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Optimisation Using Derivatives I

Optimisation Using Derivatives I. Positive and negative values of the derivative as an indication of the points at which the function is increasing or decreasing zero values of the derivative as an indication of stationary points Greatest and least values of functions

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Optimisation Using Derivatives I

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  1. Optimisation Using Derivatives I Positive and negative values of the derivative as an indication of the points at which the function is increasing or decreasing zero values of the derivative as an indication of stationary points Greatest and least values of functions relationship between the graph of a function and the graph of its derivative methods of determining the nature of stationary points greatest and least values of functions in a given interval

  2. If the gradient is positive, the curve slopes upwards from left to right • If the gradient is negative, the curve slopes downwards from left to right

  3. Page 501 Ex 21.11-2(orally)

  4. State the values of x for which the curve is (a) increasing(b) decreasing(c) stationary

  5. State the values of x for which the curve is (a) increasing x<3, x>2(b) decreasing(c) stationary

  6. State the values of x for which the curve is (a) increasing x<3, x>2(b) decreasing -3<x<2(c) stationary

  7. State the values of x for which the curve is (a) increasing x<3, x>2(b) decreasing -3<x<2(c) stationary x=-3, x=2

  8. Positive and negative values of the derivative as an indication of the points at which the function is increasing or decreasing • Consider the following functions and their derivatives: • y = x2 • y = x3 • y = x3 - 2x2 – 2x + 1

  9. Page 501 Ex 21.13

  10. Greatest and Least Values • Find the greatest and least values of the function y = x3 + 2x2 - 4x + 5 for -3  x  2

  11. y = x3 + 2x2 - 4x + 5 Check bounds When x = -3, y= 8 When x = 2, y = 13 Check stationary values y = x3 + 2x2 - 4x + 5 y’ = 3x2 + 4x – 4 = (3x-2)(x+2) S.P. when y’=0  x= ⅔ or x = -2 When x = ⅔, y = 3.519 (3dp) When x= -2, y = 13  Greatest value is 13 ; Least value is 3.519 (3dp)

  12. Page 505 Ex 21.21

  13. Model : Find the stationary points on the curve y = x3+2x2 – 4x +2

  14. When x = ⅔, y = 0.519 (3dp) When x = -2, y = 10

  15. When x = ⅔, y = 0.519 (3dp) When x = -2, y = 10

  16. When x = ⅔, y = 0.519 (3dp) When x = -2, y = 10

  17. When x = ⅔, y = 0.519 (3dp) When x = -2, y = 10

  18. When x = ⅔, y = 0.519 (3dp) When x = -2, y = 10 Now check your answer on your GC Max at (-2, 10) Min at (⅔ ,0.519)

  19. Read bottom of 509 - 510

  20. Page 510 Ex 21.31 a,d,e,f4-6, 9, 3

  21. The Second Derivative The derivative of is called the second derivative  At a S.P., if is positive then that point is a minimum if is negative then that point is a maximum

  22. Model Find the nature of the stationary points on the curve y =x3+2x2 – 4x +2

  23. Model Draw the graph of y = 2 sin 3x for 0 < x < 2Π y = 2 sin 3x dy/dx = 6 cos 3x d2y/dx2 = -18 sin 3x S.P. when dy/dx = 0  6 cos 3x = 0  cos 3x = 0  3x = (2n+1)Π/2  x = (2n+1)Π/6  x = … Π/6, 3Π/6, 5Π/6, 7Π/6, …

  24. Check on GC

  25. Repeat Page 510 Ex 21.34 using the second derivative to determine the nature of stationary points + Page 518 Ex 21.42

  26. St Luke’s Anglican School Performance Faith Honour

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