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Lesson Plan #20
Date: Tuesday October 27th, 2009
Class: Intuitive Calculus
Topic: Extrema on an interval
Aim: How do we find extrema on a given interval?
Objectives:
1) Students will be able to use the Intermediate Value Theorem.
2) Students will be able to find Relative Extrema.
HW# 20:
1) Find the value of the derivative (if it exists) at the indicated extremum
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3
f ( x)  ( x  2) at (2,0)
2) Find any critical numbers of the function f ( x)  x 2 ( x  3)
3) Determine the absolute extrema of the function f ( x)  x 3  3x 2 in the interval  1,3
Note:
Intermediate Value Theorem: If f is a continuous function on [a, b] and k is any number between
f (a ) and f (b ) , then there is at least one number c between a and b such that f (c )  k
The Intermediate Value Theorem states that if the
domain of a function is a closed interval and the
function is continuous, then there are no holes or gaps
in the functions.
You may be familiar with the Intermediate
Value Theorem from last year’s work. That is, if we
are given a polynomial f and there are two numbers
a and b such that f (a ) is negative and f (b ) is
positive, then the function must have a zero (In other
words, there is some number that makes the function
zero, in other words, its graph must cross the x-axis).
Do Now:
1) Use the Intermediate Value Theorem to show that
the polynomial function f ( x)  x 3  2 x  1 has a
zero in the interval [0, 1].
The work to show there is a zero in the interval [0, 1] is shown below:
f (0)  03  2(0)  1  1
f (1)  13  2(1)  1  2
So by the Intermediate Value Theorem there is a number, x , between 0
and 1 such that f ( x)  0
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2) Use the Intermediate Value Theorem to show that for all spheres with radii in the interval [0, 5], there is one with
a volume of 275 cubic centimeters.
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Work to show there is a value of 275 in the function V   r 3 in the interval [0, 5]:
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3) Recall the definition for the derivative of a function (the long way)
Definition of the Derivative of a Function: The derivative of a f at
x is given by
f ( x  x)  f ( x)
x 0
x
provided the limit exists
f ' ( x)  lim
If the limit doesn’t exist, then we say that the function is not differentiable at x . If the limit exists, then we say the
function is differentiable at x . If the function is differentiable at x, then the function is continuous at x.
Theorem: Differentiability implies continuity: If f is a differentiable function at x  c , then f is continuous at x  c .
The reverse is not true. Continuity does not imply differentiability. Discontinuity destroys differentiability.
To find the derivative at a specific value of x, we could differentiate like we normally do and then substitute for x, but
some functions are not that easy to differentiate, like f ( x)  x  2 . In those cases, we could use the Alternate form of
a derivative to find the value of a derivative at a particular point.
: Alternate Form of the derivative: The derivative of f at c is given by
f ' (c)  lim
x c
f ( x )  f (c )
xc
Find the values of x at which the function f (x ) is not differentiable?
1) f ( x)  x  2
Solution:
lim
x2 0
f ( x)  f (2)
 lim
 1
x2
x2
x2
lim
x2 0
f ( x)  f (2)
 lim
1
x 2
x2
x2
x2
x 2
The derivatives from the left and right are not equal, therefore, f is not differentiable at x  2 and the graph does not
have a tangent line at the point (2,0)
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2) f ( x)  x
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Solution:
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f ( x)  f (0)
x3  0
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lim
 lim
 lim 2   . Since the limit does not
x 0
x 0
x 0
x0
x
x3
exist at x  0 , f is not differentiable at x  0 .
x 2  1, x  0
3) f (x ) =
 x2  4 , x  0
Procedure:
Write the Aim and Do Now
Get students working!
Take attendance
Give back work
Go over the HW
Collect HW
Go over the Do Now
Example:
Find the minimum and maximum values of the function
f ( x)  x 2  1in the
interval [1,2]
Example:
Find the minimum and maximum values of the function
f ( x)  x 2  1 in the interval  1,2 .
Definition of Extrema: Let f be defined on an interval containing c .
1) f (c ) is the minimum of f in the interval if f (c)  f ( x) for all
x in the interval
2) f (c ) is the maximum of f in the interval if f (c)  f ( x) for all
x in the interval.
The minimum and maximum values of a function on an
interval are called the extreme values, or extrema, of the function in the
interval. When you consider extrema within a closed interval,
the minimum and maximum of a function are also called absolute
minimum and absolute maximum.
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Here is a related theorem, called the Extreme Value Theorem.
The Extreme Value Theorem: If f is continuous on a closed
the interval.
interval, then f
has both a minimum and maximum in
We see that extrema can occur at interior points or endpoints of the interval. Extrema that occur at the endpoints are called
endpoint extrema
Determine from the graph whether or not f possesses a minimum on the given interval.
A)
B)
Does the function at right have a minimum in the interval
 2,4 ?
Does the function at right have a maximum in the interval
 2,4 ?
For this function, is there an open interval for which there is a maximum or minimum?
Definition of Relative Extrema:
If there is an open interval in which f (c ) is a maximum, then f (c ) is called a relative
maximum of f . (Also called local maximum)
If there is an open interval in which f (c ) is a minimum, then f (c ) is called a relative
minimum of f . (Also called local minimum)
Example:
Find the value of the derivative at each of the relative extrema shown in the graph
of
f ( x)  x 3  3 x .
Note that relative extrema occur where the derivative is zero or where the derivative
is undefined.
Definition of a critical number: If f is defined at c , then
c is called a
critical number of f if f ' (c )  0 or if f ' is undefined at c
Example #1:
Find any critical numbers of the function
A)
g (t )  x 2 ( x  3)
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B) g (t )  t 4  t
Theorem: Relative Extrema Occur Only At Critical Numbers:
If f has a relative maximum or relative minimum at x  c , then
c is a
critical number of f
Example: Find the absolute extrema (not necessarily relative since it is closed interval) of
f ( x)  3x 4  4 x 3 on the interval
1, 2 (Hint: Start by testing endpoints in the function and then critical numbers; the low value is the minimum and the high
value is the maximum)
Example:
Find the extrema of
interval)
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f ( x)  2 x  3x on the interval  1,3 . (This would indicate absolute extrema since it is on an open