Calculus: One and Several Variables

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53 SECTION 3.1 65 CHAPTER 3 SECTION 3.1 1. f ± ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 [2 3( x + h )] [2 3 x ] h = lim h 0 3 h h = lim h 0 3= 3 2. f ± ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 k k h = lim h 0 0=0 3. f ± ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 [5( x + h ) ( x + h ) 2 ] (5 x x 2 ) h = lim h 0 5 h 2 xh h 2 h = lim h 0 (5 2 x h )=5 2 x 4. f ± ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 [2( x + h ) 3 +1] [2 x 3 h = lim h 0 2( x 3 +3 x 2 h xh 2 + h 3 ) 2 x 3 h = lim h 0 6 x 2 h +6 xh 2 +2 h 3 h = lim h 0 (6 x 2 xh h 2 )=6 x 2 5. f ± ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 ( x + h ) 4 x 4 h = lim h 0 ( x 4 +4 x 3 h x 2 h 2 xh 3 + h 4 ) x 4 h = lim h 0 (4 x 3 x 2 h xh 2 + h 3 )=4 x 3 6. f ± ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 1 x + h 1 x h lim h 0 ( x +3) ( x + h h ( x + h + 3)( x = lim h 0 h h ( x + h + 3)( x lim h 0 1 ( x + h + 3)( x = 1 ( x 2 7. f ± ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 x + h 1 x 1 h = lim h 0 ( x + h 1) ( x 1) h ( x + h 1+ x 1) = lim h 0 1 x + h x 1 = 1 2 x 1 8. f ± ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 [( x + h ) 3 4( x + h )] [ x 3 4 x ] h = lim h 0 3 x 2 h xh 2 + h 3 4 h h = lim h 0 (3 x 2 xh + h 2 4) = 3 x 2 4
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53 66 SECTION 3.1 9. f ± ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 1 ( x + h ) 2 1 x 2 h = lim h 0 x 2 ( x 2 +2 hx + h 2 ) hx 2 ( x + h ) 2 = lim h 0 2 x h x 2 ( x + h ) 2 = 2 x 3 10. f ± ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 1 x + h 1 x h lim h 0 x x + h h x x + h = lim h 0 ( x x + h )( x + x + h ) h x x + h ( x + x + h ) = lim h 0 x ( x + h ) h x x + h ( x + x + h ) lim h 0 h h x x + h ( x + x + h ) = 1 2 x x 11. f ( x )= x 2 4 x ; c =3: difference quotient: f (3 + h ) f (3) h = (3 + h ) 2 4(3 + h ) ( 3) h = 9+6 h + h 2 12 4 h +3 h = 2 h + h 2 h =2+ h Therefore, f ± (3) = lim h 0 f (3 + h ) f (3) h = lim h 0 (2 + h )=2 12. f ( x )=7 x x 2 ; c =2: difference quotient: f (2 + h ) f (2) h = 7(2 + h ) (2 + h ) 2 (10) h = 14+7 h 4 4 h h 2 10 h = 3 h h 2 h =3 h Therefore, f ± (2) = lim h 0 f (2 + h ) f (2) h = lim h 0 (3 h )=3 13. f ( x x 3 +1; c =1: difference quotient: f ( 1+ h ) f ( 1) h = 2( h ) 3 +1 ( 1) h = 2 ± 1+3 h 3 h 2 + h 3 ² h = 6 h 6 h 2 h 3 h =6 6 h 2 Therefore, f ± ( 1) = lim h 0 f ( h ) f ( 1) h = lim h 0 (6 6 h h 2 )=6 14. f ( x )=5 x 4 ; c = 1: difference quotient: f (1 + h ) f (1) h = 5 (1 + h ) 4 (4) h
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53 SECTION 3.1 67 = 5 1 4 h 6 h 2 4 h 3 h 4 4 h = 4 h 6 h 2 4 h 3 h 4 h = 4 6 h 4 h 2 h 3 Therefore, f ± (3) = lim h 0 f (1 + h ) f (1) h = lim h 0 ( 4 6 h 4 h 2 h 3 ) = 4 15. f ( x )= 8 x +4 ; c = 2: difference quotient: f ( 2+ h ) f ( 2) h = 8 ( h )+4 4 h = 8 h +2 4 h = 8 4 h 8 h ( h +2) = 4 h Therefore, f ± ( 2) = lim h 0 f ( h ) f ( 2) h = lim h
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This note was uploaded on 04/30/2011 for the course MATH 1431 taught by Professor Any during the Spring '08 term at University of Houston.

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ch03[1] - P1: PBU/OVY JWDD027-03 P2: PBU/OVY...

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