SECTION 1.5: PIECEWISE-DEFINED FUNCTIONS; LIMITS AND ...

(Section 1.5: Piecewise-Defined Functions; Limits and Continuity in Calculus) 1.5.1

SECTION 1.5: PIECEWISE-DEFINED FUNCTIONS;

LIMITS AND CONTINUITY IN CALCULUS

LEARNING OBJECTIVES

• Know how to evaluate and graph piecewise-defined functions.

• Know how to evaluate and graph the greatest integer (or floor) function.

• Preview limits and continuity from calculus.

PART A: DISCUSSION

• A piecewise-defined function applies different rules, usually as formulas, to

disjoint (non-overlapping) subsets of its domain (subdomains).

• To evaluate such a function at a particular input value, we need to figure out

which rule applies there.

• To graph such a function, we need to know how to graph the pieces that

correspond to the different rules on their subdomains.

• The greatest integer (or floor) function and its graph, seen in calculus and

computer science, exhibit similar features.

• We will take a peek into calculus and preview the related topics of one- and two-

sided limits and continuity. Piecewise-defined functions appear frequently in these

sections of a calculus course.


PART B: THE ABSOLUTE VALUE FUNCTION

Let f x( ) = x . We discussed the absolute value function f in Section 1.3, Part N.

The piecewise definition of f is given by:

f x( ) = x =x, if x 0

x, if x < 0

• For instance, f 3( ) = 3 = 3, because 3 0 , and we use the top rule, which

applies to the subdomain

0, ) .

• However, f 3( ) = 3 = 3( ) = 3 , because 3< 0 , and we use the

bottom rule, which applies to the subdomain

, 0( ) .

We say that f is continuous at 0. That is, if we trace the graph of f with a pencil in

the vicinity of the point 0, 0( ) , we do not have to lift our pencil.

• We will provide a formal definition of continuity in Part G.

WARNING 1: Piecewise-defined functions are often discontinuous (i.e., they

lose continuity) where they switch rules. §


PART C: EVALUATING PIECEWISE-DEFINED FUNCTIONS

Example 1 (Evaluating a Piecewise-Defined Function)

Let the function f be defined by:

f x( ) =x2 , 2 x < 1

x +1, 1 x 2

• To evaluate f 1( ) , we use the top rule, since 2 1< 1.

f 1( ) = 1( )2

= 1

• To evaluate f 1( ) , we use the bottom rule, since 1 1 2 .

f 1( ) = 1( ) +1

= 2

• f 10( ) , for example, is undefined, because we have no rule for x = 10 .

10 is not in the domain of f , which is

2, 2 . §

PART D: GRAPHING PIECEWISE-DEFINED FUNCTIONS

Example 2 (Graphing a Piecewise-Defined Function with a Jump Discontinuity;

Revisiting Example 1)

Graph the function f from Example 1.

§ Solution

WARNING 2: Clearly indicate any endpoints and whether they are

included in, or excluded from, the graph.

Top rule: We will graph y = x2 on the subdomain

2,1) .

• We obtain a piece of a parabola.

•

2( )2

= 4 , so the left endpoint of this piece is at 2, 4( ) .


• 2 is included in the subdomain

2,1) . Therefore, the endpoint

2, 4( ) is included in this piece, and we plot it as a filled-in circle.

• 1( )

2

= 1, so the right endpoint of this piece is at 1,1( ) .

•• Why do we use the top rule, instead of the bottom rule? See Part F.

• 1 is excluded from the subdomain

2,1) . Therefore, the endpoint

1,1( ) is excluded from the piece, and we plot it as a hollow circle.

Bottom rule: We will graph y = x +1 on the subdomain 1, 2 .

• We obtain a line segment.

• 1( ) +1= 2 , so the left endpoint of this piece is at

1, 2( ) .

• 2( ) +1= 3 , so the right endpoint of this piece is at

2, 3( ) .

• Both 1 and 2 are included in the subdomain 1, 2 . Therefore, the

endpoints 1, 2( ) and

2, 3( ) are included in this piece, and we plot

them as filled-in circles.

The graph of f is below.

In calculus, we say that f has a jump discontinuity at 1 x = 1( ) .

(See Part G.) §


Example 3 (Graphing a Piecewise-Defined Function with a Removable

Discontinuity)

Graph f , if

f x( ) =x + 3, x 3

7, x = 3.

§ Solution

• We graph the line y = x + 3 , except we leave a hole at the point 3, 6( ) ,

since 3 is deleted from the subdomain of the top rule.

• The bottom rule defines f 3( ) to be 7, so we plot the point 3, 7( ) .

• In calculus, we say that there is a removable discontinuity at 3, since

the discontinuity could be removed by simply redefining f 3( ) to be 6.

We would end up with a different function, though. (See Part G.)

