Section 12.1


Recall MARGINAL Costs, Revenue, Profit &
Sales are ALL first derivatives of
C(x), R(x),
P(x), S(x)

For our purposes, marginal functions represent the approximation of producing the next “x” item.
◦ Therefore:
 marginal cost = the cost of producing (x + 1) item


Thus far with derivatives we have inferred their meanings as it might relate to costs, revenues or profits.

Now, let’s consider how a graph of these inferences can be made.

f
 x
2 
6 x

7

( ⅔ , -1.48)
(-7, 0) (1, 0)

(-7, 0) (1, 0)

f ( x )
= x
3 +
2 x
2 -
4 x

(-2, 8)
( ⅔ , -1.48)

(-2, 8)
( ⅔ , -1.48)

(-2, 8)
( ⅔ , -1.48)

(-2, 8)
( ⅔ , -1.48)

f ( x )
= x
3 +
2 x
2 -
4 x f
¢
( )
=
3 x
2 +
4 x
-
4

Definition. The values of x in the domain of f where
f ‘(x) = 0 or does not exist are called the critical values of f.
Insight: All critical values are also partition numbers, but there may be partition numbers that are not critical values (where f itself is not defined).
If f is a polynomial, critical values and partition numbers are both the same, namely the solutions of
f ‘(x) = 0.

f
¢
( )
=
3 x
2 +
4 x
-
4
0
=
3 x
2 +
4 x
-
4

0
=
3 x
2 +
4 x
-
4 x
=
-
4
±
16
-
6 x
= -
4
±
8
6
= -
12
6
( ) or
( )
= -
4
±
16
+
48
6
4
6
® -
2 or
2
3
= -
4
±
64
6

Since our critical points are -2 and ⅔ , we construct a number line these two values highlighted.
Using f ’ (x) = (3x 2 + 4x – 4), substitute values in the interval to determine how the GRAPH of f(x) behaves
++++++++++ 0 -------------0 ++++++++++
Use -4 -2 Use -1 ⅔ Use 1

Using f ′ (x) = (3x 2 + 4x – 4), substitute values in the interval to determine how the GRAPH of f(x) behaves
++++++++++ 0 ------------0 ++++++++++
Use (-4) -2 Use (-1) ⅔ Use 1

• Critical values will ALWAYS occur in the domain of the function under consideration
• Any value that causes a 0 in the denominator is considered a Critical Value.
• A continuous function can INCREASE or Decrease on an interval containing a value of x for which f
′
(x) does NOT EXIST.
• To be a critical value, the number under consideration MUST be in the domain of the function.
• Values where a function is INCREASING or
Decreasing must be written in an Open Interval.

When the graph of a continuous function changes from rising to falling, a high point or local maximum occurs.
When the graph of a continuous function changes from falling to rising, a low point or local minimum occurs.
Theorem. If f is continuous on the interval (a, b), c is a number in (a, b), and f (c) is a local extremum, then either
f ‘(c) = 0 or f ‘(c) does not exist. That is, c is a critical point.


Let c be a critical value of f
,
◦ f(c) is a relative minimum if there exists within an open interval (a, b) value c such that f(c)
≤ f(x)

 What does this really mean????
In an interval test the values of f
This is f
'
(x)
′
(x) are as follows
-------------------0 +++++++++++++++
Values to the left of “c” c Values to the right of “c”
Since the values go from negative to positive, the graph will appear to decrease before “c” and increase after “c” making a shape. (hence, the minimum)


Let c be a critical value of f
,
◦ f(c) is a relative MAXIMUM if there exists within an open interval (a, b) value c such that f(c)
≥ f(x)
 What does this really mean????
 In an interval test the values of f '(x) are as follows
This is f
'
(x)
++++++++++++++++++ 0 -------------------------
Values to the left of “c” c Values to the right of “c”
Since the values go from positive to negative, the graph will appear to increase before “c” and decrease after “c” making a shape. (hence, the MAXIMUM)


Let c be a critical value of f
,
◦ f will NOT a relative minimum nor MAXIMUM if the sign does not change around f ' (c).
 What does this really mean????
 In an interval test the values of f '(x) are as follows
This is f
'
(x)
++++++++++++++++++ 0 ++++++++++++++++++
Values to the left of “c” c Values to the right of “c”
-------------------------
This is f
'
(x)
0 -------------------------
Values to the left of “c” c Values to the right of “c”


