4.1 Applications of the First Derivative

Report
Applications of the First
Derivative
By
Dr. Julia Arnold
using Tan’s 5th edition Applied Calculus for the managerial , life, and
social sciences text
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Increasing and Decreasing Functions
A function is increasing on an interval (a,b) if for any
two numbers x1 and x2 in (a,b) f(x1) < f(x2) whenever
x1 < x2 .
Increasing Function
A function is decreasing on an interval (a,b) if for any two numbers
x1 and x2 in (a,b) f(x1) > f(x2) whenever
x1 < x 2 .
Decreasing Function
How the first derivative can tell us where a function is
increasing and where a function is decreasing.
Decreasing Function
First Derivative is
negative.
The one feature
shared by all of
these tangent
lines is that their
slopes (first
derivative of f)
are negative.
Increasing function, First Derivative is positive.
If the function is constant the first derivative will be 0.
Example 1: Determine the intervals where the function is
increasing and decreasing.
f(x)  x3  3x2  24x  32
Step 1: Find the derivative of f
f(x)  x  3x  24x  32
3
2
f(x)  3x2  6x  24
Step 2: Set the first derivative equal to 0 and solve for x.
f(x)  3x2  6x  24
0  3x2  6x  24
0  3(x2  2x  8)
0  3(x  4)(x  2)
x  4, x  2
We have 3 intervals
-2
4
 ,2 On this inverval f’ >0
 2,4  On this inverval f’ <0
4,   On this inverval f’ >0
Which is determined by the graph.
We can also determine the same result by testing:
 ,2  Test point inside interval -3.
2
f(3)  3 3  6 3  24  3(9)  18  24  21  0
 2,4 
Test point inside interval 0.
4,  
Test point inside interval 5.
2
f(0)  3 0  60  24  24  0
2
f(5)  35  65  24  3(25)  30  24  21  0
Interpreting the results.
 ,2  Since f’(x) > 0 on this interval, f(x) is increasing on this interval.
 2,4 
Since f’(x) < 0 on this interval, f(x) is decreasing on this interval.
4,  
Since f’(x) > 0 on this interval, f(x) is increasing on this interval.
1
Example 2: Find the intervals where the function f(x)  x 
x
is increasing and where it is decreasing.
First note that the domain does not contain 0. Why?
Thus the intervals over which the function is defined are
 ,00, 
Next, find f’(x).
1
x
1
f(x )  1  2
x
f( x )  x 
Set f’(x) = 0
1
x2
0  x2  1
0  x  1x  1
x  1, x  1
0 1
Set up intervals
using the domain
and the zeros of
f’(x)
 ,1
 1,0 
0,1
1,  
Now choose a test point contained inside the interval.
Intervals
 ,1
 1,0 
0,1
1,  
Test Point
-2
-.5
.5
2
Substitution into f’(x)
1
1 3

1


2
4
4
 2
1
1
f(.5)  1 

1

 3
2
.25
 .5
1
1
f(.5)  1 

1

 3
2
.25
.5
1
1 3
f(2)  1  2  1  
4 4
2
f(2)  1 
Interpretation of results:
 ,1 , since f’(x)>0 then f(x) is increasing.
On  1,0  , since f’(x)<0 then f(x) is decreasing.
On 0,1 , since f’(x)<0 then f(x) is decreasing.
On 1,   , since f’(x)>0 then f(x) is increasing.
On
Result
+
-
+
0,1 , since
On
f’(x)<0 then f(x) is
decreasing.
-1
On  ,1 , since
f’(x)>0 then f(x) is increasing.
On 1,   , since f’(x)>0
then f(x) is increasing.
1
On  1,0 , since
f’(x)<0 then f(x) is decreasing.
Example 3: Find the intervals where the function f(x)  x
is increasing and where it is decreasing.
Find f’(x).
f(x)  x
2
3
1
2 3
2

f (x )  x  3
3
3 x
This fraction does not equal 0, and the domain of f’(x) does
not contain 0, so to make the intervals, we will use the
discontinuity at x = 0.
 ,0
0, 
2
3
Choose a test point:
Test Point
 ,0
0, 
Substitution into f’(x)
-1
1
2
2

