MAT01B1: Maximum and Minimum Values

Dr Craig

16 August 2017

My details:

I acraig@uj.ac.za

I Consulting hours:

Monday 14h40 – 15h25

Thursday 09h40 – 11h15

Friday 11h20 – 12h55

I Office C-Ring 508

https://andrewcraigmaths.wordpress.com/

(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