• Although f 3( ) = 7 , we will write limx 3

f x( ) = 6 in calculus. This is

because, as x approaches 3, f x( ) approaches 6. (See Part F.) §


PART E: THE GREATEST INTEGER (OR FLOOR) FUNCTION

The greatest integer (or floor) function is defined by f x( ) = x or x ,

the greatest integer that is not greater than x.

• Think: Round x down.

• If x is nonnegative, we simply take the integer part.

•

x = max y y x{ } . “max” takes the greatest element of the set.

Example Set 4 (Evaluating the Greatest Integer or Floor Function)

2.9 = 2

3 = 3

1.1 = 2

2.9 = 2

3 = 3

1.1 = 2

§

The ceiling function is defined by f x( ) = x ,

the least integer that is not less than x.

• Think: Round x up.

• If x is nonpositive, we simply take the integer part.

•

x = min y y x{ } . “min” takes the least element of the set.

Example Set 5 (Evaluating the Ceiling Function)

2.1 = 3 , 3 = 3, and 1.9 = 1. §


We may think of these as piecewise constant functions.

For example, the greatest integer (or floor) function is given by:

f x( ) = x or x =

2, 2 x < 1

1, 1 x < 0

0, 0 x <1

1, 1 x < 2

2, 2 x < 3

Example 6 (Graphing the Greatest Integer or Floor Function)

Looking at its graph below, we can see why the greatest integer function is

called a staircase function.

• Step functions are similar, except that the range must be a finite set; there are

only a finite number of distinct function values for a step function.

TIP 1: Remember that the left endpoints at 0, 0( ) ,

1,1( ) ,

2, 2( ) , etc. are

included in the graph. The right endpoints are excluded.

• In calculus, we say that f has jump discontinuities at all integer values

of x. (See Part G.)


PART F: LIMITS IN CALCULUS

Example 7 (Limits and the Greatest Integer or Floor Function)

Let f x( ) = x or x .

A Left-Hand, One-Sided Limit

Evaluate

limx 2

f x( ) , which is read:

“the limit of f x( ) as x approaches 2 from the left.”

• We want the real number that f x( ) approaches as x approaches 2

from lesser numbers (imagine approaching x = 2 from the left along

the real number line), if such a number exists.

• A table can help:

x 1.9 1.99 1.999 2

f x( ) 1 1 1

1

We see that limx 2

f x( ) = 1, even though f 2( ) = 2 .

WARNING 3: Sometimes,

limx a

f x( ) does not equal f a( ) .

For example, the value of f 2( ) , or whether it exists at all, is

technically irrelevant (though sometimes helpful) when

evaluating limits as x 2 .

• A graph confirms that

limx 2

f x( ) = 1. Along the graph of f below,

as x approaches 2 from the left, y approaches 1.


A Right-Hand, One-Sided Limit

Evaluate

limx 2+

f x( ) , which is read:

“the limit of f x( ) as x approaches 2 from the right.”

• We want the real number that f x( ) approaches as x approaches 2

from greater numbers (imagine approaching x = 2 from the right

along the real number line), if such a number exists.


x 2+ 2.001 2.01 2.1

f x( )

2 2 2 2

We see that

limx 2+

f x( ) = 2 .

• The graph of f below confirms this.

Two-Sided Limits

The two-sided limit limx 2

f x( ) does not exist here, because the

corresponding left-hand and right-hand limits are unequal.

If they existed and were equal, the common value would be the

two-sided limit.

• In Example 3, limx 3

f x( ) = 6 . (See the Exercises.) §


Example 8 (Finding Limits; Revisiting Examples 1 and 2)

Let the function f be defined by:

f x( ) =x2 , 2 x < 1

x +1, 1 x 2

While f 1( ) = 1( ) +1= 2 , the left-hand limit lim

x 1f x( ) = 1.

• When evaluating this left-hand limit, we only care about values of x

that are close to 1 and less than 1. The top rule, f x( ) = x2 , is the only

rule that applies to those x values. Therefore, lim

x 1f x( ) = lim

x 1x2

,

which we can compute by evaluating x2 at 1:

lim

x 1x2

= 1( )2

= 1.

• This explains why our graph approaches the endpoint at 1,1( ) from

the left, even though 1,1( ) itself is excluded from the graph.


x 0.9 0.99 0.999

1

f x( )

x2

0.81 0.9801 0.998001

1

WARNING 4: Tables can sometimes be misleading.

If the limit were 0.999, for instance, that might have been

difficult to detect.

• A graph confirms the limit:

§


PART G: CONTINUITY IN CALCULUS

f is continuous at a, or x = a

1) f a( ) is defined,

2) limx a

f x( ) exists, and

3) limx a

f x( ) = f a( ) .

f is discontinuous at a f is not continuous at a.

(Some sources require that f be defined “around” a.)

Comments

1) ensures that there is literally a point at a.

2) constrains the behavior of f immediately around a.

3) then ensures safe passage through the point

a, f a( )( ) on the graph of f .

