31 extrema of a function

Aim: Extrema of a Function Course: Calculus

Do Now:

Aim: How do we find the maximum and minimum values over an interval of a function using derivatives?

A rectangle has a length x and perimeter of 36.a. Express the area A as a function of x and

determine the domain of the function.b. Graph the area functionc. Find the length and width of the rectangle of

maximum area

x

36 2x2

P = 2l + 2wA = lw

a.

A(x) x 36 2x2

0 < x < 18


Do Now

b. Graph the area function

c. Find the length and width of the rectangle of maximum area

80

70

60

50

40

30

20

10

-40 -20 20

f x = x36-2x

2A(x)

(9, 81)

Max. - (x, A(x))Max. - (9, 81)A = lw81 = 9w9 = w

where on graph?


Maximums and Minimums

Finding maximum and minimum values for various functions and relationships is very important and has many real-life applications associated with the finding of these critical values.

Definition of Extrema

Let f be defined on an interval I containing c.

1. f(c) is the minimum of f on I if f(c) < f(x) for all x in I.

2. f(c) is the maximum of f on I if f(c) > f(x) for all x in I.

3. The minimum and maximum of a function on an interval are the extreme values, or extrema of the function on the interval.

Note: A function does not necessarily have a maximum or

minimum.


Maximums and Minimums

The y value at a critical point can be called a local/relative maximum or a local/relative minimum.

Local/relative indicates that f(c) is when x is in the ‘neighborhood’ of c.

Global/absolute maximum and global/absolute minimum are the largest and smallest of the local maximua and minima.

A critical point is a point in ‘crisis’ – it stops and can then go in a different direction.

A plateau point is a critical point with a zero derivative but no max or min.


Where Extrema Occur

Max

Mina b

endpoints

Max

Min

stationery points

Max

Min

singular points

f levels off; tangent line is horizontal

f’ fails to exist


Extrema of a Function

6

4

2

f x = x2+16

4

2

f x = x2+1

f is continuous, [-1, 2]

f is continuous, (-1, 2)

g is not continuous,

[-1, 2]

6

4

2

2 1, 0( )

2, 0x x

g xx

minimum(0,1)

no maximum

[ ] minimum(0,1)

( )

maximummaximum(2,5)

[ ]no

minimum

If f is continuous on a closed interval [a, b], then f has both a min. and a max. on the interval.

Extreme Value Theorem


2

-2

-4

f(x) = x3 – 2x2

relative maximum‘hill’

2

-2

-4

5

2

3

9( 3)( ) xf x

x

6

4

2

( )f x x

Relative Extrema

(3, 2)

(0, 0)

relative minimum‘valley’


Value of Derivative at Relative Extrema

Find the derivative: 2

3

9( 3)( ) xf x

x

3 2 2

3 2

(18 ) (9)( 3)(3 )'( )( )

x x x xf x

x

2

4

9(9 )'( ) xf x

x

Evaluate the derivative for the relative maximum point (3, 2).

2

4

9(9 (3) )'(3) 03

f

2

-2

-4

5

2

3

9( 3)( ) xf x

x

(3, 2)


Value of Derivative at Relative Extrema

Derivative at relative minimum?6

4

2

( )f x x

(0, 0)

At x = 0, the derivative of f(x) = |x| does not exist.

At the relative extrema, the derivative is either zero or undefined


Criticial Number

Let f be defined at c. If f’(c) = 0 or if f’ is undefined at c,

then c is a critical number of f.

4

2

-2

-5

6

4

2

5

c

c

f’(c) is undefined f’(c) = 0

If f has a relative minimum or relative maximum at x = c, then c is a critical

number of f.


Guidelines

Guidelines for Finding Extrema on a Closed Interval

To find the extrema of a continuous function f on a closed interval [a, b], use the following steps.

1. Find the critical numbers of f on (a, b).

2. Evaluate f at each critical number in (a, b).

3. Evaluate f at each endpoint of [a, b].

4. The least of these values is the minimum. The greatest is the maximum.


Finding Extrema on a Closed Interval

Find the extrema of on the interval [-1, 2].

4 3( ) 3 4f x x x

Find the derivative: 3 2'( ) 12 12f x x x

Find the values of x for which f’(x) = 0

3 2'( ) 12 12 0f x x x 212 ( 1) 0

0,1x x

x

Left Endpt.

Critical Number

Critical Number

Right Endpt.

f(-1) = 7 f(0) = 0 f(1) = -1 f(2) = 16

15

10

5

-1 1 2

*

*

MIN MAXplateau


Model Problem

Find the extrema of on the interval [-1, 3].

2 3( ) 2 3f x x x

Find f’1 3

1 3 1 3

2 1'( ) 2 2 xf x

x x

Find the values of x for which f’(x) = 0

Left Endpt.

Critical Number

Critical Number

Right Endpt.

f(-1) = -5 f(0) = 0 f(1) = -1 f(2) -.24MIN MAX

4

2

-2

-4

1 3

1 3

1'( ) 2 0

'(0) is undef.& '(1) 0

xf x

x

f f


Model Problem

Find the extrema of on the interval [0, 2π].

( ) 2sin cos2f x x x

Find f’ '( ) 2cos 2sin2f x x x Find the values of x for which f’(x) = 0

'( ) 2cos 4cos sin 0

2 cos 1 2sin 0f x x x x

x x

sin 2x = 2 cosx sinx

3cos 0 when ,2 2

x x

7 111 2sin 0 when ,6 6

x x

5


Model Problem

Find the extrema of on the interval [0, 2π].

( ) 2sin cos2f x x x

3cos 0 when ,2 2

x x

7 111 2sin 0 when ,6 6

x x

Left Endpt.

Critical Number

Critical Number

Critical Number

Critical Number

Right Endpt

f(0) = -1 f(π/2) = 3

f(7π/6) = -3/2

f(3π/2) = -1

f(11π/6) = -3/2

f(2π) = -1

MIN MAX

5

31 extrema of a function

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