33  1  3
2
2
f(1)  3 
3 1 3
f(1) 
Result
+
Interpretation of results:
 ,0
0, 
On this interval, f’(x)<0 so f(x) is decreasing.
On this interval, f’(x)>0 so f(x) is increasing.
f(x) is decreasing
f(x) is increasing
The Rule:
1.Find all values of x for which f’(x)=0 or f’ is discontinuous
and identify the open intervals determined by these points.
2.Select a test point c in each interval found in step 1 and
determine the sign of f’(c) in that interval.
A. If f’(c) >0 in that interval, f is increasing on that interval.
B. If f’(c) <0 in that interval, f is decreasing on that interval.
Practice problem: Find the intervals where the following
function is increasing or decreasing. f(x)  x3  3x  4
The Rule:
1.Find all values of x for which f’(x)=0 or f’ is discontinuous
and identify the open intervals determined by these points.
Which of the following is the correct first step?
f(x)  x3  3x  4
f(x)  x3  3x  4
f(x)  3x2  3
f(x)  3x2  3
0  3(x2  1)
0  3(x2  1)
0  x2  1
1  x
0  x2  1
1x
 1,1 _ and _ 1,  
and _ 1,  
The _ open _ int ervals _ are _  ,1
The _ open _ int ervals _ are _  ,1
Practice problem: Find the intervals where the following
function is increasing or decreasing. f(x)  x3  3x  4
The Rule:
1.Find all values of x for which f’(x)=0 or f’ is discontinuous
and identify the open intervals determined by these points.
Which of the following is the correct first step?
If you picked this one, you are on your way!
f(x)  x3  3x  4
f(x)  3x2  3
0  3(x2  1)
0  x2  1
1  x
The _ open _ int ervals _ are _  ,1
 1,1 _ and _ 1,  
2.Select a test point c in each interval found in step 1 and
determine the sign of f’(c) in that interval.
A. If f’(c) >0 in that interval, f is increasing on that interval.
B. If f’(c) <0 in that interval, f is decreasing on that interval.
f(x)  x3  3x  4
f(x)  3x2  3
1  x
 ,1
 1,1
1,  
Now pick test points and determine the sign of
f’ at that test point. Do this before you click the
mouse again.
Test Points
-2
0
2
Value in f’
3(4) - 3 = 9
3(0) - 3 = -3
3(4) -3 = 9
Sign of f’
+
+
Did you get the same result?
f’(c) >0 means f’ is positive and f’(c) <0 means f’ is negative.
Thus using A and B above what conclusion can you make about the function in
the 3 intervals? Answer before you click the mouse.
 ,1 f’(-2) > 0 therefore f(x) is increasing on this interval
 1,1
1,  
f’(0) < 0 therefore f(x) is decreasing on this interval
f’(2) > 0 therefore f(x) is increasing on this interval
Relative Extrema
A function has a relative minimum at x = c if there exists an
open interval (a,b) containing c such that f(x)  f(c) for all x
in (a,b).
A function has a relative maximum at x = c if there exists an
open interval (a,b) containing c such that f(x)  f(c) for all
x in (a,b).
f(c1)
rel.max
c1
f(c2)
rel.min
c2
These are relative because
they are maximums or
minimums in a locale, not
necessarily for the entire
number line.
To find the relative maximum and minimum, observe that the
maximum and minimum represents places where the tangent
line would be horizontal or have a slope of 0.
Tangent lines
c1
c2
Definition of Critical Point
A critical point of a function f is any point x in the
domain of f such that f’(x)=0 or f’(x) does not exist.
Procedure for finding relative extrema
The First Derivative Test
1. Determine the critical points of f.
2. Determine the sign of f’(x) to the left and right of each
critical point.
A. If f’(x) changes sign from positive to negative as
we move across a critical point x = c, then f(c) is a
relative maximum.
B. If f’(x) changes sign from negative to positive as
we move across a critical point x = c, then f(c) is a
relative minimum.
C. If f’(x) does not change sign as we move across a
critical point x = c then f(c) is not a relative
extremum.
Find the relative maxima and relative minima of the function
f(x)  x3  3x2  24x  32
1. Determine the critical points of f.
A critical point of a function f is any point x in the domain of f such that
f’(x)=0 or f’(x) does not exist.
f(x)  x3  3x2  24x  32
f' (x)  3x2  6x  24
0  3(x2  2x  8)
0  3(x  4)(x  2)
x  4, x  2
Since 4, and -2 are in the domain of f and cause f’ to be 0, they
are critical points of f.
2. Determine the sign of f’(x) to the left and right of each critical
point.
+
Test
Points
-2
max
-3
+
4
min
0
5
f' (3)  3 3  6 3  24  27  18  24  21
2
f' (0)  30  60  24  24
2
=-
f' (5)  35  65  24  75  30  24  21
2
=+
=+
f(x)  x3  3x2  24x  32
x = -2 is a relative maximum because as you check the signs
across the critical point from left to right you go from + to which means increasing then decreasing.
Creates a max.
The maximum is at (-2,60)
f(2)   2  3 2  24 2  32  8  12  48  32  60
3
2
x = 4 is a relative minimum because as you check the signs across the
critical point from left to right you go from - to + which means
decreasing then increasing.
Creates a min.
The minimum is at (4,-48)
f(4)  4  34  244  32  64  48  96  32  48
3
2
Remember Example 2: Find the intervals where the function
1
f(x)  x 
is increasing and where it is decreasing.
x
We can use the work we did there for finding relative extrema
+

