Middle Tennessee State University Wk 5 Absolute Maximum & Minimum Questions please read the instructions and follow them to answer the 5 questions at the l

Middle Tennessee State University Wk 5 Absolute Maximum & Minimum Questions please read the instructions and follow them to answer the 5 questions at the last page. Your answers should be explained step by step in words. You have to follow the same strategies shown in the week instructions. Let me know if you have any question. Week 5 Instruction
06/22/2020
You are highly recommended to read and follow in detail the instruction delivered in the
current lecture. This will be very helpful in solving the actual problems of the Final Exam. We go
now to Chapter 5 of our textbook and discuss it in some detail.
The first subject, to be focused on during this week, was already dealt with in the closing
segment of our previous week discussion. It touches upon the decisive role that the first and
the second derivatives of a given function play in sketching its graph. That is why, before going
to the material of the current lecture, you are highly recommended to return to the closing
segment of our Week 4 lecture. From that segment, one learns that the first derivative makes a
powerful instrument in the analysis of increase and decrease of the graph’s shape, and in the
analysis of critical points at which the graph might have relative extreme values. The second
derivative is instrumental in the analysis of concavity and inflection points of the graph.
To illustrate the point, we take a look at specific features of the graph of the function
??(??) = ?? 4 ? 8?? 2
(1)
In other words, we want to determine points at which the graph of this function has relative
maxima and minima, and also points at which the graph changes its concavity. Clearly, if all this
information is available then we have a good idea about the graph’s shape.
Of course, the derivatives ?? ? (??) and ?? ?? ??) are required to answer all the questions raised
above. These derivatives for the function in (1) are
?? ? (??) = 4?? 3 ? 16??
(2)
and
?? ?? (??) = 12?? 2 ? 16
(3)
?
Equating ?? (??) to zero, we have the following equation
4?? 3 ? 16?? = 0
or equivalently
??(?? 2 ? 4) = 0
(4)
Clearly, this equation has three distinct real solutions. They are: ??1 = ?2, ??2 = 0, and ??3 = 2 .
So, these are critical points for ??(??). They represent the x-coordinates of actual points on the
graph of ??(??) at which the graph might have relative extrema. To clarify the issue, we consider
each of these points separately, and find out if the derivative changes its sign passing through
that point. If yes, then the point is indeed the relative extreme value point. If not – then not.
As to the first of the three just found critical points (?? = ?2), the derivative in (2) is
negative a little on the left of it (?? ? (?2.1) ? ?3.44), but it is positive a little on the right
(?? ? (?1.9) ? 2.96). So, ?? = ?2 represents the relative minimum value point, because ??(??)
changes its behavior from going down to going up at ?? = ?2. As to the second of the three
critical points (?? = 0), the derivative in (2) is positive a little on the left of it (?? ? (?0.1) ? 1.60),
but it is negative a little on the right (?? ? (0.1) ? ?1.60). So, ?? = 0 represents the relative
maximum value point for ??(??). By a similar reasoning, we figure out that ?? = 2 is the relative
minimum value point for the function in (1).
Another important issue is also covered in Chapter 5, where we learn how to find
absolute extreme values of a function on a closed interval. To do so, you proceed through the
following three stage procedure:
(a) find all critical points for the given function on the indicated interval;
(b) compute values of the given function at every critical point;
(c) compute values of the given function at the endpoints of the interval;
The biggest of the values found at stages (b) and (c) represents the absolute maximum value
while the smallest of them represents the absolute minimum value of the given function.
To be specific, an illustrative example is presented. Let us consider the function
??(??) = ?? 3 + 2?? 2 ? 5?? ? 6
(5)
and determine its absolute extrema on the interval [?3, 3].
Following through stage (a), we differentiate our function
?? ? (??) = 3?? 2 + 4?? ? 5,
then set the derivative equal zero
3?? 2 + 4?? ? 5 = 0,
(6)
and then solve this quadratic equation with the quadratic formula. This results in the following
two solutions of the quadratic equation
??1 ? ?2.12 and ??2 ? 0.79
which represent the critical points of the function given in (5).
Proceeding then through stages (b) and (c), we obtain the list of values of our function:
??(?2.12) ? 4.06, ??(0.79) ? ?8.21, ??(?3) = 0, and ??(3) = 24,
the biggest of which is 24 representing the absolute maximum of our function, while the
smallest ? ?8.21 represents the absolute minimum of our function on the interval [-3, 3].
As to the concavity and inflection points of the function in (5), its second derivative is
?? ?? (??) = 6?? + 4
(7)
??
Setting it equal zero, we obtain ?? = ?2/3. Since ?? (??) is negative a little on the left of that
point (indeed, ?? ?? (?1) = ?2), but it is positive a little on the right of that point (indeed,
?? ?? (0) = 4), the concavity of the graph of the function in (5) changes from downward to
upward at ?? = ?2/3. This point, therefore, represents the inflection point of the graph.
