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Unit 13: Maxima and Minima




                                                                                                Notes
                                            Figure  13.3

















          x = 1 and negative  when  x  = –1.  Thus the  function has  minima at  x  =  1  and maxima  at
          x = –1. The minimum value of the function is 2 and the maximum value = –2 which is less than
          the minimum value. These values are shown in Figure 13.3.

                                         ax b
                 Example: The function  y        has an extreme point at A(2, –1). Find the values
                                       x  1 x  4
          of a and b. What is the nature of the extreme point?
          Solution:
          Since point A(2, –1) lies on the function, we can write

                                2a b
                                        = –1 or 2a + b = 2                         ... (1)
                              2 1 2 4

                                     dy    a x  2  5x  4  ax b  2x  5
          Further,                      =                          0  for extrema
                                     dx           x 2  5x  4  2

                 a[4 – 10 + 4] – (2a + b)(4 – 5) = 0 or – 2a + 2a + b = 0 or b = 0
          Substituting this value in (1), we get a = 1
                                                         2
                                                        d y
          To check the nature of extreme point at A(2, –1), we find   2
                                                        dx

                         dy    x  2  5x  4  2x 2  5x  x  2  4
          Now               =
                         dx        x  2  5x  4  2  x 2  5x  4  2

                                        2           2  2
                                2
                         2
                        d y    x   5x  4  2x   2 4 x  x  5x  4 2x  5
                            =
                        dx  2                x  2  5x  4  4
                            x  2  5x  4  2  2x   2x       4
                          =            4   =   2     2  =     1 0  at x = 2
                              x 2  5x  4     x  5x  4    4

          Thus the extreme point is a maxima.








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