 = 1 4 · 1 x 5 = 1 4 x - 5 and so y 0 = ( 1 4 x - 5 ) 0 = 1 4 ( x - 5 ) 0 = 1 4 ( - 5) x - 5 - 1 = - 5 4 x - 6 ± 74. Proof. f ( x ) = 7 x 3 + x 2 x = 7 x 3 2 x + x 2 x = = 7 2 · x 3 x 1 2 + 1 2 · 1 x 1 2 = = 7 2 x 3 - 1 2 + 1 2 x 1 - 1 2 = 7 2 x 5 2 + 1 2 x 1 2 Date : 01/21/2000. 1

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2 MATH 132 WI01
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Unformatted text preview: and so y = ( 7 2 x 5 2 + 1 2 x 1 2 ) = = ( 7 2 x 5 2 ) + ( 1 2 x 1 2 ) = = 7 2 Â· ( x 5 2 ) + 1 2 Â· ( x 1 2 ) = = 7 2 Â· 5 2 x 5 2-1 + 1 2 Â· 1 2 x 1 2-1 = = 35 4 x 3 2 + 1 4 x-1 2 Â±...
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This note was uploaded on 07/26/2011 for the course MATH 132 taught by Professor Staff during the Spring '08 term at Ohio State.

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sol12-2 - and so y = 7 2 x 5 2 1 2 x 1 2 = = 7 2 x 5 2 1 2...

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