Calculus

# 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 ( ) f x x x = a. sec (2) (1) 2 1 6 0 1 6 f f m = = = b. sec (3) (2) 3 2 24 6 1 18 f f m = = = c. sec (2) (1.5) 2 1.5 6 1.875 .5 8.25 f f m = = = d. sec (2.5) (2) 2.5 2 13.125 6 .5 14.25 f f m = = = e. sec (2) (1.9) 2 1.9 6 4.959 .1 10.41 f f m = = = f. sec (2.1) (2) 2.1 2 7.161 6 .1 11.61 f f m = = = g. Tangent line slope estimate: 11 15. 2 ( ) cos f x x = a. sec (2) (1) 2 1 .65 .54 1 1.19 f f m = = = − b. sec (3) (2) 3 2 .91 ( .65) 1 .26 f f m = − − = = − c. sec (2) (1.5) 2 1.5 .654 ( .628) .5 .05 f f m = − − = = − d. sec (2.5) (2) 2.5 2 1.00 ( .65) .5 3.3 f f m = − − = = e. sec (2) (1.9) 2 1.9 .65 ( .89) .1 2.4 f f m = − − = = f. sec (2.1) (2) 2.1 2 .30 ( .65) .1 3.5 f f 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 f x x = , a = 1 tan 0 2 2 0 2 0 2 0 0 (1 ) (1) lim (1 ) 2 (1 2) lim 1 2 2 1 lim 2 lim lim 2 2 h h h h h f h f m h h h h h h h h h h + = + = + + + = + = = + = 2 (1) 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 f x x x = , a = –2. tan 0 2 2 0 2 0 2 0 0 ( 2 ) ( 2) lim ( 2 ) 3( 2 ) (( 2) 3( 2)) lim 4 4 6 3 10 lim 7 lim lim 7 7 h h h h h m f h f h h h h h h 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 f x x = + , a = 1 ( ) ( ) tan 0 2 2 (1 ) 1 1 1 0 2 2 0 2 (2 ) 2 0 2 0 0 (1 ) (1) lim lim 1 lim lim lim 1 lim 2 1 2 h h h h h h h h h h h h f h f m h h h h h h + + + + + + + + = = = = = = + = − 2 (1) 1 1 1 f = = + So the equation of the tangent line is 1 ( 1) 1 2 y x = − + or 1 3 2 2 y x = − + 27. ( ) 3 f x x = + , a = –2 tan 0 0 0 0 0 0 ( 2 ) ( 2) lim ( 2 ) 3 2 3 lim 1 1 lim 1 1 1 1 lim 1 1 1 1 lim ( 1 1) 1 lim 1 1 1 2 h h h h h h f h f m h h h h h h h h h h h h h + = + + + = + = + + + = + + + = + + = + + = ( 2) 2 3 1 1 f = + = = So the equation of the tangent line is 1 ( 2) 1 2 y x = + + or 1 2 2 y x = +
Chapter 2 Differentiation 49 29. f ( x ) = | x – 1| a = 1 0 0 0 0 1 1 1 1 (1 ) (1) lim lim 0 lim lim 1 h h h h h f h f h h h h h h + + + + + + = = = = 0 0 0 0 1 1 1 1 (1 ) (1) lim lim 0 lim lim 1 h h h h h f h f h h h h h h + + = = = = − Since the one sided limits are different, 0 (1 ) (1) lim h f h f h + does not exist and thus the tangent line does not exist. 31. 2 3 2 if 0 ( ) if 0 x x f x x x < = at a = 0 3 3 0 0 2 0 2 0 0 0 (0 ) (0) 0 lim lim lim 0 (0 ) (0) 2 0 lim lim lim 2 0 h h h h h h f h f h h h h f h f h h h h + + + + = = = + = = = So the slope of the tangent line is 0. 33. 2 3 1 if 0 ( ) 1 if 0 x x f x x x < = + at a = 0 3 0 0 2 0 0 lim ( ) lim 1 1 lim ( ) lim 1 1 x x x x f x x f x x + + = + = = = − Since the function is not continuous at a = 0, the tangent line does not exist.

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