Business Calc Homework w answers_Part_27

Business Calc Homework w answers_Part_27 - 131 Section 4.1...

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39. [ 2 2.35, 2.35] by [ 2 3.5, 3.5] y 95 x ? } 2 ˇ 4 w 1 2 w x w 2 w } ( 2 2 x ) 1 (1) ˇ 4 w 2 w x w 2 w 5 } 2 x ˇ 2 1 4 w ( 2 w 4 2 x w 2 w x 2 ) }5 } ˇ 4 4 w 2 2 w 2 x x w 2 2 w } 40. [ 2 4.7, 4.7] by [ 2 1, 5] y 5 x 2 ? } 2 ˇ 3 w 1 2 w x w } ( 2 1) 1 2 x ˇ 3 w 2 w x w 5 } 2 x 2 2 1 ˇ 3 w 4 x 2 w (3 x w 2 x ) } 2 2 5 ˇ x 2 3 w 1 2 w 1 x w 2 x } 41. [ 2 4.7, 4.7] by [0, 6.2] y h 42. [ 2 4, 4] by [ 2 1, 6] y h 43. [ 2 4, 6] by [ 2 2, 6] y h 44. [ 2 4, 6] by [ 2 5, 5] We begin by determining whether f 9 ( x ) is defined at x 5 1, where f ( x ) 5 h Left-hand derivative: lim h 0 2 } f (1 1 h h ) 2 f (1) lim h 0 2 5 lim h 0 2 } 2 h 4 2 h 2 h } 5 lim h 0 2 } 1 4 } ( 2 h 2 4 h ) 52 1 Right-hand derivative: lim h 0 1 } f (1 1 h h ) 2 f (1) } 5 lim h 0 1 5 lim h 0 1 } h 3 2 3 h h 2 2 h } 5 lim h 0 1 ( h 2 2 3 h 2 1) 1 Thus f 9 ( x ) 5 h x # 1 x . 1 2} 1 2 } x 2 } 1 2 } , 3 x 2 2 12 x 1 8, (1 1 h ) 3 2 6(1 1 h ) 2 1 8(1 1 h ) 2 3 }}}} h 2} 1 4 } (1 1 h ) 2 2 } 1 2 } (1 1 h ) 1 } 1 4 5 } 2 3 }}}} h x # 1 x . 1 2} 1 4 } x 2 2 } 1 2 } x 1 } 1 4 5 } , x 3 2 6 x 2 1 8 x , x , 1 x . 1 2 2 x 2 2, 2 2 x 1 6, x , 0 x . 0 2 1, 2 2 2 x , x , 1 x . 1 2 2, 1, Section 4.1 131 crit. pt. derivative extremum value x 5 1 undefined minimum 2 crit. pt. derivative extremum value x 5 0 0 minimum 0 x 5 } 1 5 2 } 0 local max } 1 1 4 2 4 5 } 15 1/2 < 4.462 x 5 3 undefined minimum 0 crit. pt. derivative extremum value x 2 undefined local max 0 x ˇ 2 w 0 minimum 2 2 x 5 ˇ 2 w 0 maximum 2 x 5 2 undefined local min 0 crit. pt. derivative extremum value x 1 0 maximum 5 x 5 1 undefined local min 1 x 5 3 0 maximum 5 crit. pt. derivative extremum value x 5 0 undefined local min 3 x 5 1 0 local max 4
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44. continued Note that 2} 1 2 } x 2 } 1 2 } 5 0 when x 52 1, and 3 x 2 2 12 x 1 8 5 0 when x 5 5 } 12 6 6 ˇ 4 w 8 w }5 2 6 } 2 ˇ 3 3 w } . But 2 2 } 2 ˇ 3 3 w } < 0.845 , 1, so the only critical points occur at x 1 and x 5 2 1 } 2 ˇ 3 3 w } < 3.155. 45. Graph (c), since this is the only graph that has positive slope at c . 46. Graph (b), since this is the only graph that represents a differentiable function at a and b and has negative slope at c . 47. Graph (d), since this is the only graph representing a function that is differentiable at b but not at a . 48. Graph (a), since this is the only graph that represents a function that is not differentiable at a or b . 49. (a) V ( x ) 5 160 x 2 52 x 2 1 4 x 3 V 9 ( x ) 5 160 2 104 x 1 12 x 2 5 4( x 2 2)(3 x 2 20) The only critical point in the interval (0, 5) is at x 5 2. The maximum value of V ( x ) is 144 at x 5 2. (b) The largest possible volume of the box is 144 cubic units, and it occurs when x 5 2. 50. (a) P 9 ( x ) 5 2 2 200 x 2 2 The only critical point in the interval (0, ) is at x 5 10. The minimum value of P ( x ) is 40 at x 5 10. (b) The smallest possible perimeter of the rectangle is 40 units and it occurs at x 5 10, which makes the rectangle a 10 by 10 square. 51. (a) f 9 ( x ) 5 3 ax 2 1 2 bx 1 c is a quadratic, so it can have 0, 1, or 2 zeros, which would be the critical points of f . Examples: [ 2 3, 3] by [ 2 5, 5] The function f ( x ) 5 x 3 2 3 x has two critical points at x 1 and x 5 1.
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Business Calc Homework w answers_Part_27 - 131 Section 4.1...

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