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Section 5.1 – Increasing and Decreasing Functions

Section 5.1 – Increasing and Decreasing Functions. The First Derivative Test (Max/Min) and its documentation. 5.2. The Theory First……. THE FIRST DERIVATIVE TEST. If c is a critical number and f ‘ changes signs at x = c, then f has a local minimum at x = c if f ‘ changes from neg to pos.

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Section 5.1 – Increasing and Decreasing Functions

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  1. Section 5.1 – Increasing and Decreasing Functions The First Derivative Test (Max/Min) and its documentation 5.2

  2. The Theory First…… THE FIRST DERIVATIVE TEST • If c is a critical number and f ‘ changes signs at x = c, then • f has a local minimum at x = c if f ‘ changes from neg to pos. • f has a local maximum at x = c if f ‘ changes from pos to neg

  3. -3 5 1 3 NO CALCULATOR _ _ + There is a rel min at x = 1 because f ‘ changes from neg to pos There is a rel max at x = 3 because f ‘ changes from pos to neg

  4. The Theory…Part II EXTREME VALUE THEOREM If a function f is continuous on a closed interval [a, b] then f has a global (absolute) maximum and a global (absolute) minimum value on [a, b]. GLOBAL (ABSOLUTE) EXTREMA • A function f has: • A global maximum value f(c) at x = c if f(x) < f(c) for every x in • the domain of f. • A global minimum value f(c) at x = c if f(x) > f(c) for every x in • the domain of f.

  5. The Realities….. • On [1, 8], the graph of any continuous function HAS to • Have an abs max • Have an abs min

  6. _ + There is an abs min at x = -1/2

  7. 3 -1 -2 Justify your answer. _ _ + +

  8. _ +

  9. Justify your answer. _ _ +

  10. Justify your answer. _ +

  11. GRAPHING CALCULATOR REQUIRED

  12. x = 1.684 x = 0.964 x = 0

  13. [0, 0.398), (1.351, 3]

  14. The absolute max is 1.366 and occurs when x = 3 The absolute min is –0.098 and occurs when x = 1.351

  15. Let k = 2 and proceed

  16. 6 3 _ + +

  17. 1 0 _ _ +

  18. CALCULATOR REQUIRED t = 3.472

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