Another issue, touched upon in Chapter 5, is the Antiderivative of a function.
A function ??(??) is called an Antiderivative of ??(??) on an interval ?? if ??? (??) = ??(??) in ??.
Consider an illustrative example, and show that ??(??) = 10?? 4 ? 5?? 3 + ?? ? 2 represents
an antiderivative of ??(??) = 40?? 3 ? 15?? 2 + 1.
Differentiating the function ??(??), one obtains
?? ? (??) = 40?? 3 ? 15?? 2 + 1 = ??(??)
and the desired result follows.
For another illustration, let us show that the function ??(??) = ?3??4 ? ?? + 4 is an
antiderivative for ??(??) =
12??3 ?1
2?3??4 ???+4
.
To differentiate the function ??(??), we use the Chain Rule, and obtain
?? ? (??) =
12??3 ?1
2?3??4 ???+4
= ??(??)
verifying the result.
Note that a very important statement directly follows from the definition of the
antiderivative. That is, if ??(??) is an antiderivative of ??(??), then any function from the family
??(??) + ??,
(8)
where ?? is an arbitrary constant, represents an antiderivative of ??(??).
The process of finding all antiderivatives of a function is called Antidifferentiation or
Integration. A special integral sign symbol is used in calculus to indicate that the operation of
integration is to be performed. That is
(9)
? ??(??)???? = ??(??) + ??
which reads “the Indefinite Integral of ?? of ?? with respect to ?? equals ?? of ?? plus ??”. It can be
interpreted as “the Indefinite Integral of a function with respect to its variable is the family of
antiderivatives of that function”. The function ??(??) in (9) is referred to as the integrand
function, and the whole expression ??(??)???? is called the integrand expression.
Keep always in mind that when you write down the integral symbol, never ever miss
the dx symbol. These two are just parts of a single symbol and they are meaningless without
each other in the integration.
The following rules are what we refer to in calculus as the Basic Rules of Integration.
Rule 1: The indefinite integral of a constant ??(??) = ??, where k is a constant, is
? ?????? = ???? + ??
Rule 2: The Power Rule of Integration
?? ??+1
??
? ?? ???? =
+??
??+1
?? ? ?1
Rule 3: The indefinite integral of a Constant Multiple of a Function
? ????(??)???? = ?? ? ??(??)????
?? is a constant
Rule 4: The Sum Rule
?[??(??) ± ??(??)]???? = ? ??(??)???? ± ? ??(??)????
Rule 5: The Indefinite Integral of the Exponential Function
? ?? ?? ???? = ?? ?? + ??
Rule 6: The Indefinite integral of the reciprocal function ??(??) = 1/??
1
? ???? = ln |??| + ??
??
???0
To illustrate the Basic Rules of Integration, let us find the indefinite integral
3
? (2?? 3 ? 11?? ?? + ?? 2 ) ????
Before finding this integral, we transform it into an equivalent form, and then apply Rules 2, 3,
4, and 5. That is,
??4
?? ?1
?(2?? 3 ? 11?? ?? + 3?? ?2 )???? = 2 4 ? 11?? ?? + 3 ?1 + ?? =
For another illustration, let us find the indefinite integral
?
(1????)2
???
???? =
1?2???+??
? ??? ????
= ? (??
?
1
2
1
2
? 2 + ?? ) ???? =
??4
2
1
3
??2
??2
1
2
? 2?? +
3
2
3
? 11?? ?? ? ?? + ??
+ ?? = 2?s-2?? +
2?????
3
+??
Consider another illustration
?
??3 +??2 ?5??+4
??2
???? = ?(?? + 1 ? 5???1 + 4???2 )???? =
??2
2
4
+ ?? ? 5 ln |??| ? ?? + ??
For many cases when we integrate complex looking functions, a special method called
the Method of Substitution appears to be a very powerful tool. This method suggests to
simplify the given integral before tackling it. This can be done by making a change of variable
(by introducing a new variable).
Let us, for example, find the indefinite integral
(10)
? ??(3?? 2 ? 1)10 ????
If we try to tackle this integral by expanding the integrand function, then we would be involved
in an extremely tedious algebra that is very time consuming and becomes therefore difficult to
do. If, however, instead of that, we make a substitution by introducing a new variable w as
?? = 3?? 2 ? 1
(11)
and differentiate then (11), we have
???? = 6??????
(12)
Note that the left side and the right side of the equation in (12) are differentials (not
derivatives). And it is worth reminding, at this moment, that the differential ????(??) of a given
function ??(??) is the product of its derivative ?? ? (??) and the differential ???? of its variable x. Thus,
the differential ????(??) of ??(??) is
????(??) = ?? ? (??)????
