Download Calculus 30 | Curve Sketching | Determining Extremes

E-2 Determining Extremes
To determine extremes on a closed interval, we must identify any peaks and valleys that create a local
maximum value or a local minimum value respectively.
To explore the development of determining extremes, go to the concept attainment activity
Extremes take two
forms, absolute and
relative
An absolute
maximum (global
maximum) occurs
at the highest yvalue on a graph, in
the example to the
left it occurs at
(3,4).
An absolute
minimum (global
minimum) occurs
at the lowest yvalue on a graph, in
the example it
occurs at (7,-3).
A relative
maximum (local
maximum) occurs
if you can move to
the right and left of
a point and in both
cases decrease the
value. In other
words, a hill is
created. Examples
occur at (3,4) and (1,3).
A relative
minimum (local
minimum) occurs
if you can move to
the right and left of
a point and in both
cases increase the
value. In other
words, a valley is
created. Examples
occur at (1,0) and
(7,-3).
Extreme Value Theorem
If f(x) is a continuous function on a closed interval, then f(x) has
both an absolute minimum and an absolute maximum on that
closed interval.
This theorem means that if a function has distinct endpoints (closed) and has no breaks (continuous), it
MUST have an absolute maximum and an absolute minimum value.
Clarification
This function is continuous on a This function is continuous on This function is discontinuous
closed interval
an open interval
on a closed interval
There is an absolute maximum
and minimum as this example
fits the Extreme Value Theorem
perfectly.
There is no absolute minimum
as there is not a defined point
that we can label as the
minimum.
The following diagrams display derivative that DO NOT EXIST.
There is no absolute maximum
as there is not a defined point
that we can label as the
maximum.
f '(a) does not exist because
there is a corner at x = c.
f '(a) does not exist at x = a
f '(a) does not exist because
because the tangent line is vertical.
f(x) is not continuous at x = a. (The slope of the tangent line is
1/0)
Critical Number
x = c is a critical number for the function f(x) if f(c) is defined and
f '(c) = 0 or f '(c) does not exist.
Another way of explaining a critical number, is that it occurs anywhere that the slope on the graph is
either horizontal or vertical.
Example 1 Determine any critical points for f (x) = 2x2 + 10x + 12
Example 2 Determine any critical points for
Example 3 Determine any critical points for
Determining Absolute Extremes
Complete the following steps to determine extremes
1. Determine f '(x).
2. Determine any critical numbers.
3. Determine if any corners exist.
4. Calculate the values of f(x) at all endpoints, critical numbers, and corners and assign a maximum and
minimum value.
Example 4 Determine the absolute extremes for f(x) = x3 + 3x2 + 3x + 1 on the interval [-2,3]
Step One - Determine f '(x)
f '(x) = 3x2 + 6x + 3
Step Two - Determine any critical numbers.
0 = 3x2 + 6x + 3
0 = 3(x2 + 2x + 1)
0 = x2 + 2x + 1
0 = (x + 1)(x + 1)
The critical value occurs at x = -1
Step Three - Determine if any corners exist.
No corners exist on a polynomial function, only on an absolute value or a split function.
Step Four - Calculate the values of f(x) at all endpoints, critical numbers, and corners and assign a
maximum and minimum value.
The easiest way to do this is to list all endpoints and the critical value in a chart to make your
conclusion.
Item
x
f(x)
left endpoint
-2
(-2)3 + 3(-2)2 + 3(-2) + 1 =
-1
critical number -1
(-1)3 + 3(-1)2 + 3(-1) + 1 =
0
right endpoint 3
(3)3 + 3(3)2 + 3(3) + 1 = 61
The minimum value occurs at (-2, -1). The maximum value occurs at (3, 61).
Example 5 Determine the absolute extremes for
on the interval [0,8]
Item
x
f(x)
left endpoint
0
6(0)2/3 - 4(0) + 3 = 3
critical number 1
6(1)2/3 - 4(1) + 3 = 5
right endpoint 8
6(8)2/3 - 4(8) + 3 = -21
The minimum value occurs at (8,-21). The maximum value occurs at (1,5).
Many additional examples are found in the question and answer section, including absolute value and
split functions.