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Pre-Calc Homework Solutions 131

# Pre-Calc Homework Solutions 131 - Section 4.1 39 42 131...

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39. [ 2 2.35, 2.35] by [ 2 3.5, 3.5] y 9 5 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 ) } 5 } 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 9 5 h 42. [ 2 4, 4] by [ 2 1, 6] y 9 5 h 43. [ 2 4, 6] by [ 2 2, 6] y 9 5 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) } 5 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 ) 5 2 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) 5 2 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
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