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On Range and Reflecting Functions About the Line y = mx
Scott J. Beslin
Dept.of Mathematics and
Computer Science
Nicholls State University
Thibodaux, LA 70310
[email protected]
Brian K. Heck
Dept. of Mathematics and
Computer Science
Nicholls State University
Thibodaux, LA 70310
[email protected]
Jeremy J. Becnel
Dept. of Mathematics and
Statistics
Stephen F. Austin State University
Nacogdoches, TX 75962
[email protected]
Abstract: The authors explore the importance of “range” and its relationship to
continuously differentiable functions that have inverses when their graphs are reflected about
lines other than y = x . Some open questions are posed for the reader.
Mathematics Subject Classification: 26A09
Key Words: reflection, matrix representation, range, inverse
On Range and Reflecting Functions About the Line y = mx
I. Introduction
In college algebra and beyond, students study functions and their graphs. In almost every
exercise, they are asked to specify the domain and range of a function. Most students have
accepted the importance of the domain, given their work in solving equations involving roots and
fractions, and later in addressing the intricacies of graphing and finding limits. However, many
students avoid finding the range of the function; it is for them a more difficult process and many
do not see the point in it. Depending on the level of the class, the instructor has several
elementary avenues for illustrating the importance of range. We briefly mention a few.
(i) Solving Equations
The fact that x 2 = −1 and sin x = 2 have no solutions is clearly related to the ranges of
the functions involved.
(ii) Finding Domains of Composite Functions
Since the domain of the composite function f o g is the set
{x | x ∈ Domain( g ) and g ( x) ∈ Domain( f )} , the range of the function g is particularly important
in determining the domain of this composite function. For example, if g :
+
→
is given by
g ( x) = − x , then the domain of the composite g o g is empty.
(iii) Functions restricted in domain so that their inverses exist
The trigonometric functions provide numerous examples of such functions. The fact that
Arcsin(sin 34π ) ≠
3π
4
is due to the fact that
3π
4
is not in the range of the Arcsin function.
In [1], the authors generalize a problem from elementary calculus. In that paper, a “new”
type of derivative is introduced as a method of solving pursuit problems. One of the principal
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results concerning the solvability of such problems is directly related to the range of the NewtonRaphson function encountered in root-finding analysis.
In this paper, we explore a generalization related to “inverse function.” The key result of
this investigation illustrates the importance of range in a situation different from those cited
above.
II. Reflections and Reflection Matrices
In any introductory algebra class, inverse functions are studied. Usually, in addition to
the algebraic method for computing the inverse of a one-to-one function, a graphical method is
also explored. Namely, if a one-to-one function is reflected about the line y = x , the result is its
inverse. Obviously, if the function is not one-to-one, it does not have an inverse, for the result of
such a reflection is not even a function. But what if the reflection takes place about a different
line? For example, the function y = x 2 is not one-to-one, and if reflected about y = x , we get
the graph of y 2 = x which is not a function (see Figure 1).
Figure 1.
The graph of y 2 = x
2
However, if we reflect y = x 2 about the line y = −1 , we do in fact get a function, namely
y = − x 2 − 2 (see Figure 2). To see this informally, observe that y = x 2 reflected about the x axis yields the function y = − x 2 . For reflection about the line y = −1 , a line parallel to the x axis, note that (0, 0) , the vertex of y = x 2 , clearly maps to (0, −2) .
Figure 2.
The graph of y = − x 2 − 2
For a better example, consider the function y = e x . A reflection about y = x yields the
natural logarithm function. However, we can also reflect it about the lines y = 12 x and y = − x
and yield functions (see Figures 3 and 4 respectively). We will further address the notion of
reflecting functions in Section III.
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Figure 3.
The function y = e x reflected about y = 12 x
Figure 4.
The function y = e x reflected about y = − x
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Recall that by the reflection of a point P = ( x, y ) about the line y = mx , we mean the
point P′ = ( x′, y′) such that the given line is the perpendicular bisector of the segment PP′ (see
Figure 5).
Figure 5.
The definition of reflection about a line
If P is on the line y = mx , then P = P′ . Intuitively, then, if f is any function graphed
with “wet ink” in the plane, and we fold the plane along the line y = mx , the reflection of f
about this line is the imprint of f after folding.
Suppose that the line L defined by y = mx makes an angle θ with the positive x -axis,
where m = tan θ . One way of effecting the reflection of P = ( x, y ) about L is given by the
following procedure from linear algebra:
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Motion
2 x 2 Matrix of Motion
(1)
Rotate clockwise through angle θ
 cos θ
A=
 − sin θ
(2)
Reflect about the x -axis
1 0 
B=

0 −1
(3)
Rotate counterclockwise through θ
cos θ
C=
 sin θ
sin θ 
cos θ 
− sin θ 
cos θ 
Thus the final reflection matrix is
cos 2 θ − sin 2 θ
Rθ = A ⋅ B ⋅ C = 
 2 sin θ cos θ
2sin θ cos θ  cos 2θ
=
sin 2 θ − cos 2 θ   sin 2θ
sin 2θ 
− cos 2θ 
Using the appropriate trigonometric identities with m = tan θ , we have
Rθ = R(m) =
1 1 − m2

m 2 + 1  2m
2m 

m 2 − 1
Note that
1 − m2
2m
lim
= −1 and lim 2
=0
m →±∞ m 2 + 1
m →±∞ m + 1
so that
 −1 0 
Rπ = R (±∞) = 

