calcnotes0101.pdf

7/28/2019 CalcNotes0101.pdf

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(Chapter 1: Review) 1.01

CHAPTER 1:

Review(See also the Precalculus notes at http://www.kkuniyuk.com)

TOPIC 1: FUNCTIONS

PART A: AN EXAMPLE OF A FUNCTION

Consider a functionf whose rule is given by f x( ) = x2; f u( ) = u2 also works.

WARNING 1: f x( ) is read as f ofx or f atx. It doesnot mean

f timesx.

x is the input (or argument) for f, and x2 is the output or function value.

x f x2

This functionsquares its input, and the result is its output.

For example, f 3( ) = 3( )2

= 9.

3 f 9

Think of a function as acalculator button. In fact, your calculator should have a

squaring button labeled x2 .

f is a function, because no legal input yields more than one output.

A function button on a calculator never outputs two or more values at thesame time. We never get: I dont know. The answer could be 3 or 10.

A function is a special type of relation. Relations that are not functionspermit multiple outputs for a legal input.


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PART B: POLYNOMIAL FUNCTIONS

A polynomial inxcan be written in the form:

anx

n

+ an1

xn1

+ ... + a1x + a

0,

wheren is a nonnegative integer called the degree of the polynomial,the a

kcoefficients are typically real numbers, and the leading coefficient a

n 0.

A polynomial function has a rule that can be written as: f x( ) = polynomial in x.

For example:

4x3

5

2x

2+1 is a 3rd-degreepolynomial inxwith leading coefficient 4.

The rule f x( ) = 4x3 52

x2+1corresponds to apolynomial function f.

PART C: RATIONAL FUNCTIONS

A rational expression inxcan be written in the form:polynomial in x

nonzero polynomial in x.

Examples include:1

x

,5x

31

x2+ 7x 2

, and x7 + x which equalsx

7+ x

1

.

Observe in thesecondexample that irrational numbers such as 2 are

permissible.

Thelast example correctly suggests thatall polynomials are rationalexpressions.

A rational function has a rule that can be written as:

f x( ) = rational expression inx.

PART D: ALGEBRAIC FUNCTIONS

An algebraic expression inxresembles arational expression, except that radicals

andexponents that are noninteger rational numbers such as5

7

are also

permitted, even whenxappears in a radicand or in a base (but not in an exponent).


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Examples include: x andx

3+ 7x

5/7

x x+ 53

+ 4.

All rational expressions are algebraic.Although sources such asMathWorld

allow only algebraic numbers (such as rational numbers and 2 ) to be coefficients

in an algebraic expression, we will typically allow all real numbers (including ,for instance) in this work.

An algebraic function has a rule that can be written as:

f x( ) = algebraic expression inx.

A Venn diagram for expressions inxcorresponding to functions is below.Each disk represents asubset of every larger disk; for example, every polynomialis a rational expression and an algebraic expression (based on the definition in this

work).


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PART E: DOMAIN and RANGE

The domain of a functionf, which we will denote by Dom f( ) (though this isnot standard), is the set of all legal inputs.

The range of f, which we will denote by Range f( ) , is then the set of allresultingoutputs.

Unless otherwise specified (or in the context of a word problem), we typicallyassume that thedomain of a function is the set ofall real input values that yield anoutput that is areal number. This set is the implied (or natural) domain.

The implied domain of analgebraic function consists of all real numbersexceptthose that lead to (the equivalent of):

1)a zero denominator Think:0

, or

2)an even root of a negative-valued radicand Think: even( ) .As we study more types of functions, the list ofrestrictionswill grow.We will also exclude real numbers that lead to:

3) logarithms of nonpositive values Think: logb 0( )( ) , or

4)arguments of trigonometric functions that correspond to verticalasymptotes.

Word problems may imply other restrictions: nonnegativity, integer values, etc.

Example 1 (Domain and Range of the Squaring Function)

Let f x( ) = x2. Find thedomain and therangeof f.