(13)
With this in mind, let us return to the integral in (10) and transform it (in light of the relations of
(11) and (12)) to the integral written it terms of the new variable w as
1
? ??(3?? 2 ? 1)10 ???? = ? 6 ?? 10 ????
which can easily be integrated by using the Basic Rules in Integration (Rules 2 and 3). Indeed,
1
1
1 ?? 11
? 6 ?? 10 ???? = 6 ? ?? 10 ???? = 6 ?
11
+?? =
?? 11
66
+??
And now, to complete the integration in (10), we need to return to our original variable ?? by
making the backward substitution in light of (11). This results ultimately into
(3?? 2 ?1)11
+??
? ??(3?? 2 ? 1)10 ???? =
66
For another illustrative example, let us consider another indefinite integral
2?? 3??
? 1??? 3?? ????
(14)
Before tackling this integral, we use Rule 3 according to which we move the constant multiple
“2” out of the integral symbol
2?
?? 3??
1??? 3??
????,
and make now a substitution
?? = 1 ? ?? 3?? ,
Differentiating the above equation, we have
???? = ?3?? 3?? ????
This transforms the original integral into a new one written it terms of the new variable w as
1
1
2
?2 ? ? ???? = ? ln|??| + ??
3 ??
3
and applying the backward substitution, the integral in (14) is found as
2?? 3??
2
? 1??? 3?? ???? = ? 3 ln|1 ? ?? 3?? | + ??.
Along with the Indefinite Integral that was introduced earlier in this lecture (see
equation (9)), calculus is also dealing with another type of integral which is called the Definite
Integral. To introduce the definite integral, let ??(??) be a continuous function defined on the
interval [??, ??], and let also ??(??) represent any antiderivative of ??(??), that is, ?? ? (??) = ??(??).
Then the Definite Integral of ??(??) ???? [??, ??], denoted as
??
??? ??(??)???? ,
represents a number computed, in terms of any antiderivative ??(??) of ??(??), as
??
??? ??(??)???? = ??(??) ? ??(??)
(15)
The relation in (15) is called the Fundamental Theorem of Calculus.
It is very important to emphasize that, as it follows from the Fundamental Theorem of
Calculus, the Definite Integral represents a number, whereas the Indefinite Integral,
? ??(??)???? = ??(??) + ??
according to its definition in (9), represents a family of functions.
This note makes the two types of integrals quite different of each other, but in
compliance with their definitions to compute either a Definite Integral or an Indefinite Integral
we need an antiderivative ??(??) of the integrand function ??(??).
For illustration, let us evaluate the definite integral
3
2
(16)
?2 ???? ?? ????
The Substitution Method will be used to compute this integral. But, before we go ahead and
apply it for our integral, it is worth to describe how this method works in a general case for a
definite integral
??
??? ??(??)????
Once a new variable, say w, is introduced
(17)
?? = ??(??)
the integral of (17) transforms to another definite integral
??
??? ?(??)????
(18)
where ?? = ??(??) and ?? = ??(??).
Note that the values of the integrals in (17) and (18) are exactly the same, and the key
purpose of the substitution method is to make the resultant integral in (18) easier to compute
compared to the original one in (17).
When we return to the definite integral in (16) of our illustrative example, we introduce
the new variable as
?? = ??2,
(19)
and differentiate the above equation
???? = 2??????
(20)
Clearly, the limits of the integral, which (16) transforms to (according to the relation in (19)),
are: ?? = 22 = 4 and ?? = 32 = 9. So, the integral in (16) transforms as
3
2
1
9
?2 ???? ?? ???? = 2 ?4 ?? ?? ????
and, in compliance with the Fundamental Theorem of Calculus, we ultimately obtain its value as
3
1
2
?2 ???? ?? ???? = 2 (?? 9 ? ?? 4 ) ? 4,024.24 .
_________________________________________
Final Exam
1. Find the absolute extrema of the function
??(??) = 3?? 4 ? 4?? 3 on the interval [?1, 2].
2. Verify directly that
3
??(??) = ?? 2 ?? ?? , ??(??) = ??2 ln ?? + ??2 , and ??(??) = ??? 4 + 2??
are the antiderivatives of
2?? 3 +1
3
??(??) = ??(2 + 3?? 3 )?? ?? , ??(??) = ??(3 + 2 ln ??), and ?(??) = ???4
3. Find the following indefinite integrals by the Basic Rules in Integration
?
??4 ?3??2 +??
2??
???? and ?(2?? 2 + 1)(3?? ? 4)????
4. Use the Substitution Method, to find the indefinite integrals
??2
? 2??(3?? 2 ? 1)5 ???? and ? ??3 ?1 ????
5. Evaluate the definite integrals
1
1
?0 ?? 2???1 dx and ??2(?? 4 + 1)2 ????
+2??
, respectively

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