Section 3.1 – Extrema on an
Interval
Maximum Popcorn Challenge
You wanted to make an open-topped box out of a rectangular sheet of
paper 8.5 in. by 11 in. The student must cut congruent squares out
x
of each corner of the sheet and then bend the edges of the sheet
varies upward to form the sides of the box. For what dimensions does the
from box have the greatest possible volume?
8.5 – 2x
box
to
box
V  11  2 x  8.5  2 x  x
V  4 x3  39 x 2  93.5 x
Draw a Picture
11 – 2x
x
x
x
x
Eliminate
Variable(s) Use Calculus to Solve the Problem
8.5 with other
V '  12 x 2  78 x  93.5
Conditions The slope of a tangent is 0 at a max
11
What needs to be Optimized?
You can’t cut an
Volume needs to be maximized:
4.9x4.9 in.
V  lwh
square out of an
8.5x11 in. paper
0  12 x  78 x  93.5
2
78
 782  41293.5 Quad.
x
212
x  1.585, 4.915
Form.
1.585 in x 7.829 in
l  11  2 1.585  7.829 w  8.5  2 1.585  5.329
x 5.329 in
A Beginning to Optimization Problems
One of the principal goals of calculus is to investigate the
behavior of various functions. There exists a large class
of problems that involve finding a maximum or minimum
value of a function, if one exists. These problems are
referred to as optimization problems and require an
introduction to terminology and techniques.
Example of an optimization problem:
A manufacturer wants to design an open box having a
square base and a surface area of 108 square inches.
What dimensions will produce a box with maximum
volume?
Extrema of a Function
Let f be a function defined on an interval I that contains the
number c. Then:
f(c)
f(c) is an absolute maximum of f
on I if f(c) ≥ f(x) for all x in I.
c
I
I
f(c) is an absolute minimum of f on
I if f(c) ≤ f(x) for all x in I.
c
f(c)
These values are also referred to as maximum/minimum,
extreme values, or absolute extrema.
Example 1
The graph of a function f is shown below. Locate the
extreme values of f defined on the closed interval [a,b].
f(x)
The highest point occurs
at x=b
a
c
d
e
f
b
x
The lowest point occurs at
x=d
Absolute Maximum: f(b)
Absolute Minimum: f(d)
Example 2
The graph of a function f is shown below. Locate the
extreme values of f defined on the open interval (0,1).
The function may
have a limit at the
highest point BUT
there is no
absolute
maximum value
f(x)
0.5
1
x
The function may
have a limit at the
lowest point BUT
there is no
absolute minimum
value
x
y
x
y
.9
.9
.1
.1
.99
.99
.01
.01
.999 .999
.001 .001
.9999
.9999
.0001
.0001
…
…
…
…
Absolute Maximum: None Absolute Minimum: None
Example 3
The graph of a function f is shown below. Locate the
extreme values of f defined on the closed interval [-1,1].
The highest point occurs
at x=1 & -1
f(x)
There is an issue
because this function is
not continuous on the
closed interval [-1,1]
1
-0.5
0.5
Absolute Maximum: 2
x
There is no lowest point
because a discontinuity
exists at the border
Absolute Minimum: None
White Board Challenge
Sketch a graph of the function with the
following characteristics:
It is defined on the open interval (-7,-1).
It is not differentiable at x=-4
It has a maximum of 5 and a minimum of -4.
The Extreme Value Theorem
A function f has an absolute maximum and an absolute
minimum on any closed, bounded interval [a,b] where it
is continuous.
Key Word.
f(x)
Absolute Maximum
a
c
d
e
f
b
x
This function is
continuous and defined
on the intervals.
Absolute Minimum
Absolute Maximum: f(b)
Absolute Minimum: f(d)
Example 1
In each case, explain why the given function does not
contradict the Extreme Value Theorem:
a.
f(x)
2x if 0  x 1
f x   
1 if 1  x  2
b. gx  x 2 on 0  x  2
g(x)

