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Additional Mathematics  SPM  Chapter 2 Differentiation
              Determining the nature of the turning points:   Solution
              Tangent sketching method                        y = x  + ax + b
                                                                  3
              At (3, 2):                                      At (0, 4),  4  = (0)  + a(0) + b
                                                                            3
                    Choose one value   Choose one value               b  = 4
                      where x , 3        where x . 3
                                                                            2
                   x        2        3        4               Also,   dy   = 3x  + a
                                                                     dx
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                  dy                                                  0  = 3(0)  + a
                                                                             2
                  dx       –3        0        9                       a  = 0
                 Sign       –       zero      +                     \  y  = x  + 4
                                                                           3
                Tangent                                              dy     2
               sketching                                             dx   = 3x
                                                                      2
                 Graph                                          and     d y   = 6x
               sketching                                            dx 2   = 6(0)

              \ (3, 2) is a minimum point.                               = 0
                                                              \  The point (0, 4) is an inflection point.
              At (1, 6):
                   x        0        1        2                  Try Questions 6 – 8 in ‘Try This! 2.4’
                  dy
                  dx        9        0       –3                 E  Solving problems involving
                                                                    maximum and minimum values and
                 Sign       +       zero      –
                                                                    interpreting the solutions
                Tangent
               sketching                                        1.  Problems involving the maximum value and
                                                                  minimum value can be solved by using the
                 Graph                                            following steps:
               sketching                                          (a)  Based  on  the  information  given  in  the
                                                                      problem, recognise the main function
              \ (1, 6) is a maximum point.                            (area, volume etc) that needs to be formed
                                                                      or derived so that the differentiation can be
                                                                      carried out. This main function can be easily
                Alternative Method
                                                                      recognised through the word  “maximum”
              Second derivative method                                or “minimum” stated in  the problem.
               d y  = 6x – 12                                     (b)  Obtain this main function and express it in
                 2
               dx 2                                                   terms of one variable. Normally, there would
              x = 3,                                                  be one condition stated in the problem so that
               d y  = 6(3) – 12                                       the variables can be mutually substituted.
                 2
                                                                                dy
         Form 5
                dx 2  = 6 . 0                                     (c)  Let  f'(x) or   dx  = 0 to obtain the value of
              x = 1,                                                  the variable which causes the function to be
                                                                      maximum or minimum.
               d y  = 6(1) – 12                                   (d)  If there is more than one variables obtained,
                 2
                dx 2  = –6 , 0                                        investigate the nature of the values by using
                                                                      d y  (second derivative method) or  the
                                                                        2
                         17                                           dx 2            dy
              Given that the graph of the function y = x  + ax + b,   gradient of tangent  dx .
                                                 3
              where a and b are constants, has a turning point (0, 4).   (e)  Find the maximum value or the minimum
              Determine the values of a and b. Hence, determine the   value of the function according to the
              nature of the turning point.                            requirement of the problem.


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