SolutionThe implied domain of apolynomial function (such as this f) is , the set

of all real numbers. In interval form, is ,( ) . Its graph is the entirereal number line:


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WARNING 2: We use parentheses in the interval form, because (infinity) and (negative infinity) arenot real numbers and are

thereforeexcluded from the set of numbers. We will discuss infinity furtherin Chapter 2. Ifxapproaches , it (generally) increases without bound.Ifxapproaches , it (generally) decreases without bound.

Note: It is debatable whether an expression likex2+ x

x

is a polynomial. It

simplifies to x+1, but its domain excludes 0.

The resultingrangeof f is the set of all nonnegativereal numbers (all realnumbers that aregreater than or equal to 0), becauseeverysuch number isthesquareof some real number, andonly those numbers are.

WARNING 3: Squares of real numbers are never negative.

The graph of the range is:

Thefilled-in circle indicates that 0 is included in the range. We couldalso use a left bracket ([) at 0; the bracket opens towards theshading. The graph helps us figure out the interval form.

In interval form, the range is 0, ) . Thebracket next to the 0 indicates that0 is included in the range.

In set-builder form, the range is: y y 0{ } , or y : y 0{ } , whichis read the set of all real numbersysuch that y 0. Usingy instead ofx is

more consistent with our graphing conventions in thexy-plane (since wetypically associatefunction values in the range withy-coordinates), and it

helps us avoid confusion with the domain.

denotes set membership.


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Example 2 (Domain of a Function)

Let f x( ) = x 3, find Dom f( ), the domain of f.

Solution

f x( ) isreal x 3 0 x 3.

WARNING 4: We solve theweak inequalityx 3 0, not the

strict inequalityx 3> 0. Observe that 0= 0, a real number.

The domain of f

in set-builder form is: x x 3{ } , or x : x 3{ }

in graphical form is:

in interval form is: 3, )

Note: Range f( ) = 0, ) . Consider the graph ofy = f x( ).


Let f x( ) = 3 x4 . Find Dom f( ).

Solution

Solve the weak inequality: 3 x 0.

Method 1

3 x 0 Now subtract 3 from both sides. x 3 Now multiply or divide both sides by 1.

WARNING 5: We must thenreversethe direction of theinequality symbol.

x 3Method 2

3 x 0 Nowaddxto both sides.3 x Now switch the left side and the right side.

WARNING 6: We must thenreversethe direction of theinequality symbol.

x 3


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The domain of f

in set-builder form is: x x 3{ } , or x : x 3{ }


in interval form is: ,3(


Let f x( ) =1

x 3

. Find Dom f( ).

Solution

This is similar to Example 2, but we mustavoid a zero denominator.We solve thestrict inequalityx 3> 0, which gives us x> 3.

The domain of f

in set-builder form is: x x> 3{ } , or x : x> 3{ }


in interval form is: 3,( )

The hollow circle on thegraph indicates that 3 isexcluded from thedomain. We could also use a left parenthesis (() here; the parenthesisopens towards the shading. Likewise, we have aparenthesisnext to the 3 inthe interval form, because 3 isexcluded from the domain.

Types of Intervals

5,7

( )and 3,

( )are examples of open intervals, because theyexcludetheir

endpoints. 5,7( ) is a bounded interval, because it is trapped between tworeal numbers.

3,( ) is an unbounded interval.

5,7 is a closed interval, because it includes its endpoints, and it is

bounded.


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Let f x( ) = x 33 . Find Dom f( ).

Solution

Dom f( ) = , because:

The radicand, x 3, is apolynomial, and

WARNING 7: The taking ofodd roots (such as cube roots) doesnot impose any new restrictions on the domain. Remember that thecube root of a negative real number is a negative real number.


Let g t( ) =t+3

t10. Find Dom g( ) .

Solution

Thesquare root operation requires: t+ 3 0 t 3.

Weforbid zero denominators, so we also require: t10 0 t10.