In Part B, the absolute value function was continuous everywhere on .

f has a removable discontinuity at a, or x = a

1) limx a

f x( ) exists, but

2) f is still discontinuous at a

Then, the graph of f has a “hole” at the point a, lim

x af x( )( ) .

In Example 3,

f x( ) =x + 3, x 3

7, x = 3. f had a removable discontinuity at 3.


f has a jump discontinuity at a, or x = a

1)

limx a

f x( ) exists, and (call this limit L

1)

2)

limx a+

f x( ) exists, but (call this limit L

2)

3)

limx a

f x( ) limx a+

f x( ) . ( L

1L

2)

Therefore, the two-sided limit limx a

f x( ) does not exist.

It is irrelevant how f a( ) is defined, or if it is at all.

Example 2 and the greatest integer function gave us jump discontinuities.

f has an infinite discontinuity at a, or x = a

lim

x af x( ) = or , or

lim

x a+f x( ) = or

Then, the graph of f has the line x = a as a vertical asymptote (VA).

It is irrelevant how f a( ) is defined, or if it is at all.

For example, if f x( ) =1

x, then f has an infinite discontinuity at 0.

(Section 1.6: Combining Functions) 1.6.1

SECTION 1.6: COMBINING FUNCTIONS

LEARNING OBJECTIVES

• Know how to add, subtract, multiply, and divide functions.

• Be able to identify linear combinations of functions.

• Know how to construct and decompose composite functions.

• Find the domains of the functions that result from these operations.

• Be able to model functions using constraint equations and composite functions.

PART A: DISCUSSION

• In Section 1.5, we constructed functions from pieces of functions on disjoint

(non-overlapping) subdomains.

• In this section, we will combine functions by adding, subtracting, multiplying,

and dividing them over basically the same domain.

• Linear combinations are usually not formally introduced to students until a linear

algebra course, but linear combinations of functions will enable us to take a

broader view of symmetry in Section 1.7.

• We will also combine functions by composing them. This corresponds to the

successive application of functions. We did this when we transformed functions

and graphs in Section 1.4.

• Calculus theorems, including linearity theorems and the Chain Rule for differentiating

composite functions, will use the notation and ideas of this section. Composite functions are also

related to the u-substitution technique of integration.

• In applied calculus problems such as related rates and optimization problems, we will need to

devise functions, compose them, and combine them with constraint equations.


PART B: ARITHMETIC COMBINATIONS OF FUNCTIONS

Let f and g be functions.

If their domains overlap, then the overlap (intersection) Dom f( ) Dom g( )

is the domain of the following functions with the specified rules, with one

possible 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

gx( ) =

f x( )g x( )

(*) WARNING 1:

Domf

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

Example 1 (Adding Functions)

Let f x( ) = x2 and g x( ) = x3 . We will use tables and graphs for f and g to

develop a table and a graph for f + g .

• Observe that the domain is for f , g, and f + g .

x

f x( )

x2

g x( )

x3

f x( ) + g x( )

x2+ x3

3 9

27

18

2 4

8

4

1 1

1 0

0 0 0

0

1 1 1

2

2 4 8

12

3 9 27

36


f x( ) = x2 yields the blue graph below;

g x( ) = x3 yields the red graph.

f x( ) + g x( ) = x2+ x3 yields the brown graph; we add y-coordinates.

§

Example 2 (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 2: Use grouping symbols when expanding g x( )

here. Be careful when subtracting an expression with more than

one term.

= 4x x1

x

= 3x1

x

Dom f( ) = . We omit only 0 from

Dom g( ) and also

Dom f g( ) .

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


Example 3 (Dividing Functions)

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

x

x 2. Find

f

gx( ) and

Domf

g.

§ Solution

f

gx( ) =

f x( )g x( )

=1

x

x 2

=x 2

xx 2( )

WARNING 3: Watch the domain when you take reciprocals. We state

x 2( ) , because that restriction is not implied by the expression x 2

x.

• Dom f( ) = .

• Dom g( ) = \ 2{ } = x x 2{ } .

• g x( ) = 0 x = 0 , so we require: x 0 .

• Therefore, we must exclude both 0 and 2 from

Domf

g.

Domf

g= \ 0, 2{ } in set-difference form.

Also,

Domf

g …

… in set-builder form is: x x 0 and x 2{ } , or

x : x 0 and x 2{ }

… in graphical form is:

… in interval form is: , 0( ) 0, 2( ) 2,( )

§


PART C: LINEAR COMBINATIONS OF FUNCTIONS

Let f and g be functions. Let c and d be real numbers, possibly 0.

cf + dg is then called a linear combination of f and g.

• Its domain is the overlap (intersection) Dom f( ) Dom g( ) .

• More generally, a linear combination of objects is a sum of constant

multiples of those objects.