Determine the CRITICAL POINT(S) & intervals of increase/decrease for the following: f ( x )
=
2 x
3 x
4


Determine the CRITICAL POINT(S) & intervals of increase/decrease for the following:

Step 1: Find derivative.
f f ( x )
=
2 x
3 x
4
¢
( )
=
6 x
2 -
4 x
3


Determine the CRITICAL POINT(S) & intervals of increase/decrease for the following:

Step 2: Set derivative = 0, find critical points.
f
¢
( )
=
6 x
2 -
4 x
3
0
=
6 x
2 -
4 x
3 ®
2 x
2
(
3
-
2 x
)
=
0
2 x
2 =
0 or 0
=
3
-
2 x
® x
=
0 or x
=
3
2



Step 3
Create an interval test table for the critical points.
x
=
0 or x
=
3
2 f
¢
( )
=
6 x
2 -
4 x
3
This is f ’ (x)
+++++++++ 0 ++++++ -------------
-1 0 1 3
2
0
2



Step 4
Create intervals using critical values.
( )
( )
,
0,
( )
3
® and
( ) increasing
( )
= decreasing
+++++++++ 0 ++++++ 0 -------------
-1 0= ND 1 3
2
2 = MAX

NOTICE: at 0 no extrema occurs



Step 5
Make determinations: on
( )
,
( )
2 x
3
x 4 is INCREASING
( )
0, 3
® increasing
( )
= decreasing
+++++++++ 0 ++++++ 0 -------------
-1 0= ND 1 3
2
2 = MAX


Graph:
( )
=
2 x
3 x 4


Graph:
( )
=
2 x
3 x 4
Notice the tangents in the intervals we made in
STEP 3 go in the direction of the signs:
(
-¥
,0
)
= increasing
( )
( )
=
= increasing decreasing
3
2


According to some studies, oil companies are investing less in exploration since they continue to find less oil than they pump out of the ground.
The function
A(t) = 0.0265t
3 – 0.453t
2 + 1.796t + 7.47
(0 ≤ t ≤ 10) , represents in billions of dollars , the amount invested by large oil companies in exploration for new reserves t years after 1995.

Find the relative extrema and graph the function.


Find the relative extrema and graph the function.


A(t) = 0.0265t
3 – 0.453t
2 + 1.796t + 7.47
A
’
(t) = 0.0795t
2 – 0.906t + 1.796




Find the relative extrema and graph the function.
Create an interval table
A
’
(t) = 0.0795t
2 – 0.906t + 1.796
 t = 2.5553 or t = 8.841
(-∞, 2.5553)
+++++++++
-1
A
’
(t) = 0.0795t
2 – 0.906t + 1.796
0
(2.5553, 8.841)
-------0
2.5553
5 8.841
(8.841, ∞)
++++++++
9




Find the relative extrema and graph the function.
Create an interval table
A(t) = 0.0265t
3 – 0.453t
2 + 1.796t + 7.47
(
(
-¥
,2.5553
)
= increasing
(
8.841,
¥
2.5553,8.841
)
= decreasing
)
= increasing


Relative MAX at x = 2.553
Relative min at x = 8.841


The Graph:
A(t) = 0.0265t
3 – 0.453t
2 + 1.796t + 7.47

Write a description of the graph of the marginal revenue function, y = R ’ (x), including a discussion of any x-intercepts.

Analyzing the graph, we can see that from (0, 12500) the marginal revenue is positive. At x=12500, the marginal revenue is
0, and on the interval (12500, 25000) the marginal revenue is negative.

Write a description of the graph of the marginal profit function, y = P ’ (x), including a discussion of any x-intercepts.

Analyzing the graph, we can see that from (0, 1000) the Marginal
Profit is positive. At x=1000, the marginal revenue is 0, and on the interval (1000, 1500) the Marginal Profit is negative.

 First Derivative Test: determines where a function is increases or decreases
 Critical Values occur where the First Derivative = 0.
 Sign charts help to see the change in signs of the First Derivative which allows for locating the local maximum/minimum points.
 Local maximum is where f ′(x) changes from positive to negative
 Local minimum occurs when f ′(x) changes from negative to positive
 If f ′(x) is discontinuous at “c,” then f ′(x) DNE (Does Not Exist)

Section 12.2


2 nd Derivative Test stated:
◦ For y = f(x), the second derivative test, if it exists, is f
  
 d f
  
◦ If the 1 st Derivative Test determines intervals of increasing/decreasing & critical points; dx
 Then the 2 nd Derivative Test determines
CONCAVITY.


