12.5: absolute maxima and minima
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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 - PowerPoint PPT PresentationTRANSCRIPT
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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) = x3 – 12xA) [-5,5]
This function is continuous for all xf’(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) = -16F(-2) = 16F(-5) = -65F(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) = 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)
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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)
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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 2nd derivative test
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Example 2
Use the 2nd derivative test if necessary to find the local maxima and
minima for each function.
A) 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
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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
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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
x -1 0 .5 1 2
F’ - 0 + 0 +decreasing increasing increasing
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/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
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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/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