Example 4 (Linear Combinations)

a) 3 f + 4g ,

1

2f 5g , g , and 0 are linear combinations of f and g.

b) 2 f + 3g 4h and 3.7 f + g are linear combinations of f , g, and h. §

• In Section 1.7, we will see that a linear combination of even functions is even.

The same goes for odd functions.

• WARNING 4: People often erroneously attempt to apply linearity properties in

precalculus. For example, remember that the “square root of a sum” is typically

not equal to the “sum of the square roots.” Think: 1+ 4 1 + 4 .

• In calculus, we will see important linearity theorems for limits, derivatives, and integrals.

For example, let’s say we have a group of functions of x. We find their limits as x a , and all

the limits exist as real numbers. We can then find the limit of any linear combination of those

functions as x a . This is because, if the limits exist, …

•• the limit of a constant multiple of a function equals the constant times the limit of

the function. That is, limx a

cf x( ) = c limx a

f x( ) .

(Constant factors can “move across” the limit operator.)

•• the limit of a sum (difference) of functions equals the sum (difference) of the limits

of the functions. That is,

limx a

f x( ) + g x( ) = limx a

f x( ) + limx a

g x( ) , and

limx a

f x( ) g x( ) = limx a

f x( ) limx a

g x( )

(Limits can be taken term-by-term.)


PART D: COMPOSITION OF FUNCTIONS

We compose functions when we apply them in sequence, as we did in

Section 1.4 on transformations.

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

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

Its domain is

x x Dom g( ) and g x( ) Dom f( ){ } .

• The domain consists of the “legal” inputs to g that yield outputs

that are “legal” inputs to f .

x g g x( ) f

f g

f g x( )( )

Think of f g as a “merged” function.

WARNING 5: The function f g applies g first and then f .

Think of pressing a g button on a basic calculator followed by an f button.

WARNING 6: f g may or may not be the same function as g f , which

applies f first. Composition of functions is not commutative the way that, say,

addition is. (See the Exercises.)

Example 5 (Composition of Functions)

Let f u( ) =

1

u and

g x( ) = x 1 .

WARNING 7: Writing f x( ) =

1

x together with

g x( ) = x 1 would

be acceptable, but that may lead to confusion.

Find

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


§ Solution

f g( ) x( ) = f g x( )( )= f x 1( )

Substitute u = x 1 into f u( ) =

1

u.

Conceptually, f takes the reciprocal of its input.

=1

x 1

Find Dom f g( ) .

The only restriction implied by the final expression is: x > 1.

In fact, Dom f g( ) …

… in set-builder form is:

x x > 1{ } , or

x : x > 1{ }


… in interval form is: 1,( )

WARNING 8: As we will see in Example 7, we might not be able to

obtain all the necessary domain restrictions on f g from our final

expression for f g( ) x( ) , as we did here.

• It is typically sufficient to analyze the final expression and the

domain of the first function applied, but there are counterexamples:

• If, instead of f u( ) =

1

u, we had been given

f u( ) =u 2

u u 2( ), then x 5

would be another restriction on top of x > 1. (Check this.)


A more rigorous analysis of Dom f g( ) follows.

xg :

g x( ) = x 1x 1, our u

g x( )

f :

f u( ) =1

u

f g

1

x 1

f g x( )

x > 1x 1 from Dom g( )x = 1 0 is an "illegal" input to f

0 Dom f( )( )

(See the bottom of the diagram in blue.)

• The set of “legal” inputs to g, Dom g( ) = 1, ) .

• The set of outputs from g, Range g( ) = 0, ) , and

• The set of “legal” inputs to f , Dom f( ) = \ 0{ } .

•• The outputs from g (u values) are “legal” inputs to f ,

except 0, which is the output for only the input 1 in Dom g( ) .

x 1 = 0 x = 1.( )

• Therefore, Dom f g( ) = Dom g( ) \ 1{ } = 1,( ) . §


Example 6 (Evaluating Composite Functions; Revisiting Example 5)

Again, let f u( ) =

1

u and g x( ) = x 1 .

Evaluate

f g( ) 3( ) ,

f g( ) 0( ) , and

f g( ) x + h( ) , where x + h > 1.

§ Solution

In Example 5, we developed the formula:

f g( ) x( ) =1

x 1 on

1,( ) .

Evaluate

f g( ) 3( ) . Evaluate

f g( ) x + h( ) .

Observe: 3 1,( ) . We assume: x + h > 1.

f g( ) 3( ) =1

3 1

=1

2

=2

2

f g( ) x + h( ) =1

x + h( ) 1

=1

x + h 1

=x + h 1

x + h 1

Evaluate

f g( ) 0( ) .

Observe: 0 1,( ) .

f g( ) 0( ) is undefined.

