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
in the bloodstream after an injection, or the
minimum pollution in some areas.
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.
Local Extrema and Absolute Extrema
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.
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.
Example 1
Find the absolute maximum and minimum values of f(x) = x3 – 12x
A) [-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)
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)
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)
Functions with no absolute extrema
• 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.
Important Note: Need to find out the critical values first before we
can apply the 2nd derivative test
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
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
Example 2 (continue)
Use the 2nd derivative test if necessary to find he local maxima and
minima for each function.
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
C)
x
F’
-1
-
decreasing
0
0
.5
+
1
0
increasing
2
+
increasing
Therefore, there is a
local minimum at x=0
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.
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
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