Calculus - Chapter 2 Differentiation Chapter 2 Section 2.1...

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Chapter 2 Differentiation 47 Chapter 2 Section 2.1 5. y 0.5 1.0 x 0 1 7. y 1 x 0 1 9. At 1 x = the slope of the tangent line is approximately –1. 11. C, B, A, D 13. 3 () fx x x =− a. sec (2) (1) 21 60 1 6 ff m = = = b. sec (3) (2) 32 24 6 1 18 m = = = c. sec (2) (1.5) . 5 61 . 8 7 5 .5 8.25 m = = = d. sec (2.5) (2) 2.5 2 13.125 6 .5 14.25 m = = = e. sec (2) (1.9) . 9 64 . 9 5 9 .1 10.41 m = = = f. sec (2.1) (2) 2.1 2 7.161 6 .1 11.61 m = = = g. Tangent line slope estimate: 11 15. 2 ( )c o s fx x = a. sec (2) .65 .54 1 1.19 m = −− = b. sec (3) (2) .91 ( .65) 1 .26 m = = c. sec (2) (1.5) . 5 .654 ( .628) .5 .05 m = = d. sec (2.5) (2) 2.5 2 1.00 ( .65) .5 3.3 m = = = e. sec (2) (1.9) . 9 .65 ( .89) .1 2.4 m = = = f. sec (2.1) (2) 2.1 2 .30 ( .65) .1 3.5 m = = = g. Tangent line slope estimate: 3
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Chapter 2 Differentiation 48 17. 0 0.5 1 1.5 2 2.5 3 3.5 4 5 10 15 20 25 x y 19. The answer is an animation. 21. 2 () 2 fx x =− , a = 1 tan 0 22 0 2 0 2 0 0 (1 ) (1) lim ) 2 2) lim 12 21 lim 2 lim lim 2 2 h h h h h fh f m h h h hh h h h +− = = ++−+ = + = =+ = 2 1 2 1 f =−= So the equation of the tangent line is y = 2( x – 1) + (–1) or y = 2 x – 3 23. 2 3 x , a = –2. tan 0 0 2 0 2 0 0 (2 ) ) lim ) 3 ) ( ) ) lim 44 63 1 0 lim 7 lim lim 7 7 h h h h h m f h h h h h h −+ − − = −+ − −+ − − − − = −++−− = = + 2 ( 2) ( 2) 3( 2) 10 f −=− −−= So the equation of the tangent line is y = –7( x + 2) + 10 or y = –7 x – 4 25. 2 1 fx x = + , a = 1 tan 0 ) 1 1 1 0 2 2 0 2( 2 ) 2 0 2 0 0 ) lim lim 1 lim lim lim 1 lim 2 1 2 h h h h h h h h h h h h f m h h h h h h ++ + + + + = = = = = = + 2 1 11 f == + So the equation of the tangent line is 1 ) 1 2 yx − + or 13 + 27. 3 x , a = –2 tan 0 0 0 0 0 0 ) ) lim 2 3 lim lim lim lim 1 ) 1 lim 1 2 h h h h h h f m h h h h h h h h h = −+ + − −+ = = =⋅ = = = ) f −=−+= = So the equation of the tangent line is 1 ) 1 2 + or 1 2 2
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Chapter 2 Differentiation 49 29. f ( x ) = | x – 1| a = 1 00 0 0 11 1 1 (1 ) (1) lim lim 0 lim lim 1 hh h h h fh f h h h h ++ + + →→ +−−− +− = = = = 0 0 1 1 ) lim lim 0 lim lim 1 h h h f h h h h −− = = = =− Since the one sided limits are different, 0 ) lim h f h does not exist and thus the tangent line does not exist. 31. 2 3 2i f 0 () if 0 xx fx −< = at a = 0 33 2 0 2 0 (0 ) (0) 0 lim lim lim 0 (0 ) (0) 20 lim lim lim 2 0 h h f h h f h h + = = = = = So the slope of the tangent line is 0. 33. 2 3 1if 0 0 = +≥ at a = 0 3 2 lim ( ) lim 1 1 lim ( ) lim 1 1 x x =+ = = Since the function is not continuous at a = 0, the tangent line does not exist. 35. 2 ( ) 16 10 ft t a. 22 (2) (0) 16(2) 10 (16(0) 10) 64 32 2 avg ff v = + = == b. (2) 21 16(2) 10 (16(1) 10) 48 avg v = + = = c. (2) (1.9) . 9 16(2) 10 (16(1.9) 10) . 9 6.24 .1 62.4 avg v = + = = = d. (2) (1.99) . 9 9 16(2) 10 (16(1.99) 10) . 9 9 .6384 .01 63.84 avg v = + = = = e. 0 0 2 0 2 0 0 (2 ) (2) (2) lim 16(2 ) 10 (16(2 ) 10) lim 16(4 4 ) 10 64 10 lim 64 16 lim lim 64 16 64 h h h h h f v h h h h h h = + = ++ +−− = + = =
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Chapter 2 Differentiation 50 37. 2 () 8 ft t t =+ a. (2) (0) 20 41 6 0 2 20 2 5 2.236068 avg ff v = +− = = = = b. (2) (1) 21 20 1 8 1 20 3 1.472136 avg v = −+ = =− = c.
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This homework help was uploaded on 04/19/2008 for the course MATH 105 taught by Professor Schroder during the Spring '08 term at Millersville.

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Calculus - Chapter 2 Differentiation Chapter 2 Section 2.1...

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