We can also evaluate

f g( ) 3( ) and

f g( ) x + h( ) without the formula.

f g( ) 3( ) = f g 3( )( )= f 3 1( )= f 2( )

=1

2

=2

2

f g( ) x + h( ) = f g x + h( )( )

= f x + h( ) 1( )= f x + h 1( )

=1

x + h 1

=x + h 1

x + h 1

§


Example 7 (“Hidden” Domain Restrictions)

Let f x( ) =1

x2 7 and g x( ) = x 3 . Find

f g( ) x( ) and

Dom f g( ) .

§ Solution

Rewriting the rule for f as, say, f u( ) =

1

u2 7, may clarify the solution.

f g( ) x( ) = f g x( )( )= f x 3( )

• If we had kept the formula f x( ) =1

x2 7, we would

informally (and awkwardly) substitute the x in

1

x2 7

with

x 3( ) . Substituting the u in

1

u2 7 with

x 3( )

is less awkward.

=1

x 3( )2

7

=1

x 3( ) 7

=1

x 10x 3; we will explain this( )

Find Dom f g( ) .

It is typically sufficient to analyze the final expression and the

domain of the first function applied, although there are

counterexamples (see Example 5). Here, 1

x 10 imposes the

restriction x 10 , and Dom g( ) imposes the restriction x 3 .


In fact, Dom f g( ) …

… in set-builder form is: x x 3 and x 10{ } , or

x : x 3 and x 10{ }


… in interval form is:

3,10) 10,( )

We write: f g( ) x( ) =1

x 10x 3( ) . The restriction

x 10( ) is

evident.

A more rigorous analysis of Dom f g( ) follows.

xg :

g x( ) = x 3x 3, our u

g x( )

f :

f u( ) =1

u2 7

f g

1

x 10

f g x( )

x 3 from Dom g( ) 7 "illegal" input to f , but not an output from g

x = 10 7 "illegal" input to f

7 Dom f( )( )

• The set of “legal” inputs to g, Dom g( ) = 3, ) .

• The set of outputs from g, Range g( ) = 0, ) , and

• The set of “legal” inputs to f , Dom f( ) = \ 7, 7{ } .

•• This is because: u2 7 = 0 u = ± 7 .

•• The outputs from g (u values) are “legal” inputs to f ,

except 7 , which is the output for only the input 10 in

Dom g( ) .

x 3 = 7 x = 10.( )

• Therefore, Dom f g( ) = Dom g( ) \ 10{ } = 3,10) 10,( ) . §


PART E: “DECOMPOSING” COMPOSITE FUNCTIONS

Example 8 (“Decomposing” a Composite Function)

Find component functions f and g such that

f g( ) x( ) = 3x +1 .

We want to “decompose” f g .

• Neither f nor g may be an identity function.

For example, do not use: g x( ) = x and

f u( ) = 3u +1 .

This would not truly be a decomposition. f does all the work!

xg :

g x( ) = xx, our u

g x( )

f :

f u( ) = 3u +1

f g

3x +1

f g x( )

§ Solution

• We need: f g x( )( ) = 3x +1 .

• We can think of f and g as buttons we are designing on a basic calculator.

We need to set up f and g so that, if x is an initial input to Dom f g( ) , and

if the g button and then the f button are pressed, then the output is 3x +1 .

xg :

g x( ) = ?u = ?

f :

f u( ) = ??

f g

3x +1

f g x( )

• A common strategy is to let g x( ) , or u, be an “inside” expression

(for example, a radicand, an exponent, a base of a power, a denominator, an

argument, or something being repeated) whose replacement simplifies the

overall expression.

• Here, we will let g x( ) = 3x +1 .


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

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

acceptable. However, f u( ) = x is not acceptable.

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

xg :

g x( ) = 3x +13x +1, our u

g x( )

f :

f u( ) = u

f g

3x +1

f g x( )

There are infinitely many possible answers.

For example, we could let g x( ) = 3x and

f u( ) = u +1 .

xg :

g x( ) = 3x3x, our u

g x( )

f :

f u( ) = u +1

f g

3x +1

f g x( )

§


PART F: APPLICATIONS

Example 9 (Conical Water Tank)

Water is flowing into a tank shaped as a right circular cone with base radius

5 meters and height 10 meters. Express the volume of the “water cone” in

the tank as a function of the base radius of the water cone.

§ Solution

• Define variables.

• Draw a diagram.

• Indicate known information.

Let r = the base radius of the water cone, in feet.

Let h = the height of the water cone, in feet.

Let V = the volume of the water cone, in cubic feet.

• Express the key formula.

We need the formula for the volume of a right circular cone of height

(or altitude) h and base radius r.

V =

1

3r 2h

As it stands, we are expressing V as a function of two variables,

r and h. However, r and h are not independent here, because the

shape of the tank forces a relationship between them.


• Express any other relationships between variables.

The height of the tank is twice its base radius.

This proportion must hold for the water cone, as well.

h = 2r

This may be thought of as a constraint equation.

It expresses h as a function of r.