Understanding:
 When the 1 st Derivative is increasing and the
2 nd Derivative is positive, a graph of f will be concave UP.
 However, when the 1 st Derivative is decreasing and the 2 nd Derivative is negative, a graph of f will be concave Down.


Like with the 1 st Derivative Test, the 2 nd Derivative
Test has critical points called:



f ( x )
= x
3 +
2 x
2 -
4 x

f ( x )
= x
3 +
2 x
2 -
4 x f
  
3 x
2


4 x
3 x
2

4


4 x
0


4

3 x

2
 x

2


0


3 x

2


0 or
 x

2


0 x

2
3 or x
 
2
2
3 or

2

f ( x )
= x
3 +
2 x
2 -
4 x f f

  
 


3 x
2
6 x


4
4 x

4

6 x

4

0 x
 
2
3

2
3

f ( x )
= x
3 +
2 x
2 -
4 x f ′(x)=(3x – 2)(x + 2) ++++ 0
---------------
-5
-2 0 f ″(x) = 6x + 4
---------------
0
⅔
++++
2
0 +++++++++++++++++++++
-⅔


The graph increases from (-∞, -2) and (2, ∞) and decreases from (-2, ⅔).
The graph is concave Up from (-⅔, ∞) and concave down (-∞, -⅔)

Critical point x = -2 is a MAXIMUM value
Critical Point x= ⅔ is a local Minimum value
The inflection point x = - ⅔, is where the graph turns from concave Down to concave UP x = - ⅔

On what intervals is this graph concave Up?
On what intervals is this graph concave down?
Where is f ″(x) < 0?
Where is f ″(x) > 0?
Where is f '(x) increasing
?
Where is f '(x) decreasing
?
Where are the extrema?

On what intervals is this graph concave Up? (a, b) (b, c) (d, e)
On what intervals is this graph concave down? (c, d)
Where is f ″(x) < 0? (c, d)
Where is f ″(x) > 0?
(a, b) (b, c)
(d, e)
Where is f '(x) increasing
?
(a, b) (d, e)
Where is f '(x) decreasing
?
(b, c) (c, d)
Where are the extrema? x = b x = d

Section 12.2


With FIRST DERIVATIVE TEST: Interval Tables provide information about a function’s intervals of
Increase & Decrease

With Second DERIVATIVE TEST : Interval Tables provide information about a function’s CONCAVITY x f(x)
-3 0 1 2
-4 0 2 1
++ND +++++0 -------------0
4 5
-1 0
++++++++
0 1 2 4
+ +ND --------------0+++++++++++++++++++


In the EXAMPLE below:
◦ the Intervals of increase are (-∞, 0) and (4, ∞)
◦ Decrease Interval (1, 4)
◦ The Function is CONCAVED Up on (-∞, 0) & (2, ∞)
◦ Concaved down on (-∞, 2) x f(x)
-3 0 1 2
-4 0 2 1
++ND +++++0 -------------0
4 5
-1 0
++++++++
0 1 2 4
+ +ND --------------0+++++++++++++++++++

f ( x )
=
2 x
3 -
4 x
2 +
5 x
-
6 f ( x )
= e
x
2

f ( x )
=
2 x
3 -
4 x
2 +
5 x
-
6

f ( x )
= e
x
2

This graph below is f
’
(x)
.
Create a table based on this graph determine intervals of increase and decrease, any inflection points, intervals of concavity, then graph f(x)
.

x
-∞ < x < -2 x = -2
-2 < x < 0 x = 0
0 < x < 2 x = 2
-2 < x < ∞ f
’
(x) f(x)
Negative & increasing Decreasing & concave UP
Local max Inflection Point
Negative & decreasing Decreasing & concave down
Local min Inflection Point
Negative & increasing Decreasing & concave UP
Local max Inflection Point
Negative & decreasing Decreasing & concave down
Negative means the value of f
’
(x) is negative
Positive means the value of f
’
(x) is positive.