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File: Course Review Solutions
chapters 1 to 8 course review chapters 1 to 8 course review question 1 page 509 a i 21 4 1 3 7 ii 1 1 5 1 0 5 ...

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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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...Chapters to course review question page a i ii iii b the trend is for average slope at x decrease rate since speed over period of time that ali drove instantaneous velocity was measured specific moment c temperature dropped hours night d leakage h ah this an approximation value tangent f xat mhr calculus and vectors solutions change m s choose interval t alternatively let in expression part no limit sequence does not converge lim exist graph function has vertical asymptote limits on left right sides are unequal dy dx xh y equation limx g...

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