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12.5: Absolute Maxima and Minima

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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.

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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.

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Local Extrema and Absolute Extrema

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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.

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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. a.Check to make sure f is continuous over [a, b]. b.Find the critical values in the interval [a, b]. c.Evaluate f at the end points a and b and at the critical values found in step b. d.The absolute maximum on [a, b] is the largest of the values found in step c. e.The absolute minimum on [a, b] is the smallest of the values found in step c.

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Example 1 Find the absolute maximum and minimum values of f(x) = x 3 – 12x A)[-5,5] This function is continuous for all x f’(x) = 3x 2 – 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)

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Example 1 (continue) Find the absolute maximum and minimum values of f(x) = x 3 – 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)

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Example 1 (continue) Find the absolute maximum and minimum values of f(x) = x 3 – 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)

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Functions with no absolute extrema

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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.

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

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Example 2 Use the 2 nd derivative test if necessary to find the local maxima and minima for each function. A)f(x) = x 3 – 9x x -10 f’(x) = 3x 2 – 18x + 24 = 0 3(x 2 – 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

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

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Example 2 (continue) Use the 2 nd derivative test if necessary to find he local maxima and minima for each function. C) f(x) = 10x 6 – 24x x 4 f’(x) = 60x 5 – 120x x 3 = 0 60x 3 (x 2 – 2x + 1) = 0 60x 3 (x – 1) 2 = 0 so x = 0 or 1 both are critical points f’’(x) = 300x 4 – 480x x 2 f’’(0) = 0 can’t determine, the 2 nd derivative test is failed because f’’=0 f’’(1) = 0 can’t determine If this occurs, use the 1 st derivative test x0.512 F’-0+0+ decreasingincreasing Therefore, there is a local minimum at x=0

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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.

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Example 3 A)Find the absolute maximum value of f(x) = 12 – x – 9/x on the interval (0, ∞) F’(x) = -1+ 9/x 2 = 0 - x = 0 multiply both sides by x 2 x 2 – 9 = 0, (x-3) (x+3) = 0, so x = -3 or 3, only 3 is a critical value F’’(x) = -18/x 3 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

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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/x 2 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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