Aim: How do we find the maximum and
minimum values over an interval of a
function using derivatives?
Do Now:
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 function
c. Find the length and width of the rectangle of
maximum area
P = 2l + 2w
A = lw
x
36  2x
2
36  2x 
 0 < x < 18
a. A(x)  x
a Function
2 
Aim: Extrema of
Course: Calculus
Do Now
x36-2 x
b. Graph the area A(x)
f  x =
2
function
c. Find the length
and width of the
rectangle of
maximum area where on graph?
(9, 81)
80
70
60
50
Max. - (x, A(x))
Max. - (9, 81)
40
30
A = lw
81 = 9w
9=w
20
10
-40
Aim: Extrema of a Function
-20
20
Course: Calculus
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.
Note: A function does not
a maximum
or for all x
2. f(c) is thenecessarily
maximumhave
of f on
I if f(c) > f(x)
minimum.
in I.
3. The minimum and maximum of a function on an
interval are the extreme values, or extrema of the
Aim: Extrema of a Function
Course: Calculus
function on the interval.
Maximums and Minimums
A critical point is a point in ‘crisis’ – it stops
and can then go in a different direction.
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 plateau point is a critical point with a zero
derivative but no max or min.
Aim: Extrema of a Function
Course: Calculus
Where Extrema Occur
Max
f levels off; tangent
line is horizontal
Max
Min
Min
a
b
endpoints
stationery points
Max
f’ fails to exist
Min
Aim: Extrema of a Function
singular points
Course: Calculus
Extrema of a Function
f  x  = x 2+1
f  x  = x 2+1
6
6
(2,5)
maximum
4
no
4
maximum
2
2
(0,1)
[minimum ]
 x 2  1, x  0
g( x )   6
 2, x  0
maximum
4
2
(0,1)
(minimum )
no
[ minimum]
g is not
f is continuous,
f is continuous,
continuous,
[-1, 2]
(-1, 2)
[-1, 2]
Extreme Value Theorem
If f is continuous on a closed interval [a, b], then
Aim: Extrema of a Function
Course: Calculus
f has both a min. and a max. on the interval.
Relative Extrema
f(x) = x3 – 2x2
2
relative maximum
‘hill’
relative minimum
‘valley’
-2
-4
6
2
(3, 2)
f ( x)  x
4
5
2
-2
9( x 2  3)
f ( x) 
x3
(0, 0)
-4
Aim: Extrema of a Function
Course: Calculus
Value of Derivative at Relative Extrema
Find the derivative:
9( x 2  3)
f ( x) 
x3
x 3 (18 x )  (9)( x 2  3)(3 x 2 )
f '( x ) 
( x 3 )2
2
9(9  x 2 )
f '( x ) 
x4
(3, 2) Evaluate the derivative
for the relative maximum
point (3, 2).
5
-2
9( x 2  3)
f ( x) 
x3
9(9  (3)2 )
f '(3) 
0
4
3
-4
Aim: Extrema of a Function
Course: Calculus
Value of Derivative at Relative Extrema
Derivative at relative minimum?
6
f ( x)  x
4
2
(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
Aim: Extrema of a Function
Course: Calculus
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.
f’(c) = 0
f’(c) is undefined
4
6
2
4
-5
c
2
-2
c
5
If f has a relative minimum or relative
maximum at x = c, then c is a critical
number of f.
Aim: Extrema of a Function
Course: Calculus
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.
Aim: Extrema of a Function
Course: Calculus
Finding Extrema on a Closed Interval
4
3
f
(
x
)

3
x

4
x
Find the extrema of
on the interval [-1, 2].
3
2
f
'(
x
)

12
x

12
x
Find the derivative:
Find the values of x
for which f’(x) = 0
f '( x )  12 x 3  12 x 2  0
12 x 2 ( x  1)  0
x  0,1
15
10
5
-1
*
1
Left
Endpt.
f(-1) = 7
2
Critical
Critical
Number Number
f(0) = 0
f(1) = -1
*
plateau
MIN
Aim: Extrema of a Function
Right
Endpt.
f(2) = 16
MAX
Course: Calculus
Model Problem
f ( x)  2 x  3 x2 3
Find the extrema of
on the interval [-1, 3].
13

2
x 1
Find f’ f '( x )  2  1 3  2 

13
x
x


Find the values of x
for which f’(x) = 0
4
2
-2
-4
 x1 3  1 
f '( x )  2 
0

13
 x

f '(0) is undef. & f '(1)  0
Left
Critical
Endpt.
Number
f(-1) = -5 f(0) = 0
MINAim: ExtremaMAX
of a Function
Critical
Number
f(1) = -1
Right
Endpt.
f(2) 
-.24
Course: Calculus
Model Problem
Find the extrema of f ( x )  2sin x  cos 2 x
on the interval [0, 2π].
Find f’
4
2
2
5
f '( x )  2cos x  2sin 2 x
Find the values of x
for which f’(x) = 0
sin 2x = 2 cosx sinx
f '( x )  2cos x  4cos x sin x  0
2  cos x  1  2sin x   0
 3
cos x  0 when x  ,
2 2
7 11
1  2sin x  0 when x 
,
6 6
Aim: Extrema of a Function
Course: Calculus
Model Problem
Find the extrema of f ( x )  2sin x  cos 2 x
on the interval [0, 2π].
 3
cos x  0 when x  ,
2 2
4
2
5
7 11
1  2sin x  0 when x 
,
6 6
2
Left
Endpt.
Critical
Number
Critical
Number
Critical
Number
3
f(7π/6)
= -3/2
f(3π/2) f(11π/6) f(2π) =
= -1
= -3/2
-1
MIN
MAX
f(0) = -1 f(π/2) =
Aim: Extrema of a Function
Critical
Number
Course: Calculus
Right
Endpt