ch09 - Chapter 9 Cur ve Sketching Review of Prerequisite...

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3. a. y = x 7 – 430 x 6 – 150 x 3 } d d y x } =7 x 6 – 2580 x 5 – 450 x 2 If x = 10, } d d y x } <0 . If x = 1000, } d d y x } >0 . The curve rises upward in quadrant one. c. y = x ln x x 4 } d d y x } = x 1 } 1 x } 2 + ln x – 4 x 3 =1+ln x – 4 x 3 If x = 10, } d d y x } . If x = 1000, } d d y x } . The curve is decreasing downward into quadrant four. 5. b. f ( x )= x 5 – 5 x 4 + 100 f' ( x )=5 x 4 – 20 x 3 Let ( x )=0 : 5 x 4 – 20 x 3 =0 5 x 3 ( x – 4) = 0 x or x =4 . xx 0 0< x <4 4 x >4 f'(x) +0–0 + Graph Increasing Decreasing Increasing d. f ( x } x x 2 + 1 3 } ( x } x 2 +3 ( x 2 + 2 x 3 ( ) x 2 –1) } Let ( x )= 0, therefore, – x 2 +2 x +3=0 . Or x 2 – 2 x – 3 = 0 ( x – 3)( x +1)=0 x =3 x =–1 <–1 –1 –1< x <3 3 x >3 –0+ 0 Graph Decreasing Increasing Decreasing Chapter 9: Curve Sketching 207 Chapter 9 • Curve Sketching Review of Prerequisite Skills 2. c. t 2 – 2 t t 2 – 2 t – 3<0 ( t – 3)( t +1)<3 Consider t and t = –1. The solution is –1 < t . d. x 2 x – 4>0 ( x + 4)( x – 1) > 0 Consider t = –4 and t =1 . The solution is x < –4 or x >1 . 4. b. lim x 2 } x 2 + x 3 x 2 –10 } =lim x 2 } ( x + x 5) ( x 2 –2) } x 2 ( x +5) Exercise 9.1 1. c. f ( x )=(2 x – 1) 2 ( x 2 – 9) ( x )= 2(2 x – 1)(2)( x 2 – 9) + 2 x (2 x – 1) 2 Let ( x )=0: 2(2 x – 1)(2( x 2 – 9) + x (2 x – 1)) = 0 2(2 x – 1)(4 x 2 x – 18) = 0 2(2 x – 1)(4 x – 9)( x +2)=0 x = } 1 2 } or x = } 9 4 } or x = –2. The points are 1 } 1 2 } ,0 2 , (2.24, –48.2), and (–2, –125). –4 1 > 0 < 0 > 0 –1 3
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e. f ( x )= x ln( x ) f' ( x )= ln x + } 1 x } ( x ) = ln x +1 Let ( x )=0: ln x +1=0 ln x =–1 x = e –1 = } 1 e } = 0.37. xx 00 < x < 0.37 0.37 x > 0.37 f'(x) No values 0 Graph Decreasing Increasing 6. ( x )=( x – 1)( x + 2)( x +3) Let ( x then ( x – 1)( x + 2)( x +3)=0 x =1 or x = –2 or x = –3. <–3 –3 –3< x <–3 –2 –2< x <1 1 x >1 –0+0– 0 + Graph Decreasing Increasing Decreasing Increasing 7. g' ( x )=(3 x – 2) ln(2 x 2 – 3 x +2) Let ( x then (3 x – 2) ln(2 x 2 – 3 x +2) =0 3 x – 2 = 0 or ln(2 x 2 – 3 x +2 ) =0 x = } 2 3 } or 2 x 2 – 3 x +2 = e 0 2 x 2 – 3 x +2 =1 2 x 2 – 3 x +1 =0 (2 x – 1)( x – 1) = 0 x = } 1 2 } or x . < } 1 2 }} 1 2 } < x < } 2 3 2 3 } < x <1 x –+– + Graph Decreasing Increasing Decreasing Increasing 9. f ( x x 3 + ax 2 + bx + c ( x )=3 x 2 ax + b Since f ( x ) increases to (–3, 18) and then decreases, (3)=0 . Therefore, 27 – 6 a + b =0 or6 a b = 27. (1) Since f ( x ) decreases to the point (1, –14) and then increases, (1)=0 . Therefore, 3 + 2 a + b or2 a + b = –3. (2) Add (1) to (2): 8 a = 24 and a =3 . When a , b =6+ b = –3 or b = –9. Since (1, –14) is on the curve and a , b = –9, then –14=1+3 9+ c c = –9. The function is f ( x x 3 +3 x 2 – 9 x – 9. 10. 11. a. 2 1 –1 1 f (x) x –5 1 f (x) x 1 y x –2 (–1, 0) 1 3 5 208 Chapter 9: Curve Sketching
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14. Let x 1 , x 2 be in the interval a x b , such that x 1 < x 2 . Therefore, f ( x 2 )> f ( x 1 ), and g ( x 2 g ( x 1 ). In this case, f ( x 1 ), f ( x 2 ), g ( x 1 ), and g ( x 2 )<0. Multiplying an inequality by a negative will reverse its sign. Therefore, f ( x 2 ) ? g ( x 2 )< f ( x 1 ) ? g ( x 1 ). But L.S. > 0 and R.S. > 0. Therefore, the function fg is strictly increasing. Exercise 9.2 3. b. f ( x )= } x 2 2 + x 9 } f' ( x } 2( x 2 + ( x 9 2 ) + 9 2 ) 2 x (2 x ) } = } 1 ( x 8 2 + 4 9 x ) 2 2 } Let ( x )=0: therefore, 18 – 2 x 2 =0 x 2 =9 x = ±3.
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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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ch09 - Chapter 9 Cur ve Sketching Review of Prerequisite...

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