The domain ofg

in set-builder form is:t t 3 andt 10{ } , ort : t 3 andt 10{ }


in interval form is: 3,10 ) 10,( )

Weinclude3butexclude10. (Some instructors believe that 0 should

also be indicated on the number line.)

The union symbol ( ) is used to separate intervals in the event that anumber or numbers need to be skipped.


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PART F: GRAPHS OF FUNCTIONS

The graph of y= f x( ) , or the graph of f, in the standardxy-plane consists of

all points [representing ordered pairs] of the form x, f x( )( ), wherex is in thedomain of f.

In a sense, the graph of f= x, f x( )( ) xDom f( ){ } .

We typically assume

x is the independent variable, because it is the input variable.

y is the dependent variable, because it is theoutput variable.Its value (thefunction value) typically depends on the value of the input x.

Then, it is customary to say thaty is a function ofx, even thoughy is a variable

here. The form y= f x( ) implies this.

A brute force graphing method follows.

Point-Plotting Method for Graphing a Function f in thexy-Plane

Chooseseveral xvalues in Dom f( ).

For each chosenxvalue, find f x( ) , its corresponding function value.Plot the corresponding points x, f x( )( ) in thexy-plane. Try to interpolate(connect the points, though often not with linesegments) andextrapolate(go beyond the scope of the points)as necessary, ideally based on some apparent pattern.

Ensure that the set ofx-coordinatesof the points on the graph is,

in fact, Dom f( ).


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Example 7 (Graph of the Squaring Function)

Let f x( ) = x . Graph y = f x( ).

Solution

TIP 1: As usual, we associatey-coordinateswith function values.

Whenpoint-plotting, observe that: Dom f( ) = 0, ).

For instance, if we choose x= 9, we find that f 9( ) = 9= 3,

which means that thepoint 9, f 9( )( ), or 9,3( ), lies on the graph.

On the other hand, f 9( ) isundefined (as a real number), because

9Dom f( ) . Therefore, there isnocorresponding point on the graphwith x= 9.

A (partial) table can help:

x f x( ) Point

0 0 0,0( )

1 1 1,1( )

4 2 4,2( )

9 3 9,3( )

Below, we sketch the graph of f, or y = f x( ).

WARNING 8: Clearly indicate anyendpointson a graph, such asthe origin here.

The lack of a clearly indicated right endpoint on our sketch implies that thegraphextendsbeyond the edge of our figure. We want to draw graphs insuch a way that these extensions are as one would expect.

WARNING 9: Sketches of graphs produced bygraphing utilitiesmight notextend as expected. The user must still understand the math involved.Point-plotting may be insufficient.


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PART G: THE VERTICAL LINE TEST (VLT)

The Vertical Line Test (VLT)

A curve in a coordinate plane passes the Vertical Line Test (VLT)

there isnovertical line that intersects the curve more than once.

Anequation inxandydescribesyas afunction ofx itsgraph in thexy-planepasses the VLT.

Then, there isno inputxthat yields more than one outputy.

Then, we can write y= f x( ) , wheref is a function.

Example 8 (Square Root Function and the VLT; Revisiting Example 7)

The equation y= x explicitly describesyas afunction ofx, since it is of

the form y= f x( ) , wheref is the square root function from Example 7.

Observe that the graph of y= xpasses the VLT.

Eachvertical linein thexy-plane either

misses the graph entirely, meaning that the correspondingxvalue is

not in Dom f( ), or

intersects the graph in exactly one point, meaning that thecorrespondingxvalue yields exactly oneyvalue as its output.


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Example 9 (An Equation that Does Not Describe a Function)

Show that the equation x2 + y2 = 9 doesnot describeyas a function ofx.

Solution (Method 1: VLT)

The circular graph ofx2 + y2 = 9 below fails the VLT, because there exists

a vertical line that intersects the graphmore than once. For example, wecan take the red line (x= 2) below:

Therefore, x2 + y2 = 9 doesnot describeyas a function ofx.

