VIEWS: 2 PAGES: 11 POSTED ON: 9/14/2012 Public Domain
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