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12.5: Absolute Maxima and Minima

12.5: Absolute Maxima and Minima. Finding the absolute maximum or minimum value of a function is one of the most important uses of the derivative. For example: Finding the maximum profit, the time it takes for a drug to reach its maximum concentration

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12.5: Absolute Maxima and Minima

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  1. 12.5: Absolute Maxima and Minima

  2. Finding the absolute maximum or minimum value of a function is one of the most important uses of the derivative. For example: Finding the maximum profit, the time it takes for a drug to reach its maximum concentration in the bloodstream after an injection, or the minimum pollution in some areas.

  3. Absolute Maxima and Minima Definition: f (c) is an absolute maximum of f if f (c) >f (x) for all x in the domain of f. f (c) is an absolute minimum of f if f (c) <f (x) for all x in the domain of f.

  4. Local Extrema and Absolute Extrema

  5. Extreme Value Theorem Theorem 1. (Extreme Value Theorem) A function f that is continuous on a closed interval [a, b] has both an absolute maximum value and an absolute minimum value on that interval.

  6. Finding Absolute Maximum and Minimum Values Theorem 2. Absolute extrema (if they exist) must always occur at critical values of the derivative, or at end points. • Check to make sure f is continuous over [a, b] . • Find the critical values in the interval [a, b]. • Evaluate f at the end points a and b and at the critical values found in step b. • The absolute maximum on [a, b] is the largest of the values found in step c. • The absolute minimum on [a, b] is the smallest of the values found in step c.

  7. Example 1 Find the absolute maximum and minimum values of f(x) = x3 – 12x • [-5,5] This function is continuous for all x f’(x) = 3x2 – 12 = 0 3(x-2)(x+2) = 0 so x = 2 or -2. Both are critical values. Check: f(2), f(-2), f(-5) and f(5), we have F(2) = -16 F(-2) = 16 F(-5) = -65 F(5) = 65 Therefore, the absolute minimum value is (-5, -65) and the absolute maximum value is (5,65)

  8. Example 1 (continue) Find the absolute maximum and minimum values of f(x) = x3 – 12x B) [-3,3] Same critical values Check: f(2), f(-2), f(-3) and f(3), we have F(2) = -16 F(-2) = 16 F(-3) = 9 F(3) = -9 Therefore, the absolute minimum value is (2, -16) and the absolute maximum value is (-2,16)

  9. Example 1 (continue) Find the absolute maximum and minimum values of f(x) = x3 – 12x C) [-3,1] Same critical values -2 and 2, but since the interval is [-3, 1] so we do not check f(2) Check: f(-2), f(-3) and f(1), we have F(-2) = 16 F(-3) = 9 F(1) = -11 Therefore, the absolute minimum value is (1, -11) and the absolute maximum value is (-2,16)

  10. Functions with no absolute extrema

  11. To find the local extrema and or absolute extrema of interval that is not closed, we need to sketch the graph or use the second derivative.

  12. Important Note: Need to find out the critical values first before we can apply the 2nd derivative test

  13. Example 2 Use the 2nd derivative test if necessary to find the local maxima and minima for each function. • f(x) = x3 – 9x2 + 24x -10 f’(x) = 3x2 – 18x + 24 = 0 3(x2 – 6x + 8) = 0 3(x – 4) (x -2) = 0 so x = 2 or 4 both are critical points f’’(x) = 6x – 18 f’’(2) = -6 < 0 so f has a local maximum at x = 2 f’’(4) = 6 > 0 so f has a local minimum at x = 4

  14. Example 2 (continue) Use the 2nd derivative test if necessary to find he local maxima and minima for each function. B) f(x) = ex – 5x f’(x) = ex – 5 = 0 ex = 5 ln ex = ln 5 so x = ln 5 is a critical point f’’(x) = ex f’’(ln5) = eln5 = 5 > 0 so f has a local minimum at x = ln5

  15. Example 2 (continue) Use the 2nd derivative test if necessary to find he local maxima and minima for each function. C) f(x) = 10x6 – 24x5 + 15x4 f’(x) = 60x5 – 120x4 + 60x3 = 0 60x3 (x2 – 2x + 1) = 0 60x3(x – 1)2 = 0 so x = 0 or 1 both are critical points f’’(x) = 300x4 – 480x3 + 180x2 f’’(0) = 0 can’t determine, the 2nd derivative test is failed because f’’=0 f’’(1) = 0 can’t determine If this occurs, use the 1st derivative test Therefore, there is a local minimum at x=0 decreasing increasing increasing

  16. Second Derivative Test Theorem 3. Let f be continuous on interval I with only one critical value c in I. If f ’(c) = 0 and f ’’ (c) > 0, then f (c) is the absolute minimum of f on I. If f ’(c) = 0 and f ’’ (c) < 0, then f (c) is the absolute maximum of f on I.

  17. Example 3 • Find the absolute maximum value of f(x) = 12 – x – 9/x on the interval (0, ∞) F’(x) = -1+ 9/x2 = 0 - x2 + 9 = 0 multiply both sides by x2 x2 – 9 = 0, (x-3) (x+3) = 0, so x = -3 or 3, only 3 is a critical value F’’(x) = -18/x3 F’’(3) = -2/3 < 0 Since we only have 1 critical value and f’’ < 0 so this function has an absolute maximum at x = 3

  18. Example 3 (continue) B) Find the absolute maximum value of f(x) = 5lnx - x on the interval (0, ∞) F’(x) = 5/x - 1 = 0 5/x = 1 x = 5 is a critical value F’’(x) = -5/x2 F’’(5) = -1/5 < 0 Since we only have 1 critical value and f’’ < 0 so this function has an absolute maximum at x = 5

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