-1
-1 is a critical point
and is a relative max
0
+
1
1 is a critical point
and is a relative min

(1,2) is a relative min
-1
1
(-1,-2) is a relative max
In Example 3
f(x)  x
f(x)  x
2
3
2
3
1
2
2
f(x)  x 3  3
3
3 x
The derivative is undefined at 0 but 0 is in the domain of f,
therefore 0 is a critical point.
-

+
0

From earlier work we see that the signs tell us 0 is a relative
minimum.
f(x) is decreasing
f(x) is increasing
(0,0) is a relative minimum
Practice Problem: Find the relative max and relative min, if
x
any, for
h(x ) 
x 1
Procedure for finding relative extrema
The First Derivative Test
1. Determine the critical points of f.
2. Determine the sign of f’(x) to the left and right of each critical point.
Which of the following is the correct first step?
x
x 1
x  11  x  1  1
h(x) 
x  12
x  12
1
0
_ has _ no _ solutions
2
x  1
h( x ) 
x  1 _ makes _ f' (x) _ discontinuous,
x
x 1
x  11  x  1  1
h(x) 
x  12
x  12
1
0
_ has _ no _ solutions
2
x  1
therefore
x  1 _ makes _ f' (x) _ discontinuous.
it _ is _ a _ cri tical_ po int .
There _ are _ no _ critical_ po int .
h(x) 
Practice Problem: Find the relative max and relative min, if
x
any, for
h(x ) 
x 1
Procedure for finding relative extrema
The First Derivative Test
1. Determine the critical points of f.
2. Determine the sign of f’(x) to the left and right of each critical point.
Which of the following is the correct first step?
This is correct because there
are no zeroes and while x = -1
makes h’ discontinuous, x = -1
is not in the domain of h.
Definition of Critical Point
A critical point of a function f is any
point x in the domain of f such that
f’(x)=0 or f’(x) does not exist.
x
x 1
x  11  x  1  1
h(x) 
x  12
x  12
1
0
_ has _ no _ solutions
2
x  1
h(x) 
x  1 _ makes _ f' (x) _ discontinuous.
There _ are _ no _ critical_ po int .
Practice Problem: Find the relative max and relative min, if
x
any, for
h(x ) 
x 1
Procedure for finding relative extrema
The First Derivative Test
1. Determine the critical points of f.
2. Determine the sign of f’(x) to the left and right of each critical point.
Which of the following is the correct next step?
Since there are no
critical points, there
are no relative
extrema.
Although there are no critical
points, you still set up the
intervals using the undefined
point x = -1 and create test
points.
 ,1
 1,  
Practice Problem: Find the relative max and relative min, if
x
any, for
h(x ) 
x 1
Procedure for finding relative extrema
The First Derivative Test
1. Determine the critical points of f.
2. Determine the sign of f’(x) to the left and right of each critical point.
Which of the following is the correct next step?
Since there are no
critical points, there
are no relative
extrema.
This is the correct conclusion
concerning relative extrema.
Practice Problem: Find the relative extrema for
f(x) 
1 4
x  3x2  4x  8
2
Which is the correct first step conclusion?
1 4
x  3x2  4x  8
2
f(x)  2x3  6x  4
f(x) 
0  2x3  6x  4
0  x  3x  2
3
To solve we will need
to use the rational
