Download Section 3.1 Day 1 Extrema on an Interval

Section 3.1 Day 1
Extrema on an Interval
AP Calculus BC
Learning Targets
• Define the terms absolute extrema, relative extrema,
absolute maximum/minimum, relative
maximum/minimum, critical number
• Determine the absolute extrema and relative extrema from
a graph
• Determine the absolute extrema analytically of a function
• Define and apply the Extreme Value Theorem
• Determine the critical numbers of a function
• Communicate the relationship between the relative
extrema and critical numbers
Intermediate Value Theorem – Bonus Theorem
• If 𝑓 is continuous on the closed interval [𝑎, 𝑏] and 𝑘 is any
number between 𝑓 𝑎 and 𝑓(𝑏), then there is at least one
number 𝑐 in [𝑎, 𝑏] such that 𝑓 𝑐 = 𝑘.
BONUS Example
• Let 𝑓 𝑥 = 𝑥 3 + 2𝑥 − 1. Explain why there must be a value 𝑐
for 0 < 𝑐 < 1 such that 𝑓 𝑐 = 0.
• 1. Since 𝑓(𝑥) is a polynomial, it is continuous on the given
interval.
• 2. We need to find one value below 0 and one value above 0 to
use IVT. Let’s check the end points of the interval.
• 3. 𝑓 0 = −1 and 𝑓 1 = 2. Thus, 𝑓 0 < 0 < 𝑓 1 .
• 4. By IVT, there must be a 𝑐 in 0 < 𝑐 < 1 such that 𝑓 𝑐 = 0.
Extrema
The minimum &
maximum values of
the function on the
given interval
Absolute Maximum
The maximum
function value on
the given interval
Absolute Minimum
The minimum
function value on
the given interval
Relative Extrema
The minimum &
maximum values of
the function on an
open interval
Relative Maximum
The maximum
function value on
the open interval
Relative Minimum
The minimum
function value on
the open interval
Key Point
• It is possible for a function to not have a
maximum or a minimum.
For example:
• Absolute Extrema: the function could have a hole at
those points
• Relative Extrema: the function could be
continuously increasing or decreasing over the
entire real number line
Extreme Value Theorem
• If f is continuous on a closed interval [a, b], then
f has both an absolute minimum and absolute
maximum on [a, b].
Critical Numbers
Let 𝑓 be defined at 𝑐. If 𝑓 ′ 𝑐 = 0 or if 𝑓 is not
differentiable at 𝑐, then 𝑐 is a critical number of 𝑓.
Relative Extrema & Critical Numbers
• If 𝑓 has a relative extrema at 𝑥 = 𝑐, then 𝑐 is a critical
number of 𝑓.
• In other words, relative extrema only occur at critical
numbers.
Example 1
1. Determine if the graph will
consist of both absolute
and relative extrema or just
relative extrema. Support
your stance.
2. Determine the coordinate
points of the extrema of
the function
Example 2
1. Determine if the graph will
consist of both absolute
and relative extrema or just
relative extrema. Support
your stance.
2. Determine the coordinate
points of the extrema of
the function
Procedure for finding Absolute Extrema
on a Closed Interval
1.
2.
3.
4.
Find the critical numbers of 𝑓 in the interval
Evaluate 𝑓 at each critical number in the interval
Evaluate 𝑓 at the endpoints of the interval
Use the function values (y-values) to determine
where the absolute extrema occur within the
interval
Example 3
Find the absolute extrema of 𝑓 𝑥 = 3𝑥 2 − 4𝑥 3 on the
interval [−1, 2]
1. 𝑓 ′ 𝑥 = 6𝑥 − 12𝑥 2 = 6𝑥(1 − 2𝑥)
1
′
2. 𝑓 𝑥 = 0 ⇒ 𝑥 = 0, 𝑥 =
3. −1, 7 , 0, 0 ,
Min (2, −20)
Max (−1, 7)
1 1
,
2 4
2
, 2, −20
x
-1
0
f(x)
7
0
½
2
¼
-20
Example 4
2
3
Find the absolute extrema of 𝑓 𝑥 = 2𝑥 − 3𝑥 on the
interval [−1, 3]
1
−
3
1. 𝑓 ′ 𝑥 = 2 − 2𝑥
2. 𝑓 ′ 𝑥 = 0 ⇒ 𝑥 = 1
′
3. 𝑓 𝑥 = 𝐷𝑁𝐸 ⇒ 𝑥 = 0
Min (−1, −5)
Max (0, 0)
x
-1
0
f(x)
-5
0
1
3
-1
-0.24
Example 5
Find the absolute extrema of 𝑓 𝑥 =
[−1, 1]
′
1. 𝑓 𝑥 =
𝑡 2 2𝑡 −(𝑡 2 +3)(2𝑡)
(𝑡 2 +3)2
2. 𝑓 ′ 𝑥 = 0 ⇒ 𝑥 = 0
3. 𝑓 ′ 𝑥 = 𝐷𝑁𝐸 ⇒ 𝑛𝑜𝑛𝑒
Min (0, 0)
Max
1
−1,
4
,
1
1,
4
=
𝑡2
𝑡 2 +3
2𝑡 3 −2𝑡 3 −6𝑡
(𝑡 2 +3)2
on the interval
=
−6𝑡
(𝑡 2 +3)2
x
-1
0
1
f(x)
¼
0
¼
Exit Ticket
• Find the absolute extrema of 𝑓 𝑥 = −𝑥 2 + 3𝑥 on the interval [0, 3]