SECTION 1.5: PIECEWISE-DEFINED FUNCTIONS; LIMITS...
Transcript of SECTION 1.5: PIECEWISE-DEFINED FUNCTIONS; LIMITS...
(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.
(Section 1.5: Piecewise-Defined Functions; Limits and Continuity in Calculus) 1.5.2
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. §
(Section 1.5: Piecewise-Defined Functions; Limits and Continuity in Calculus) 1.5.3
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( ) .
(Section 1.5: Piecewise-Defined Functions; Limits and Continuity in Calculus) 1.5.4
• 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.) §
(Section 1.5: Piecewise-Defined Functions; Limits and Continuity in Calculus) 1.5.5
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.) §
(Section 1.5: Piecewise-Defined Functions; Limits and Continuity in Calculus) 1.5.6
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. §
(Section 1.5: Piecewise-Defined Functions; Limits and Continuity in Calculus) 1.5.7
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.)
(Section 1.5: Piecewise-Defined Functions; Limits and Continuity in Calculus) 1.5.8
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.
(Section 1.5: Piecewise-Defined Functions; Limits and Continuity in Calculus) 1.5.9
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.
• A table can help:
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.) §
(Section 1.5: Piecewise-Defined Functions; Limits and Continuity in Calculus) 1.5.10
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.
• A table can help:
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:
§
(Section 1.5: Piecewise-Defined Functions; Limits and Continuity in Calculus) 1.5.11
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.
(Section 1.5: Piecewise-Defined Functions; Limits and Continuity in Calculus) 1.5.12
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.
(Section 1.6: Combining Functions) 1.6.2
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
(Section 1.6: Combining Functions) 1.6.3
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,( ) . §
(Section 1.6: Combining Functions) 1.6.4
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,( )
§
(Section 1.6: Combining Functions) 1.6.5
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.)
(Section 1.6: Combining Functions) 1.6.6
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( ) .
(Section 1.6: Combining Functions) 1.6.7
§ 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 graphical form is:
… 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.)
(Section 1.6: Combining Functions) 1.6.8
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,( ) . §
(Section 1.6: Combining Functions) 1.6.9
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
§
(Section 1.6: Combining Functions) 1.6.10
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 .
(Section 1.6: Combining Functions) 1.6.11
In fact, Dom f g( ) …
… in set-builder form is: x x 3 and x 10{ } , or
x : x 3 and x 10{ }
… in graphical form is:
… 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,( ) . §
(Section 1.6: Combining Functions) 1.6.12
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 .
(Section 1.6: Combining Functions) 1.6.13
• 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( )
§
(Section 1.6: Combining Functions) 1.6.14
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.
(Section 1.6: Combining Functions) 1.6.15
• 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.
(Section 1.6: Combining Functions) 1.6.16
• 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. §
(Section 1.6: Combining Functions) 1.6.17
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
(Section 1.6: Combining Functions) 1.6.18
• 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.
(Section 1.7: Symmetry Revisited) 1.7.2
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. §
(Section 1.7: Symmetry Revisited) 1.7.3
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. §
(Section 1.7: Symmetry Revisited) 1.7.4
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.) §
(Section 1.7: Symmetry Revisited) 1.7.5
§ 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.
§ Solution 1 (Short Cut)
f x( ) is polynomial and only has terms of odd degree (5 and 1).
Therefore, f is odd. §
§ Solution 2 (Rigorous)
Dom f( ) = . x ,
f x( ) = 4 x( )5
+ x( )= 4x5 x
= 4x5+ x( )
= f x( )
Q.E.D. §
(Section 1.7: Symmetry Revisited) 1.7.6
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.
§ Solution 1 (Short Cut)
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. §
§ Solution 2 (Rigorous)
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. §
(Section 1.7: Symmetry Revisited) 1.7.7
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.
(Section 1.7: Symmetry Revisited) 1.7.8
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. §
(Section 1.7: Symmetry Revisited) 1.7.9
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.
(Section 1.7: Symmetry Revisited) 1.7.10
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.
(Section 1.7: Symmetry Revisited) 1.7.11
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
x, y( ) is on the graph
x, y( ) is on the graph.
3) Symmetry about the origin
x( )2
+ y( )2
= 9
x2+ y2
= 9
x, y( ) is on the graph
x, y( ) is on the graph.
4) Symmetry about the line y = x
y2+ x2
= 9
x2+ y2
= 9
x, y( ) is on the graph
y, x( ) is on the graph.
§