•• This may be shown using similar triangles, for which corresponding

side lengths are in equal proportion.

h

10=

r

5

h = 2r

• Combine the key formula with any constraint equations.

We will make the substitution h = 2r( ) into

V =

1

3r 2h .

V =1

3r 2 2r( )

V =2

3r3

We are now expressing V as a function of one variable, r.


• Find the domain.

We may rewrite V as f r( ) , or even

V r( ) .

r, the base radius of the water cone, can be anything from

0 meters (a degenerate case corresponding to no water at all) to

5 meters, the base radius of the tank (corresponding to a full tank).

If V = f r( ) , then

Dom f( ) = 0, 5 . Units (m) aren’t written here.

Answer: V =

2

3r3 on 0, 5 .

• Note: We can also express V as a function of h on 0,10 .

We solve the constraint equation for r and express r as a function of h.

h = 2r

r =h

2

We then make the substitution

r =h

2 into V =

1

3r 2h .

V =1

3

h

2

2

h

V =1

3

h2

4h

V =h3

12

• To find the volume of the water cone at a particular instant, we only need

its radius or its height (the water level) at that instant. Then, we can use the

appropriate formula, V =2

3r3 , or

V =

h3

12.

•• If h = 3 m , say, then V =

h3

12=

3( )3

12=

27

12=

9

4m3 7.07 m3

.

• TIP 1: Observe that, if r or h is given in meters, the unit of volume we

obtain is (correctly) cubic meters. This dimensional analysis can be a useful

check. §


Example 10 (Conical Water Tank Problem with a Time Variable;

Revisiting Example 9)

Let’s say the tank from Example 9 is empty at time t = 0 , at which time

water starts flowing in. The radius of the water cone increases at the rate of

3m

min until the tank is full.

• In calculus, we call this rate

dr

dt, the derivative of the radius with respect to

time. (Does this mean that the water is flowing in at an increasing rate or a

decreasing rate as it fills the tank?)

Express the volume of the water cone as a function of t, the time elapsed,

until the tank is full.

§ Solution

• In Example 9, we developed as our new key formula:

V = f r( ) =

2

3r3

• We express r as a function of t.

If r starts at 0 m and increases at 3

m

min for t min, we can let:

r = g t( ) = 3t

• We can now express V as a composite function of t.

V = f g t( )( )

V =2

3r3 r = 3t( )

V =2

33t( )

3

V =2

327t3( )

V = 18 t3


• Find the domain of the composite function f g .

•• As in Example 9, r, the base radius of the water cone, can be

anything from 0 meters to 5 meters, the base radius of the tank.

•• Find the corresponding values of t.

0 r 5

0 3t 5

0 t5

3

Dom f g( ) = 0,5

3. (t is measured in minutes.)

Let’s review the composition of functions here.

tg :

g t( ) = 3t3t, our r

g t( )

f :

f r( ) =2

3r3

f g

18 t3

f g t( )

A tree diagram illustrates the relationships between variables.

• In calculus, the Chain Rule for differentiating (taking the derivative of)

composite functions states:

dV

dt=

dV

dr

dr

dt

• In multivariable calculus, the tree gets bushy! §

(Section 1.7: Symmetry Revisited) 1.7.1

SECTION 1.7: SYMMETRY REVISITED

LEARNING OBJECTIVES

• Recognize short cuts for identifying even functions and odd functions.

• Be able to justify those short cuts using proofs.

• Use equations to identify symmetries in their graphs.

PART A: DISCUSSION

• In Section 1.3, we defined even functions and odd functions, whose graphs in the

xy-plane were symmetric about the y-axis and the origin, respectively.

• We will now systematize our study of these functions and their graphical

symmetries. We will discuss useful short cuts for identifying even functions and

odd functions, and we will show how to prove them rigorously. We will also

extend related symmetries to graphs of general equations in x and y.

• A zero function f , where f x( ) = 0 (on any domain), will be called a trivial function.

If its domain is symmetric about 0, then it is both even and odd.

• Symmetry will prove helpful when graphing functions and equations and finding areas, surface

areas, and volumes using definite integrals in calculus. In multivariable calculus, it will be a

huge time-saver when finding volumes, masses, and centers of mass of three-dimensional solids.

PART B: LINEAR COMBINATIONS

We introduced linear combinations in Section 1.6, Part C.

For example, 2 f + 5g is a linear combination of f and g.

Linear combinations of even functions are even.

Linear combinations of odd functions are odd.

• In particular, sums, differences, and constant multiples of even functions

are even. The same goes for odd functions. Informally, E ± E E ,

O ± O O , cE E , and cO O .

• The domain of a linear combination of functions is the overlap

(intersection) of the domains of those functions. We assume this is

nonempty.

• Domains of even functions and odd functions must be symmetric about 0.


Example 1 (A Proof: The Sum of Odd Functions is Odd)

Prove that the sum of two odd functions is also odd.

§ Solution

• Let f and g be odd functions. Let h = f + g .