1
2
1
2
x
Even though the function has no
maximum, it does not contradict
the EVT because it is no
continuous on [0,2].
1
2
x
Even though the function has no
minimum, it does not contradict
the EVT because it is not
defined on a closed interval.
White Board Challenge
The function below describes the position a
particle is moving in a horizontal straight
line.
x  100  20t  5t
2
Find the average velocity between t = 2 and
4.
v
 10
t
Plural = Relative maxima/minima
Relative Extrema of a Function
A function f has a relative maximum (or local maximum)
at c if f(c) ≥ f(x) when x is near c. [This means that
f(c) ≥ f(x) for all x in some open interval containing c.]
A function f has a relative minimum (or local minimum) at
c if f(c) ≤ f(x) when x is near c. [This means that
f(c) ≤
f(x) for all x in some open interval containing c.]
f(x)
Typically
relative extrema
of continuous
functions occur
at “peaks” and
“valleys.”
a
c
d
e
f
b
x
f(d)
f(c) is a
f(e)
f(f)
Endpoints
is a relative
are
relative
minimum
not relative
at
maximum
minimum at
extrema.
x=f
x=d
x=c
x=e
Relative Extrema and Derivatives
Since relative extrema exist at “peaks” and
“valleys,” this suggests that they occur when:
The derivative is zero
(horizontal tangent)
The derivative does not
exist (no tangent)
Critical Numbers and Critical Points
Suppose f is defined at c and either f '(c)=0 or f '(c)
does not exist. Then the number c is called a
critical number of f, and the point (c, f(c)) on the
graph of f is called a critical point.
-3 is a critical number and (-3,7)
is a critical point
2 is a critical number and (2,3)
is a critical point
Example 1
Find the critical numbers for f  x   4 x3  5x2  8x  20 .
Domain of Function: All Real Numbers
Take the Derivative
f ' x 
d
dx
3
2
4
x

5
x
 8 x  20 

f ' x 
d
dx
4 x  5x  8x 
3
d
dx
2
d
dx
f '  x   12 x2 10 x  8  0
Solve the Derivative for 0
0  12 x 2  10 x  8
d
dx
20
0  2  6 x2  5x  4
0  2  3x  4  2 x  1
Both values are in
x  43 ,  12
f '  x   12 x  10 x  8
the domain.
Now find when the derivative When is the derivative undefined?
The derivative is defined
is 0 and/or undefined for x
for all real numbers.
values in the domain.
2
Example 2
Find the critical numbers for f  x  
x2
x 2
.
Domain of Function: All Real Numbers except 2
Take the Derivative
Solve the Derivative for 0
d x2
f '  x   dx x 2
0  x2  4x
f ' x 
f ' x 
f ' x 
f ' x 
 x  2 dxd x2  x2  dxd  x  2
 x 222
 x  22 x  x 1
Both
2
 x  2
2 x2  4 x  x2
 x  2 2
x2  4 x
 x  2 2
Now find when the derivative
is 0 and/or undefined for x
values in the domain.
0  x  x  4
values are in
the domain.
x  0, 4
When is the derivative undefined?
The derivative is not
defined for x = 2.
BUT x = 2 is not in the
domain of the function.
White Board Challenge
Consider the function below:
f  x   4x  6x  19
2
Find the equation of the tangent line to the
function at the vertex.
y  21.25
Example 3
Find the critical numbers for f  x   2 x  6  x  .
Domain of Function:All Real Numbers greater than or equal to 0
Take the Derivative
f ' x 
d
dx
12
2
x
 6  x 
f ' x 
d
dx
12
32
12
x

