Chapter 3
Applications of Differentiation
Definition of Extrema
Figure 3.1
Theorem 3.1 The Extreme
Value Theorem
Extreme Value Theorem
If f is continuous on [a,b] then f must have both an absolute maximum
and absolute minimum value in the interval.
Examples:
4
4
4
2
2
2
a
a
b
-5
5
b
-5
a
5
b
-5
5
-2
-2
-2
-4
-4
-4
Extreme Value Theorem
If f is continuous on [a,b] then f must have both an absolute maximum
and absolute minimum value in the interval.
Examples:
4
Absolute
Maximum
2
2
a
Abs Max
a
b
-5
4
4
5
2
a
b
-5
5
b
-5
5
-2
-2
Abs Min-2
-4
-4
-4
Abs Min
Absolute
Minimum
Abs Max
Extreme values can be in the interior or the end points of
a function.
4
No Absolute
Maximum
3
2
1
yx
2
D   ,  
-2
-1
0
1
2
Absolute Minimum

4
No
Maximum
3
2
1
yx
2
D   0, 2 
-2
-1
0
1
2
No Minimum

Definition of Relative Extrema
Local Extreme Values:
A local maximum is the maximum value within some
open interval.
A local minimum is the minimum value within some open
interval.

Figure 3.2
Absolute maximum
(also local maximum)
Local maximum
Local minimum
Local minimum
Absolute minimum
(also local minimum)
Local extremes
are also called
relative extremes.

Absolute maximum
(also local maximum)
Local maximum
Local minimum
Local minimum
Absolute minimum
(also local minimum)
Local extremes
are also called
relative extremes.

Absolute maximum
(also local maximum)
Local maximum
Local minimum
Notice that local extremes in the interior of the function
occur where f  is zero or f  is undefined.

Definition of a Critical Number
and Figure 3.4
Theorem 3.2 Relative Extrema
Occur Only at Critical Numbers
Note:
Maximum and minimum points in the interior of a function
always occur at critical points, but critical points are not
always maximum or minimum values.
Guidelines for Finding Extrema on a
Closed Interval
EXAMPLE 3 FINDING ABSOLUTE EXTREMA
Find the absolute maximum and minimum values of
f  x   x2/3 on the interval  2,3 .
f  x  x
2/3

1
3
2
f  x  x
3
2
f  x  3
3 x
There are no values of x that will make
the first derivative equal to zero.
The first derivative is undefined at x=0,
so (0,0) is a critical point.
Because the function is defined over a
closed interval, we also must check the
endpoints.

f  x  x
2/3
D   2,3
x  0 f  0  0
At:
f  1  1
To determine if this critical point is
actually a maximum or minimum, we
try points on either side, without
passing other critical points.
f 1  1
Since 0<1, this must be at least a local minimum, and
possibly a global minimum.
At: x  2
At:
x3
2
3
f  2    2   1.5874
2
3
f  3   3  2.08008

f  x  x
2/3
D   2,3
x  0 f  0  0
At:
f  1  1
f 1
To determine if this critical point is
Absoluteor minimum, we
actually a maximum
0, 0
try points on minimum:
either side, without
passing other critical points.
Absolute
3, 2.08
1
maximum:




Since 0<1, this must be at least a local minimum, and
possibly a global minimum.
At: x  2
At:
x3
2
3
f  2    2   1.5874
2
3
f  3   3  2.08008

Critical points are not always extremes!
2
yx
3
1
-2
-1
0
1
2
f0
-1
(not an extreme)
-2

2
yx
1/ 3
1
-2
-1
0
1
2
f  is undefined.
-1
(not an extreme)
-2
p
Locate the extrema of the
function on [-5, 10]
x3 5 x 2
f ( x)  
 24 x
3
2
1. Find f’(x) = 0
f '( x)  x 2  5 x  24
0   x  8  x  3
C.P.at x  8, 3
2. Test the critical
points and the
endpoints in the
original function.
3. Identify the
highest and lowest
points.
f (5)  15.833
f (3)  40.5
f (8)  181.3
f (10)  156.7
Absolute max at (3, 40.5)
Absolute min at (8, -181.3)
max
min
f ( x)  x3  5x 2  9 x  45
1. Find f’(x) = 0
Find the extrema on [-5, -2]
f '( x)  3x 2  10 x  9
0  3x 2  10 x  9
x  .407 and .737
2. Test the critical
points and the
endpoints in the
original function.
f(-5) = 0
f(-4.07) = 7.035
f(.737) = -48.52
f(-2) = -15
3. Identify the
highest and lowest
points.
We don’t use
x = .737
since it’s not
in the
interval.
Absolute max (-4.07. 7.035)
Absolute min (-2, -15)
max
min
• Find the extrema of
f ( x)  2sin x  cos x2 x on the interval [0,2p ]
f ' ( x)  2cos x  2sin 2 x  0  2cos x  4cos x sin x  0  2cos x(1  2sin x)  0
1
 cos x  0 or sin x 
2
f (0)  1
p 3p 7p 11p
So, x  , , ,
2 2 6 6
p
f ( )  3 ----- MAXIMUM
2
7p
3
f ( )   ----- MINIMUM
6
2
3p
f ( )  1
2
11p
3
f(
)   ----- MINIMUM
6
2
f (2p )  1
2 x 2  8 x  6
f ( x)   2
 x  2x  6
1. Find f’(x) = 0
3  x  0
0 x3
Find the extrema of the
function.
4 x  8
f '( x)  
 2x  2
f '(2)  0
3  x  0
0 x3
f ' 1  0
2. Test the critical
points and the
endpoints in the
original function.
3. Identify the
highest and lowest
points.
f (3)  0
f (2)  2
f (0)  6
lim f (0)  6 f (1)  7
f (3)  3
x 0
Absolute max (-2, 2)
Absolute min (1, -7)
max
min
Or you can set the equation equal to 0 and look for the
critical number!!!!!!