MAT01B1: Maximum and Minimum Values
Dr Craig
16 August 2017
My details:
I Consulting hours:
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Thursday 09h40 – 11h15
Friday 11h20 – 12h55
I Office C-Ring 508
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(Or, just google “Andrew Craig maths”.)
Assignments and Class Tests
I Collect Assignment 1 and Class Test 1
from the collection facility.
I Scripts will be thrown away after two
weeks.
I Check the memo and your scripts to learn
from these assessments.
e-Quiz 1
I Live now.
I 60min time limit per attempt. Unlimited
attempts.
I Answers are available immediately after
submission.
I Your score does not count towards
Semester Mark.
I Use this to prepare for Semester Test 1.
Semester Test 1
I Saturday 26 August
I D1 Lab 208
I Starts at 08h30. Be seated by 08h15.
I Scope: Ch 7.1–7.5, 7.8, 4.1, 4.2
I Also examinable: Proofs of Fermat’s
Theorem, Rolle’s Theorem, Mean Value
Theorem
Saturday classes
This week: 10h00 to 13h00 in D1 Lab K09.
IT Clash tomorrow
I 08h00 to 09h35 in C-LES 202
I Check your email if you can’t make that
time.
Today
I Pop Quiz
I Maxima, Minima and Extreme Values
I Fermat’s Theorem (with proof)
I Critical numbers
Pop Quiz: write down the following
I
∫secx dx
I formula for integration by parts
I squared identity with cot θ
I formula for sinA sinB
I cos2 x in terms of cos 2x
I trig substition for√3 + x2
I form of the partial fraction decomposition
for4x + 2
(2x + 7)3
Maximum and Minimum Values
Some examples of where we might want to
be able to calculate maximum and minimum
values:
I What shape will minimize the
manufacturing cost of a tin can?
I What is the maximum acceleration of a
space shuttle during take-off?
I At what angle should blood vessels
branch so as to minimize the energy
expended by the heart in pumping blood?
Definition:
Let c be a number in the domain D of a
function f . Then f (c) is the
I absolute maximum value of f on D
if f (c) > f (x) for all x ∈ D;
I absolute minimum value of f on D
if f (c) 6 f (x) for all x ∈ D.
The absolute maximum/minimum is often
called the global maximum/minimum.
f (a) is the abs. min., f (d) is the abs. max.
Definition:
the number f (c) is a
I local maximum value of f if
f (c) > f (x) when x is near c.
I local minimum value of f if
f (c) 6 f (x) when x is near c.
“x near c” = when x is in some open
interval containing c, i.e. x ∈ (c− ε, c + ε).
Important:
When we refer to a ∈ R as being a local or
absolute maximum or minimum we are
referring to a as a y-value.
Examples of maxima and minima:
I f (x) = cosx
I f (x) = x2
I f (x) = x3
Examples of maxima and minima:
Point about end-points
The end point of a closed interval cannot be
used as the x-value for a local minimum or
local maximum.
In the picture on the previous slide the
function is defined on [−1, 4]. Therefore
neither 37 (f (−1) = 37) nor 32 (f (4) = 32)
are local maxima.
Extreme Value Theorem
If f is continuous on a closed interval
[a, b], then f attains an absolute
maximum value f (c) and an absolute
minimum value f (d) at some numbers c
and d in [a, b].
We do not cover the proof of this theorem.
What if the hypotheses do not hold?
Fermat’s Theorem: If f has a local
maximum or minimum at c, and if f ′(c)
exists, then f ′(c) = 0.
Proof: Suppose that f has a local maximum
at c and f ′(c) exists. By the definition of a
local maximum we have that f (c) > f (x)
for any x sufficiently close to c. Thus, for h
positive or negative and sufficiently close to
0, we have f (c) > f (c + h).
Proof of Fermat’s Theorem continued
From f (c) > f (c + h) we get that
f (c + h)− f (c) 6 0.
For h > 0 and sufficiently small, we get
f (c + h)− f (c)h
6 0.
Now take limits to get
limh→0+
f (c + h)− f (c)h
6 limh→0+
0 = 0.
We took limits to get
limh→0+
f (c + h)− f (c)h
6 limh→0+
0 = 0.
Since f ′(c) exists, we have
f ′(c) = limh→0
f (c + h)− f (c)h
= limh→0+
f (c + h)− f (c)h
.
Hence f ′(c) 6 0.
If h < 0 but sufficiently close to 0, then
when we divide both sides of the inequality
f (c + h)− f (c) 6 0
by h, we getf (c + h)− f (c)
h> 0.
Taking limh→0− we have
f ′(c) = limh→0
f (c + h)− f (c)h
= limh→0−
f (c + h)− f (c)h
> 0.
We have shown that f ′(c) 6 0 and f ′(c) > 0
and hence f ′(c) = 0.
We assumed at the beginning that f had a
local maximum at c. If we had assumed that
f had a local minimum at c then we can use
a similar approach to prove that we will have
f ′(c) = 0.
Fermat’s Theorem: If f has a local
maximum or minimum at c, and if f ′(c)
exists, then f ′(c) = 0.
Definition: a critical number of a
function f is a number c in the domain of
f such that either f ′(c) = 0 or f ′(c) does
not exist.
Example:
Find the critical numbers of
f (x) = x3/5(4− x)
Solution: x = 0, x = 3/2.
A different version of Fermat’s Theorem
If f has a local maximum or minimum at
c, then c is a critical number of f .
How do we get this?
(p ∨ q) ∧ r → s ≡[¬((p ∨ q) ∧ r)
]∨ s
≡[¬(p ∨ q) ∨ ¬r
]∨ s
≡ ¬(p ∨ q) ∨ (¬r ∨ s)≡ (p ∨ q) → (¬r ∨ s)
Example:
Prove that the function
f (x) = x101 + x51 + x + 1
has neither a local maximum nor a local
minimum.
Hint: Use proof by contradiction. Assume
and that f (x) does have a local maximum
and apply Fermat’s Theorem. Then reach a
contradiction by showing that f ′(c) 6= 0 for
all c ∈ R.
Closed Interval Method: to find the
absolute maximum and absolute minimum
values of a continuous function f on a closed
interval [a, b]:
1. Calculate f (c) for every critical number
c ∈ (a, b).
2. Calculate f (a) and f (b), i.e. find the
value of f at the end points.
3. The largest value from Steps 1 & 2 is the
absolute maximum. The smallest value
from Steps 1 & 2 is the absolute min.
Example:
Find the absolute maximum and minimum
values of the function
f (x) = x3 − 3x2 + 1
where −12 6 x 6 4.
Solution: critical numbers x = 0, x = 2.
f (0) = 1, f (2) = −3f (−12 ) =
18, f (4) = 17
Abs. min = −3 and abs. max. = 17
Example:
Use calculus to find the absolute minimum
and maximum values of
f (x) = x− 2 sinx
on the interval [0, 2π].
Solution: critical numbers x = π3 , x = 5π
3
f (0) = 0, f (2π) = 2π
f (π3) = π/3−√3, f (5π3 ) = 5π/3 +
√3
Abs. min. = π/3−√3
Abs. max.= 5π/3 +√3.
Next week’s lecture on Ch 4.2:
I Use Fermat’s Theorem to prove Rolle’s
Theorem
I Use Rolle’s Theorem to prove the Mean
Value Theorem
I Applications of Rolle’s Theorem and the
Mean Value Theorem