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Chapters 1 to 8 Course Review Chapters 1 to 8 Course Review Question 1 Page 509 a) i) [2(16)!12+4]![2!3+4] = 21 4!1 3 =7 ii) [2(2.25)! 4.5+4]![2!3+4] = 1 1.5!1 0.5 =2 [2(1.21)!3.3+4]![2!3+4]= 0.12 iii) 1.1!1 0.1 =1.2 b) The trend is for the average slope at x = 1 to decrease. The slope at x = 1 is 1. Chapters 1 to 8 Course Review Question 2 Page 509 a) Average rate, since the speed is over the period of time that Ali drove. b) Instantaneous rate, since the velocity was measured at a specific moment. c) Average rate, since the temperature dropped over the hours of the night. d) Instantaneous rate, since the leakage was measured at a specific moment. Chapters 1 to 8 Course Review Question 3 Page 509 3(a+h)2 !3a2 3a2 +6ah+3h2 !3a2 a) i) h = h 6ah+3h2 = h =6a+3h 6(2) + 3(0.01) = 12.03 ii) (a + h)3 ! a3 a3 +3a2h+3ah2 +h3 !a3 3a2h+3ah2 +h3 h = h = h =3a2 +3ah+h2 2 2 3(2) +3(2)(0.01) + (0.01) = 12.0601 b) i) This is an approximation of the value of the slope of the tangent to f (x) = 3x2at x = 2. MHR • Calculus and Vectors 12 Solutions 1020 ii) This is an approximation of the value of the slope of the tangent to f (x) = x3at x = 2. Chapters 1 to 8 Course Review Question 4 Page 509 a) !4.9[(2+ h)2 !22]+6h !4.9(4h+h2)+6h h = h =!19.6!4.9h+6 =!13.6!4.9h The average rate of change is (–13.6 – 4.9h) m/s. b) Choose the interval 1.9 ≤ t ≤ 2.1. 2 2 !4.9(2.1 !1.9 )+6(0.2) = !4.9(0.2)(4)+6(0.2) 0.2 0.2 =!19.6+6 =!13.6 The instantaneous rate of change is –13.6 m/s. Alternatively, let h = 0 in the expression in part a). c) Chapters 1 to 8 Course Review Question 5 Page 509 a) No limit; the sequence does not converge. b) The limit is 5. c) No limit; the sequence does not converge. d) The limit is 0. MHR • Calculus and Vectors 12 Solutions 1021 Chapters 1 to 8 Course Review Question 6 Page 509 a) lim(3x2 "4x+1)=5 x!2 b) lim 5x+40 = lim 5(x+8) x!"8 x+8 x!"8 x+8 =5 c) lim x"6=0 x!6+ d) The limit does not exist. The graph of the function has a vertical asymptote at x = 3. Chapters 1 to 8 Course Review Question 7 Page 509 a) b) i) lim f (x) =17 x!3 ii) The limit does not exist. The limits on the left and right sides are unequal. Chapters 1 to 8 Course Review Question 8 Page 509 a) dy = lim f (x + h)" f (x) dx h!0 h 4(x+h)2 "3 " 4x2 "3 =lim( ) ( ) h!0 h 8xh+h2 =lim h!0 h =lim(8x+h) h!0 =8x MHR • Calculus and Vectors 12 Solutions 1022 At x = 2, dy = 16 and y = 13. dx The equation of the tangent is: y!13=16(x!2) y =16x!19 b) f !(x) = lim f (x + h)# f (x) h"0 h ((x+h)3#2(x+h)2)#(x3#2x2) =lim h"0 h 3x2h+3xh2 +h3#4xh#2h2 =lim h"0 h =lim3x2 +3xh+h2 #4x#2h h"0 =3x2 #4x At x = 2, f !(x) = 4 and f(x) = 0. The equation of the tangent is: y!0=4(x!2) y = 4x!8 c) g!(x) = lim f (x + h)# f (x) h"0 h $ 3 #3' & x + h x) =lim% ( h"0 h $ 3x #3x#3h' & x(x+h) ) =lim% ( h"0 h =lim #3h h"0 xh(x+h) =lim #3 h"0 x(x+h) =# 3 x2 At x = 2, g (x)= 3 and g(x) = 3 . ! !4 2 The equation of the tangent is: MHR • Calculus and Vectors 12 Solutions 1023
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