Solution (Method 2: Solve for y)

This is also evident if we solve x2 + y2 = 9 for y:

x2

+ y2

= 9

y2

= 9 x2

y= 9 x2

Any input value for x in the interval 3,3( ) yields two differentyoutputs.

For example, x= 2 yields the outputs y= 5 and y= 5.


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(Chapter 1: Review) 1.13PART H: ESTIMATING DOMAIN, RANGE, and FUNCTION VALUESFROM A GRAPH

Thedomainof f is the set of all x-coordinatesof points on the graph of

y= f x( ) . (Think of projecting the graph onto thex-axis.)Therangeof f is the set of all y-coordinates of points on the graph of

y= f x( ) . (Think of projecting the graph onto they-axis.)

Domain

Think: x f

Range

Think: y

Example 10 (Estimating Domain, Range, and Function Values from a Graph)Let f x( ) = x2 +1. Estimate thedomainand therangeof f based on the

graph ofy = f x( ) below. Also, estimate f 1( ) .

Solution

Apparently, Dom f( ) = , or ,( ) , and Range f( ) =1, ) .

We will learn more about determining ranges from the graphingtechniques in Chapter 4.

It also appears that the point 1,2( ) lies on the graph, and thus f 1( ) = 2.

WARNING 10: Graph analyses can be imprecise. The point 1,2.001( ) ,for example, may be hard to identify on a graph. Not all coordinates areintegers.


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PART I : FUNCTIONS THAT ARE EVEN / ODD / NEITHER; SYMMETRY

A function f is even f x( ) = f x( ), xDom f( )

The graph ofy

=

f x( ) issymmetric about they-axis.

means for all or for every.

Example 11 (Proving that a Function is Even)

Let f x( ) = x2. Prove that f is aneven function.

Solution

Dom f( ) = . x ,

f x( ) = x( )2

= x2

= f x( )

Q.E.D. (Latin: Quod Erat Demonstrandum)

This signifies the end of a proof. It means which was to

be demonstrated / proven / shown.

TIP 2:Think: If we replacexwith x( ) as the input, we obtainequivalent(y) outputs. The point x, y( ) lies on the graph if and only if x, y( ) does.

Observe that the graph ofy = f x( ) below is symmetric about they-axis,meaning that the parts of the graph to the right and to the left of they-axisare mirror images (or reflections) of each other.


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The term even function may have come from the following fact:

If f x( ) = xn, wheren is aneven integer, thenf is anevenfunction.

These are the functions for:, x4, x2, x0, x2, x4, .

The graph of y = x2 is called a parabola. The graphs of y= x4, y= x6, etc.are similarly bowl-shaped but arenot parabolas.

A function f is odd f x( ) = f x( ), xDom f( )

The graph ofy= f x( ) issymmetric about theorigin.

In other words, if the graph is rotated180

about the origin,we obtain the same graph.

Example 12 (Proving that a Function is Odd)

Let f x( ) = x3. Prove that f is anodd function.

Solution

Dom f( ) = . x ,

f x( ) = x( )3

= x3

= x3( )

= f x( )

Q.E.D.

TIP 3:Think: If we replacexwith

x( ) as the input, we obtainopposite (y)outputs. The point x, y( ) lies on the graph if and only if x, y( ) does.


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The term odd function may have come from the following fact:

If f x( ) = xn, wheren is anodd integer, thenf is anodd function.

The graphs of y= x5, y= x7, etc. resemble the graph of y= x3 below.

WARNING 11: Zero functions are functions that only output 0 (Think: f x( ) = 0).Zero functions on domains that are symmetric about 0 on the real number line aretheonly functions that areboth even and odd. (Can you show this?)

WARNING 12: Many functions areneither even nor odd.


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PART J : ARITHMETIC COMBINATIONS OF FUNCTIONS

Let f andgbe functions.