zero theory from
precalculus which is:
if
p _ is _ a _ factor _ of _ 2
p   1,2
q _ is _ a _ factor _ of _ 1
p   1
p
  1,2
q
continued
Substitute x = 1
13  3(1)  2  0
The zeroes are 1,1 and
-2
Thus, x = 1 is a zero
Using synthetic
division to reduce
the equation
1 0 -3 2
1
+1 +1 -2
1 1 -2 0
x2  x  2  0
x  2x  1  0
x  2, x  1
continued
There are no critical
points.
The critical points are
1 and -2.
Practice Problem: Find the relative extrema for
If you said:
1 4
x  3x2  4x  8
2
f(x)  2x3  6x  4
f(x) 
f(x) 
1 4
x  3x2  4x  8
2
How do you determine whether the
critical points are a relative max or a
relative min?
0  2x3  6x  4
0  x3  3x  2
The zeroes are 1,1 and
-2
The critical points are
1 and -2.
A. Set up intervals on the number line
and test points. If you go from + to across the number then the number is a
rel max. If you go from - to + across the
number then the number is a rel min.
B. Substitute 1 and -2 into f(x),
whichever is larger is the max and
whichever is smaller is the min.
Practice Problem: Find the relative extrema for
If you said:
1 4
x  3x2  4x  8
2
f(x)  2x3  6x  4
f(x) 
0  2x3  6x  4
0  x  3x  2
3
The zeroes are 1,1 and
-2
The critical points are
1 and -2.
f(x) 
1 4
x  3x2  4x  8
2
How do you determine whether the
critical points are a relative max or a
relative min?
A is the correct answer
A. Set up intervals on the number line
and test points. If you go from + to across the number then the number is a
rel max. If you go from - to + across the
number then the number is a rel min.
Doing B may work some of the time.
B. Substitute 1 and -2 into f(x),
whichever is larger is the max and
whichever is smaller is the min.
Practice Problem: Find the relative extrema for
If you said:
1 4
x  3x2  4x  8
2
f(x)  2x3  6x  4
f(x) 
0  2x3  6x  4
The critical points are
1 and -2.
1 4
x  3x2  4x  8
2
A. Set up intervals on the number line
and test points. If you go from + to across the number then the number is a
rel max. If you go from - to + across the
number then the number is a rel min.
0  x3  3x  2
The zeroes are 1,1 and
-2
f(x) 
-
+
Rel Min
 ,2
 2,1
1,  
Test
-3
0
2
-2
+
1 Nothing
f’(test)
2(-27)-6(-3)+4=-32
2(0)-6(0)+4=4
2(8)-6(2)+4=8
There is a relative minimum at x = -2.
X = 1 is neither a relative max or a relative min.
Practice Problem: Find the relative extrema for
1 4
x  3x2  4x  8
2
f(x)  2x3  6x  4
f(x) 
0  2x  6x  4
3
0  x3  3x  2
The zeroes are 1,1 and
-2
The critical points are
1 and -2.
f(x) 
1 4
x  3x2  4x  8
2
Doing B may work some of the time.
B. Substitute 1 and -2 into f(x),
whichever is larger is the max and
whichever is smaller is the min.
1 4
1  312  4(1)  8  6.5
2
1
4
2
f(2)   2  3 2  4(2)  8  20
2
f(1) 
Which would lead you to conclude that 1 is
a relative max and -2 is a relative min
which would only be half right.
Finally, let’s look at the graph and see
what’s happening.
4y
3
2
1
4
3
2
Rel Min
1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
1
2
Neither
3
4
x
5
Go to the HW

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