• Let D = Dom f( ) Dom g( ) . Then,

D = Dom h( ) .

• Because f and g are odd on D,

x D , f x( ) = f x( ) , and g x( ) = g x( ) .

• Show that h is odd. That is, x D , h x( ) = h x( ) .

x D ,

h x( ) = f + g( ) x( )= f x( ) + g x( )

by definition of f + g; see Section 1.6( )= f x( ) g x( )

because f and g are odd( )= f x( ) + g x( )= h x( )

Q.E.D. §


Example 2 (Demonstrating Example 1)

Let f x( ) = x and

g x( ) = x3 . We will use tables and graphs for f and g to

develop a table and a graph for f + g .

• The domain is for f , g, and f + g .

x

f x( )

x g x( )

x3

f x( ) + g x( )

x + x3

3

3

27

30

2

2

8

10

1 1 1 2

0 0 0

0

1 1 1

2

2 2 8

10

3 3 27

30

• f , g, and f + g are all odd functions. This explains why opposite

x-values always yield opposite function values for all three functions.

f x( ) = x yields the blue graph below;

g x( ) = x3 yields the red graph.

f x( ) + g x( ) = x + x3 yields the brown graph; we add y-coordinates.

• All three graphs are symmetric about the origin. §


PART C: POLYNOMIAL FUNCTIONS

Short Cuts for Determining whether a Polynomial Function is

Even, Odd, or Neither

Let f x( ) be a nontrivial (or “nonzero”) polynomial written in descending

powers of x, though the terms could be reordered.

• Delete any terms with zero coefficients.

• Nonzero constant terms have degree 0, which is an even degree.

•• We could rewrite the constant term 3, for example, as 3x0.

•• We define 00 to be 1 here.

If every term of f x( ) has even degree, then f is even.

If every term of f x( ) has odd degree, then f is odd.

• WARNING 1: If f x( ) has a nonzero constant term, then

f can’t be odd.

If f x( ) has a term of even degree and a term of odd degree, then

f is neither even nor odd.

Example 3 (An Even Polynomial Function)

Let f x( ) = 3x4 7x2

+1. Prove that f is even.

§ Solution 1 (Short Cut)

f x( ) is polynomial and only has terms of even degree (4, 2, and 0).

Therefore, f is even.

WARNING 2: 1 is a (nonzero) constant term, so it has degree 0.

• 1 has even degree; the fact that 1 is an odd integer is a different idea.

• The graph of y = 3x4 7x2 is symmetric about the y-axis, and so is

the graph of y = 3x4 7x2+1. (How do the graphs differ? Review

Section 1.4.) §


§ Solution 2 (Rigorous)

Dom f( ) = . x ,

f x( ) = 3 x( )4

7 x( )2

+ 1

= 3x4 7x2+ 1

= f x( )

Q.E.D. §

Example 4 (An Odd Polynomial Function)

Let f x( ) = 4x5

+ x . Prove that f is odd.


f x( ) is polynomial and only has terms of odd degree (5 and 1).

Therefore, f is odd. §


Dom f( ) = . x ,

f x( ) = 4 x( )5

+ x( )= 4x5 x

= 4x5+ x( )

= f x( )

Q.E.D. §


Example 5 (A Polynomial Function that is Neither Even nor Odd)

Let g t( ) = t3

+ t +1. Is g even, odd, or neither? Justify your answer.


g t( ) is polynomial and has terms of odd degree (3 and 1) and a term

of even degree (0). Therefore, g is neither even nor odd. §


Dom g( ) = . Evaluate g t( ) and compare it to g t( ) and g t( ) .

g t( ) = t( )3

+ t( ) +1

= t3 t +1

• g is not even, because g t( ) is not equivalent to

g t( ) on .

• g is not odd, because g t( ) is not equivalent to g t( ) on .

• A counterexample suffices to justify both statements.

For instance, g 1( ) = 1, and

g 1( ) = 3.

Then, g 1( ) g 1( ) , and

g 1( ) g 1( ) .

WARNING 3: If it had been true that g 1( ) = g 1( ) or

g 1( ) = g 1( ) , that would not have been enough to prove that

g was even or odd. Examples cannot prove that a function g is

even or odd, unless Dom g( ) is a finite set.

g is neither even nor odd. §

Let f be a (nontrivial) even function, and let g be a (nontrivial) odd function.

Then, f + g and f g must be neither even nor odd.

(Experiment with graphs and numbers. See the Exercises.)

Example 6 (Revisiting Example 5)

Let g t( ) = t3

+ t +1; this is not g in the box above. Since g t( ) = 1( ) + t3+ t( ) ,

g can take the form even( ) + odd( ) , and thus g is neither even nor odd. §


PART D: PRODUCT AND QUOTIENT RULES

Product and Quotient Rules for Even Functions and Odd Functions

The rules for multiplying and dividing (nontrivial) even functions and

odd functions are similar to the sign rules for multiplying and dividing

positive numbers and negative numbers, respectively.