If their domains overlap, then the overlap (intersection) Dom f( )Dom g( ) is thedomain of the following functions with the specified rules, with onepossible exception (*):

f+ g, where f+ g( ) x( ) = f x( ) + g x( )

f g, where f g( ) x( ) = f x( ) g x( )

fg, where fg( ) x( ) = f x( ) g x( )

f

g, where

f

g

x

( )=

f x( )g x( )

(*) WARNING 13:Domf

g

= xDom f( )Dom g( ) g x( ) 0{ }.

Example 13 (Subtracting Functions)

Let f x( ) = 4x and g x( ) = x+1

x. Find f g( ) x( ) and Dom f g( ).

Solution

f g( ) x( ) = f x( ) g x( )

= 4x( ) x+1

x

WARNING 14: Usegrouping symbolswhen expanding g x( ) here, since we aresubtractingan expression withmore thanone term.

= 4x x1

x

= 3x1

x

Dom f( ) = . Weomit only 0 fromDom g( ) and also Dom f g( ) .

Dom f g( ) = \ 0{ } = x x 0{ } = ,0( ) 0,( ) .


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PART K : COMPOSITIONS OF FUNCTIONS

We compose functions when we apply themin sequence.

Let f andgbe functions. The composite function f g is defined by:

f g( ) x( ) = f g x( )( )

Its domain is x xDom g( ) andg x( )Dom f( ){ } .

The domain consists of the legal inputstog that yieldoutputsthat are legal inputs to f.

x g g x( ) f

fg

f g x( )( )

Think of f g as a merged function.

WARNING 15: The function f g appliesgfirst and thenf. Think of pressing a

gbutton on a calculator followed by anf button.

WARNING 16: f gmay or may not represent the same function as g f

(in whichf is applied first). Composition of functions isnot commutativetheway that, say, addition is. Think About It: Try to think of examples where f g

and g f represent the same function.

Example 14 (Composition of Functions)

Let f u( ) =1

uand g x( ) = x1. Find f g( ) x( ) and Dom f g( ) .

Solution

f g

( )x

( )= f g x

( )( )= f x1

( )=

1

x1

. In fact, Dom f g

( )

in set-builder form is: x x>1{ } , or x : x>1{ }


in interval form is: 1,( )


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Example 15 (Decomposing a Composite Function)

Find component functions f andgsuch that f g( ) x( ) = 3x+1.We want to decompose f g.

Neither f nor gmay be an identity function.

For example, donot use: g x( ) = x and f u( ) = 3u+1.This wouldnot truly be a decomposition. f does all the work!

x g:

g x( ) = x x, our u

g x( )

f :

f u( ) = 3u+1

fg

3x+1

f g x( )

Solution

We need: f g x( )( ) = 3x+1.

We can think of f andgasbuttonswe are designing on a calculator.

We need to set upf andgso that, ifx is aninitial input to Dom f g( ) ,and if thegbutton and then thef button are pressed, then the output is

3x+1.

x g:

g x( ) = ? u= ?

f:

f u( ) = ??

fg

3x+1

f g x( )

A common strategy is to let g x( ) , or u, be aninside expression(for example, a radicand, an exponent, a base of a power, a denominator, anargument, or something being repeated) whose replacement simplifies theoverall expression.

Here, we will let g x( ) = 3x+1.


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We then needf to apply the square root operation. We will let f u( ) = u .

The use ofu is more helpful in calculus, but f x( ) = x is also

acceptable. However, f u

( )= x isnot acceptable.

Possible Answer: Let g x( ) = 3x+1 and f u( ) = u .

x g:

g x( ) = 3x+1 3x+1, ouru

g x( )

f:

f u( ) = u

fg

3x+1

f g x( )

There are infinitely many possible answers.

For example, we could let g x( ) = 3x and f u( ) = u+1.

x g:

g x( ) = 3x 3x, ouru

g x( )

f:

f u( ) = u+1

fg

3x+1

f g x( )

These ideas will be critical to the Chain Rule of Differentiation in Section 3.6 andtheu-substitution technique of integration in Section 5.2.

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