12.5: Absolute Maxima and Minima

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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) gt f (x) for all x in the domain of f.
f (c) is an absolute minimum of f if f (c) lt
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.

1. Check to make sure f is continuous over a, b
   .
2. Find the critical values in the interval a, b.
3. Evaluate f at the end points a and b and at the
   critical values found in step b.
4. The absolute maximum on a, b is the largest of
   the values found in step c.
5. 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 12x
• -5,5
• This function is continuous for all x
• f(x) 3x2 12 0
• 3(x-2)(x2) 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) 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.
• 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 lt 0 so f has a local maximum
  at x 2
• f(4) 6 gt 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 gt 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 cant determine, the 2nd
  derivative test is failed because f0
• f(1) 0 cant determine
• If this occurs, use the 1st derivative test

x -1 0 .5 1 2
F - 0 0
Therefore, there is a local minimum at x0
decreasing
increasing
increasing
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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) gt 0, then f (c) is the absolute
minimum of f on I. If f (c) 0 and f
(c) lt 0, then f (c) is the absolute maximum of f
on I.
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Example 3

• Find the absolute maximum value of f(x) 12 x
  9/x on the interval (0, 8)
• F(x) -1 9/x2 0
• - x2 9 0 multiply both sides by
  x2
• x2 9 0, (x-3) (x3) 0, so x
  -3 or 3, only 3 is a critical value
• F(x) -18/x3
• F(3) -2/3 lt 0
• Since we only have 1 critical value and f lt 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, 8)
• F(x) 5/x - 1 0
• 5/x 1
• x 5 is a critical value
• F(x) -5/x2
• F(5) -1/5 lt 0
• Since we only have 1 critical value and f lt 0
  so this function has an
• absolute maximum at x 5