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# Chapter 2 - SECTION 2.1(PAGE 98 R A ADAMS CALCULUS CHAPTER...

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SECTION 2.1 (PAGE 98) R. A. ADAMS: CALCULUS CHAPTER 2. DIFFERENTIATION Section 2.1 Tangent Lines and Their Slopes (page 98) 1. Slope of y = 3 x 1a t ( 1 , 2 ) is m = lim h 0 3 ( 1 + h ) 1 ( 3 × 1 1 ) h = lim h 0 3 h h = 3 . The tangent line is y 2 = 3 ( x 1 ) ,o r y = 3 x 1. (The tangent to a straight line at any point on it is the same straight line.) 2. Since y = x / 2 is a straight line, its tangent at any point ( a , a / 2 ) on it is the same line y = x / 2. 3. Slope of y = 2 x 2 5a t ( 2 , 3 ) is m = lim h 0 2 ( 2 + h ) 2 5 ( 2 ( 2 2 ) 5 ) h = lim h 0 8 + 8 h + 2 h 2 8 h = lim h 0 ( 8 + 2 h ) = 8 Tangent line is y 3 = 8 ( x 2 ) or y = 8 x 13. 4. The slope of y = 6 x x 2 at x =− 2is m = lim h 0 6 ( 2 + h ) ( 2 + h ) 2 4 h = lim h 0 3 h h 2 h = lim h 0 ( 3 h ) = 3 . The tangent line at ( 2 , 4 ) is y = 3 x + 10. 5. Slope of y = x 3 + 8a t x m = lim h 0 ( 2 + h ) 3 + 8 ( 8 + 8 ) h = lim h 0 8 + 12 h 6 h 2 + h 3 + 8 0 h = lim h 0 ± 12 6 h + h 2 ² = 12 Tangent line is y 0 = 12 ( x + 2 ) or y = 12 x + 24. 6. The slope of y = 1 x 2 + 1 at ( 0 , 1 ) is m = lim h 0 1 h ³ 1 h 2 + 1 1 ´ = lim h 0 h h 2 + 1 = 0 . The tangent line at ( 0 , 1 ) is y = 1. 7. Slope of y = x + t x = 3is m = lim h 0 4 + h 2 h · 4 + h + 2 4 + h + 2 = lim h 0 4 + h 4 h ( h + h + 2 ) = lim h 0 1 4 + h + 2 = 1 4 . Tangent line is y 2 = 1 4 ( x 3 ) r x 4 y 5. 8. The slope of y = 1 x at x = 9is m = lim h 0 1 h ³ 1 9 + h 1 3 ´ = lim h 0 3 9 + h 3 h 9 + h · 3 + 9 + h 3 + 9 + h = lim h 0 9 9 h 3 h 9 + h ( 3 + 9 + h ) 1 3 ( 3 )( 6 ) 1 54 . The tangent line at ( 9 , 1 3 ) is y = 1 3 1 54 ( x 9 ) r y = 1 2 1 54 x . 9. Slope of y = 2 x x + 2 at x = m = lim h 0 2 ( 2 + h ) 2 + h + 2 1 h = lim h 0 4 + 2 h 2 h 2 h ( 2 + h + 2 ) = lim h 0 h h ( 4 + h ) = 1 4 . Tangent line is y 1 = 1 4 ( x 2 ) , or x 4 y 2. 10. The slope of y = 5 x 2 at x = 1is m = lim h 0 µ 5 ( 1 + h ) 2 2 h = lim h 0 5 ( 1 + h ) 2 4 h ± µ 5 ( 1 + h ) 2 + 2 ² = lim h 0 2 h µ 5 ( 1 + h ) 2 + 2 1 2 The tangent line at ( 1 , 2 ) is y = 2 1 2 ( x 1 ) r y = 5 2 1 2 x . 40

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INSTRUCTOR’S SOLUTIONS MANUAL SECTION 2.1 (PAGE 98) 11. Slope of y = x 2 at x = x 0 is m = lim h 0 ( x 0 + h ) 2 x 2 0 h = lim h 0 2 x 0 h + h 2 h = 2 x 0 . Tangent line is y x 2 0 = 2 x 0 ( x x 0 ) , or y = 2 x 0 x x 2 0 . 12. The slope of y = 1 x at ( a , 1 a ) is m = lim h 0 1 h ± 1 a + h + 1 a ² = lim h 0 a a h h ( a + h )( a ) =− 1 a 2 . The tangent line at ( a , 1 a ) is y = 1 a 1 a 2 ( x a ) ,o r y = 2 a x a 2 . 13. Since lim h 0 | 0 + h |− 0 h = lim h 0 1 | h | sgn ( h ) does not exist (and is not or −∞ ), the graph of f ( x ) = | x | has no tangent at x = 0. 14. The slope of f ( x ) = ( x 1 ) 4 / 3 at x = 1is m = lim h 0 ( 1 + h 1 ) 4 / 3 0 h = lim h 0 h 1 / 3 = 0 . The graph of f has a tangent line with slope 0 at x = 1. Since f ( 1 ) = 0, the tangent has equation y = 0 15. The slope of f ( x ) = ( x + 2 ) 3 / 5 at x 2is m = lim h 0 ( 2 + h + 2 ) 3 / 5 0 h = lim h 0 h 2 / 5 =∞ .
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Chapter 2 - SECTION 2.1(PAGE 98 R A ADAMS CALCULUS CHAPTER...

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