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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 40

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online