2
x


f '  x   12 x  2 x
d
dx
12
d
dx
32
Solve the Derivative for 0
0  6 x 1 2  3x1 2
x1 2  2 x 1 2
2 is in the
domain.
f '  x   6 x1 2  3x1 2
Now find when the derivative
is 0 and/or undefined for x
values in the domain.
x  2 and 0
When is the derivative undefined?
The derivative is not defined
for 0 or negative numbers.
Since 0 is in the domain, it is
also a critical point.
Example 4
Find the critical points for f  x    x  1
2
 x  2 .
Domain of Function: All Real Numbers
Take the Derivative
Solve the Derivative for 0
2
f '  x   dxd  x  1  x  2 
0   x  1 3x  3
f '  x    x  1
x  1,  1
 x  2    x  2   x  1
Both values are in the
2
domain.
f '  x    x  1 1   x  2   2  x  1
f '  x    x 1   x 1   x  2  2 When is the derivative undefined?
The derivative is defined
f '  x    x 1 3x  3
for all real numbers.
2 d
dx
Now find when the derivative is 0 and/or
undefined for x values in the domain.
 1, 4 and 1,0
d
dx
2
Find the y-value(s)
f  1   1  1  1  2   4
2
f 1  1  1 1  2   0
2
Example 5
Find the critical numbers for f  x   x  1 .
Domain of Function: All Real Numbers
Take the Derivative
 x  1 x  1
f  x  
 x  1 x  1
1 x  1
f ' x  
1 x  1
Now find when the derivative
is 0 and/or undefined for x
values in the domain.
Solve the Derivative for 0
The derivative never
equals 0.
When is the derivative undefined?
The derivative is
undefined for x=-1.
Since -1 is in
the domain
x  1
Critical Number Theorem
If a continuous function has a relative
extremum at c, then c must be a critical
number of f.
NOTE: The converse is not necessarily
true. In other words, if c is a critical
number of a continuous function f, c is
NOT always a relative extremum.
Important Note
Not every critical point is a relative extrema.
yx
3
Take the Derivative
y '  3x 2
Solve the Derivative for 0
0  3x 2
x0
Find the y-value(s)
y  03  0
 0, 0 
is NOT a relative extrema
White Board Challenge
Find the derivative of the function below:
f  x   sin  sec x 
f '  x   cos  sec x  sec  x  tan  x 
How do we Find Absolute
Extrema?
Suppose we are looking for the absolute extrema of a
continuous function f on the closed, bounded interval [a,b].
Since the EVT says they must exist, how can we narrow
the list of candidates for points where extrema exist?
f(x)
Absolute Maximum
a
c
d
e
f
b
x
On a closed interval,
extrema exist at endpoints
or at relative extrema.
Absolute Minimum
Absolute Maximum: f(b)
Absolute Minimum: f(d)
Procedure for Finding Absolute
Extrema on an Closed Interval
To find the absolute maximum and
minimum values of a continuous function f
on a closed interval [a,b]:
1. Find the values of f at the critical numbers
of f in (a,b).
2. Find the values of f at the endpoints of
the interval.
3. The largest of the values from Steps 1
and 2 is the absolute maximum value; the
smallest of these values is the absolute
minimum value.
Summary of Procedure
Find the absolute maximum and minimum of the function
graphed below.
Find the values of f at critical numbers
The value of the function at the
critical number 2 is:
-3 smallest
Find the values of f at the endpoints
The value of the The value of the
function at the function at the
enpoint 0 is:
enpoint 3 is:
largest 1
-2
Find the largest and smallest
values from the above work
1 is the maximum and -3 is the
minimum
Example 1
Find the absolute extrema of the function defined by the
equation f  x   x 4  2 x 2  3 on the closed interval [-1,2].
Find the values of f at critical numbers Domain of f: All Reals
Find the values of f at the endpoints
f '  x   dxd x 4  2 x 2  3


f '  x   4x  4x
3
0  4 x3  4 x
0  4 x  x 2  1
0  4 x  x 1 x  1 Not a
critical
x  0, 1,  1
point
f  0   0  2  0  3  3
since it’s
4
2
an
f 1  1  2 1  3  2
enpoint
smallest
4
2
f  1   1  2  1  3  2
smallest
4
2
f  2    2   2  2   3  11
4
2
largest
Answer the Question
The maximum occurs at x=2
and is 11; the minimum
occurs at x=-1 and 1 and is 2
Example 2
Find the absolute extrema of the function defined by the
equation f  x   x  2sin x on the closed interval [0,2π].
Find the values of f at critical numbers Domain of f: All Reals
Find the values of f at the endpoints
f '  x   dxd  x  2sin x 
f  0  0  2sin 0  0
f '  x   1  2cos x
0  1  2cos x
2cos x  1
cos x  12
x  3 , 53
f
f
 3   3  2sin 3
 53   53  2sin 53
smallest
 0.684853
 6.968039
largest
f  2   2  2sin 2  6.28
Answer the Question
The maximum occurs at
x=5π/3 and is 6.97; the
minimum occurs at x= π/3 and
is -0.68
Example 3
Find the absolute extrema of the function defined by the
equation f  x   x2 3  5  2 x  on the closed interval [-1,2].
Find the values of f at critical numbers Domain of f: All Reals
Find the values of f at the endpoints
f '  x   dxd 5 x 2 3  2 x 5 3

f ' x  x  x
0  103 x  x
0  103 x 1  x 
x 1

f  1   1
1 3
10 2 3
3
1 3
10 2 3
3
1 3
10
3
f  0   0
 5  2 1 
3
5  2 0
0
23
23
 5  2  1   7
largest
f  2   2
x=0 is a critical number too since it
makes the derivative undefined.
f 1  1
23
smallest
23
5  2  2
 1.5874
Answer the Question
The maximum occurs at x=-1
and is 7; the minimum occurs
at x=0 and is 0