In the table below, we denote even functions by P (for “Positive”) and

odd functions by N (for “Negative”). The resulting statements in the

“Sign Patterns” column correspond to the sign rules for multiplying and

dividing real numbers.

Even Function and

Odd Function Patterns Sign Patterns

even( ) even( )= even( )

P( ) P( ) = P( )

even( ) odd( )= odd( )

P( ) N( ) = N( )

odd( ) odd( )= even( )

N( ) N( ) = P( )

even( )even( )

= even( )

P( )P( )

= P( )

even( )odd( )

= odd( )

P( )N( )

= N( )

odd( )even( )

= odd( )

N( )P( )

= N( )

odd( )odd( )

= even( )

N( )N( )

= P( )

Special Cases: Reciprocals of Even Functions and Odd Functions

The reciprocal of a (nontrivial) even function is even.

The reciprocal of a (nontrivial) odd function is odd.

Patterns:

1

even( )= even( ) , and

1

odd( )= odd( ) .

• Remember that f x( ) = 1 defines an even function.


Example 7 (Proving a Quotient Rule)

Prove: even( )odd( )

= odd( ) . That is, prove that, if a (nontrivial) even function is

divided by a (nontrivial) odd function, the resulting function is odd.

§ Solution

• Let f be an even function, and let g be an odd function. Let h =f

g.

• Let D = Dom f( ) Dom g( ) \ x g x( ) = 0{ } . Then, D = Dom h( ) .

• Because f is even on D,

x D , f x( ) = f x( ) .

• Because g is odd on D,

x D , g x( ) = g x( ) .

• Show that h is odd. That is, show that:

x D , h x( ) = h x( ) .

x D ,

h x( ) =f

gx( )

=f x( )g x( )

by definition of f

g; see Section 1.6

=f x( )g x( )

because f is even, and g is odd( )

=f x( )g x( )

= h x( )

Q.E.D. §


Example 8 (Applying a Product Rule)

Let f x( ) = 7x6 x2

+ 4( ) x3 5x( ) . Is f even, odd, or neither?

§ Solution

f x( ) = even( ) odd( )= odd( )

Sign Rule: P( ) N( ) = N( ) .

f is an odd function. §

Example 9 (Applying Product and Quotient Rules)

In Chapter 4, we will see that the sine function (abbreviated “sin” when

applied to an argument) is odd, and the cosine function (abbreviated “cos”)

is even. Let q( ) =cos

sin. Is q even, odd, or neither?

§ Solution

WARNING 4: The “ ” sign in front of the fraction is inconsequential in

our analysis; multiplication by the constant 1 corresponds to multiplication

by an even function. When considering corresponding sign rules, the

notations (P) and (N) may be preferable to +( ) and ( ) due to this

potentially confusing matter.

Rewrite the given formula as:

q( ) =( ) cos( )

sin( ).

q( ) =

odd( ) even( )odd( )

= even( )

Sign Rule: N( ) P( )

N( )= P( ) .

q is an even function. §

TIP 1: When in doubt, use the definitions of even function and odd function.


PART E: SYMMETRY AND EQUATIONS

Consider an equation in x and y and its graph in the usual xy-plane.

1) The graph is symmetric about the y-axis

Replacing all occurrences of x with x( ) yields an equivalent equation.

Special Case: y = f x( ) , where f is an even function.

The equations below are then equivalent:

y = f x( )y = f x( )

2) The graph is symmetric about the x-axis

Replacing all occurrences of y with

y( ) yields an equivalent equation.

3) The graph is symmetric about the origin

Replacing x with x( ) and y with y( ) yields an equivalent equation.

Special Case: y = f x( ) , where f is an odd function.

The equations below are then equivalent:

y( ) = f x( )y = f x( ) because f is odd( )y = f x( )

4) The graph is symmetric about the line y = x

The equation is symmetric in x and y, meaning that

switching x and y throughout the equation yields an equivalent equation.

• In multivariable calculus, symmetry in x and y allows us to trim down

repetitive work such as finding partial derivatives of a function with respect

to x and y.


Example 10 (Identifying Graphical Symmetries from an Equation)

The graph of x2+ y2

= 9 exhibits all four aforementioned symmetries.

1) Symmetry about the y-axis

x( )2

+ y2= 9

x2+ y2

= 9

x, y( ) is on the graph

x, y( ) is on the graph.

2) Symmetry about the x-axis

x2+ y( )

2

= 9

x2+ y2

= 9



3) Symmetry about the origin

x( )2

+ y( )2

= 9

x2+ y2

= 9



4) Symmetry about the line y = x

y2+ x2

= 9

x2+ y2

= 9


y, x( ) is on the graph.

§

SECTION 1.5: PIECEWISE-DEFINED FUNCTIONS; LIMITS AND ...

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