ch04 - Chapter 4 Derivatives Review of Prerequisite Skills...

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Review of Prerequisite Skills 1. f. i. 6. d. f. 9. c. d. 32 43 18 24 6 48 18 48 66 24 6 30 11 4 6 5 ––– + = + × = + = = 232 342 61 72 2 4 93 2 30 17 2 23 30 17 2 23 + = + × + + = ++ = + = + –– x x x x xx x x x x + = + () + + + = + = + + 1 2 2 3 13 22 23 43 4 47 xyxy xy + ÷ + = + × + = + ×≠ 51 0 5 10 2 3 3 2 32 12 6 12 2 3 52 5 6 aa b ab b a () () [] = = 46 12 24 12 2 79 15 16 15 pp p p p p × = = Exercise 4.1 3. 4. b. () = + ( ) = + + = = + = →→ f fh f h hh h hh h h h h h 3 33 3 1 9 9 1 63 9 9 0 0 2 0 2 0 2 lim lim lim lim Now f f h f h 9 9 3 9 39 2 0 + ( ) =+ = + = lim Since af h h h f + =++ + = 3 3 3 1 91 9 333 31 19 2 2 2 , fxx x a fa fa h fa h h ++ = = + ( ) 2 0 31 3 ; lim y x 1 1 y x Chapter 4: Derivatives 73 Chapter 4 • Derivatives
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c. ; 5. a. b. () = + ( ) = ++ + = + + = + = + fx fx h fx h xh x h xx h hx h x x x h h h h lim lim lim ––– lim 0 0 0 0 2 3 2 3 2 36 336 22 3 3 2 x = + 3 2 =+ x 23 = + ( ) = + + = + h xh x x h hx h h h h h h h lim lim lim lim 0 0 2 2 0 2 0 33 fx x x 2 3 = + ( ) = + = + × = + = = f fh f h h h h h h h h hh h f h h h h h 0 00 11 1 0 1 2 0 0 0 0 0 lim lim lim lim lim a = 0 x 1 c. d. 6. b. dy dx x x h hxxh h xxh x h x dy dx x h h h = + + = + + + ( ) = + ( ) = lim lim –––– – –– lim 0 0 0 2 1 1 1 1 2 2 1 y x x = + 1 1 = + ( ) = + = + = + = = h x h xx x hx h x x x x h h h h lim lim lim –– – lim 0 0 2 2 0 2 2 2 0 2 2 4 3 2 2 2 2 x = 1 2 x h x h x x x hxh x x h h h = + ( ) = + × +++ + = = + 32 332 32 332 32 332 32 33232 3 0 0 0 lim lim lim 2 74 Chapter 4: Derivatives
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7. Since Slopes of the tangents at x = 0, 1, and 2 are ,. 8. s ( t ) = – t 2 + 8 t; t = 0, t = 4, t = 6 st h st h t h h hth + () ( ) =+ =− + –– 28 2 th t ht h + ++ + –– – 2 22 8 8 vt s t h h =′ = + ( ) lim 0 y x 0 12 y = 2 x 2 – 4 x () = ff 10 24 , and f 04 = + = fx hx h h xh x h h lim lim 0 0 424 44 fx h fx h fx h x h x hx h x x h h h = + ( ) + + =++ = + ( ) lim 0 2 2 2 2424 4 yf x x x = = 2 yx x = 2 Velocities at , 4, and 6 are and . 9. , parallel to The slope of the tangent to is parallel to . The point of tangency will be . The equation of the line will be or . xy –6 0 += 3 1 6 8 = 88 8 3 ,, f = ∴′ + = = x x x 1 6 1 21 1 6 13 8 4 0 x 1 = + ( ) = + = + × +++ + = + = h x h x h x x x hxh x x h h h h h lim lim lim lim lim 0 0 0 0 0 11 1 = + 1 x x 1 v 64 vv 0840 () = () = t = 0 h t h h = −+ lim lim – 0 0 Chapter 4: Derivatives 75
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13. at (2, 2) At , . Slope of the tangent at (2, 2) is –1. 14. And exists for all and . 15. But + () = + = lim lim . h h fa h h h 0 0 6 2 3 and fa () = 0 = + ( ) = fa h fa h h lim 0 6 f a =′ 06 , = f 00 xR fx ∴′ =< =≥ xx x x –, , 20 For x f x x x f x x <= ≥= 0 0 2 2 ,– , fx xx = = –1 x = 2 + ( ) = + + = +− + + = + = fx h fx h xh x x h h x h x xh x hx x h h x h h h h lim lim lim ––– –– lim 0 0 0 22 0 2 11 1 1 1 y x x x y x x == = 1 1 1 xy += 16.
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This note was uploaded on 02/10/2010 for the course MATHS MCV-01 taught by Professor Mr.m during the Spring '10 term at Seneca.

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ch04 - Chapter 4 Derivatives Review of Prerequisite Skills...

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