LESSON 4 maxima and minima by fhSiUpI

VIEWS: 2 PAGES: 11

									  Differentiation (4)
- maxima and minima


•Finding   and Identifying Maxima
            and minima
Maxima, minima and stationary points
               B   Minimum
                                        Terminology
y
     Maximum                         B is a local maximum
                           D
                                     D is a local minimum
                               x
       Stationary points           Points where dy/dx=0 are
               dy                    call stationary points
dy                =0
               dx
dx                                 B & D are also known as
                                        turning points
                               x     (where they gradient
                                   changes from +ve to -ve)
     Finding Stationary Points
At a maximum            At a minimum

                dy
                   >0
                dx


+           -           -           +
                dy
                   <0
                dx

    dy
       =0
    dx                      dy
                               =0
                            dx
                        Example
  f(x)= x3 - 12x + 1
    “Find stationary points and determine their types”
  f’(x)= 3x2 - 12
Stationary            When x = -2           When x = +2
points occur        Pick point to left    Pick point to left
when gradient       (x = -2.1)            (x = 1.9)
[derivative] is 0   f`(x) = 3x(-2.1)2 -12 f`(x) = 3x(1.9)2 -12
                          = 1.23 [+ve]          = -1.17 [-ve]
3x2 - 12 = 0        Pick point to right    Pick point to right
3x2 = 12            (x = -1.9)             (x = 2.1)
x2 = 4              f`(x) = 3x(-1.9)2 -12  f`(x) = 3x(2.1)2 -12
                          = -1.17 [-ve]          = 1.23 [+ve]
x = 2 or x =    -2                        -ve +ve
                   +ve      -ve
                                Maximum              Minimum
              Example (continued)
   f(x)=   x3   - 12x + 1        f’(x)= 3x2 - 12
      “… and sketch the graph”
 Stationary points

   When x = -2         When x = +2          When x=0
                                            [y-axis]
+ve      -ve            -ve   +ve
            Maximum                Minimum y = 03 -12x0 +1
                                             =1
y = (-2)3 -12(-2) +1   y = (2)3 -12(2) +1
  = -8 + 24 + 1          = 8 - 24 + 1         (0,1)
  = 17                   = -15

Maximum at (-2,17)     Minimum at (2,-15)
           Example (continued)
f(x)= x3 - 12x + 1
                                   Crosses y-axis at (0,1)
 Cubic function
                               Y
                         20
   Maximum at (-2,17)




                                                       X
  Try                   -2         2
Page 125
  Ex A
                         -20           Minimum at (2,-15)
   Q1
The second derivative                f(x)


f(x)= x3 - 12x + 1
                                                  x
  First derivative
  f’(x) = 3x2 -12                    f’(x)

 [=0 at maximum and minimum]
                                                  x
                                -2            2

  Second derivative                  f’’(x)
  f’’(x) = 6x
 Gives the
 “rate of change of gradient”
                                                  x
    -ve at maximum
    +ve at minimum
        The second derivative
KEY POINTS
   The second derivative is denoted by d2 y
                                       dx2

  .. for a function p(x) it is denoted by p’’(x)
It is used at a simple way of telling if a
stationary point is a maximum or a minimum

 If d2y is negative then the stationary point is a maximum
    dx2
    d2y
 If    2 is positive then the stationary point is a minimum
    dx
The second derivative - example
f(x) =   1/
              3x
                3   - 2x2 + 3x +1 “Find stationary points and
                                    determine their nature”


 f’(x) = x2 - 4x + 3  Stationary points when f’(x)=0
 x2 - 4x + 3 = 0
 (x - 3)(x - 1) = 0
 x = 3 and x = 1 are maxima or minima
 Look at the second derivative to determine their nature
f’’(x) = 2x - 4
When x=3                            When x=1
f’’(x) = 2x3 - 4 = 2                f’’(x) = 2x1 - 4 = -2
which is positive                   which is negative
… hence a minimum                   … hence a maximum
The second derivative - ex C, 4e
f(x) = 4x3 - 9x2 - 30x +1      “Find stationary points and
                               determine their nature”
f’(x) = 12x2 - 18x - 30
  12x2 - 18x - 30 = 0   Stationary points when f’(x)=0
  2x2 - 3x - 5 = 0
  (2x - 5)(x + 1) = 0
  x = 2.5 and x = -1 are maxima or minima
 Look at the second derivative to determine their nature
f’’(x) = 24x - 18
When x=2.5                        When x=-1
f’’(x) = 24x2.5 - 18 = 42         f’’(x) = 24x-1 - 18 = -42
which is positive                 which is negative
… hence a minimum                 … hence a maximum
              Activity -
         now and homework

•   Page 128; exercise C
    • Q1, Q3, Q4 (a-c)
•   Page 122; ‘mixed questions’
    • Q1, Q2, Q9

								
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