2
 0 1
is the usual matrix for reflection about the y -axis. In addition,
1 0 
R0 = R (0) = 

0 −1
is the matrix of reflection about the x -axis, and
Rπ = R (1) =
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1  0 2  0 1 
=
2  2 0  1 0 
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is the matrix associated with inversion (reflection about y = x ).
The following list of elementary linear algebraic properties of the matrix R = R (m) is not
difficult to verify. Some, like number (2), are easily seen geometrically. We leave these
properties as student exercises.
(1) The determinant of R = R = −1 .
(2) R is invertible and R = R −1 .
(3) trace ( R ) = 0 .
(4) R is symmetric; that is R = RT .
(5) R is an orthogonal matrix.
(6) The minimal polynomial of R is t 2 − 1 ; its eigenvalues are ±1 .
Clearly the product of two R -matrices is not necessarily an R -matrix. For example,
 0 1
R (0) ⋅ R (1) = 

 −1 0 
which is not symmetric. However, geometrically, if we “reflect” the product about the x -axis
once more, we obtain
 0 −1
R (0) ⋅ R (1) ⋅ R (0) = 
 = R (−1)
 −1 0 
or equivalently,
R (0) ⋅ R (1) = R (−1) ⋅ R (0) −1 = R (−1) ⋅ R (0).
Algebra and trigonometry can be used to show that this relationship holds more generally. If
m = tan α and n = tan β represent slopes of lines through the origin, then
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tan α − tan β
1 + tan α tan β
m−n
=
1 + mn
tan(α − β ) =
as long as the two lines are not perpendicular, in which case α − β = π2 and mn = −1 .
1 0 
From (somewhat tedious) algebra, letting J = R (0) = 
 , we have
0 −1
R(m) ⋅ R(n) ⋅ J = R ( 1m+−mnn ) if mn ≠ −1 . If the lines are perpendicular, then R (m) ⋅ R (n) ⋅ J must
 −1 0 
effect a simple reflection about the y -axis; hence in that case R (m) ⋅ R (n) ⋅ J = 
 = R (±∞) .
 0 1
Since J = J −1 , we can sum up by writing for all real numbers m and n ,
 R ( 1m+−mnn ) ⋅ J if mn ≠ −1,
R ( m ) ⋅ R ( n) = 
 R ( ±∞ ) ⋅ J if mn = −1.
Taking appropriate limits, we may extend this result to include the cases:
R(m) ⋅ R(±∞) = R ( − m1 ) ⋅ J ;
R(±∞) ⋅ R(n) = R ( 1n ) ⋅ J ;
1 0 
R (±∞) ⋅ R (±∞) = 
.
0 1 
If the set M is defined by
M = {R (m) | m is a real number or m = +∞ or m = −∞} ,
we may define an operation * on M via: A * B = A ⋅ B ⋅ J . In the language of abstract algebra,
the operation * is a binary operator on M which is neither commutative nor associative.
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III. When Does Reflection Give a Function?
We now explore conditions on m such that the reflection of a function f by R (m)
yields another function.
Geometrically, it is easy to see that the reflection of a linear function f ( x) = ax + b by
R (m) is almost always another linear function. (This reflection will result in a vertical line if
m = a ± a 2 + 1 .) But generally speaking, if we abandon graphing as a “proof” that the
reflection is or is not a function, it is not algebraically evident what the case may be. For
example, reflecting the function f ( x) = e x about the line y = 12 x results in the parametric
representation (for −∞ < t < +∞ ):
x = 15 (3t + 4et ) and y = 15 (4t − 3et ) .
3 4 
To see this, observe that R ( 12 ) = 51 
 , so that
 4 −3
 x  1 3 4   t 
 y  = 5  4 −3 et  . From the parametric
 

 
equations alone, it is not immediately clear whether y can be written as a function of x .
In what follows, we let f :
→
be a continuously differentiable ( C 1 ) function, and
ask: When is the reflection of f by R (m) a function? The idea of “range” plays a significant
part in the answer to this question.
Reflection of f by R (m) yields:
x
1 1 − m 2
 y  = m2 + 1 
 
 2m
x(t ) =
1
m2 +1
2m   t 

 , or
m 2 − 1  f (t ) 
(1 − m 2 )t + 2mf (t )  and y (t ) =
1
m 2 +1
 2mt + (m 2 − 1) f (t )  .
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If x′(t ) ≠ 0 , then x′(t ) > 0 for all t or x′(t ) < 0 for all t , by the continuity of x′ and the
Intermediate Value Theorem. In any case, x −1 exists and t = x −1 . So y = y (t ) = y o x −1 is a
function of x . Observe that,
dy
=
dx
dy
dt
dx
=
dt
2m + (m 2 − 1) f ′(t )
; in particular,
1 − m 2 + 2mf ′(t )
x′(t ) = 0 if f ′(t ) =
m2 − 1
.
2m
Result 1 follows:
Result 1: Let f :
→
be a C 1 function. If m is such that
m2 −1
2m
∉ Range( f ′) , then reflection
of f by R (m) yields a function.
Note that reflection of a function by R (0) always yields a function.
Corollary 1: If f :
→
is a C 1 non-decreasing function (with f ′ ≥ 0 ), and
m2 −1
2m
< 0 , then
reflection of f by R (m) yields a function.
Example: We return to the increasing function f ( x) = e x . Since f ′( x) = e x , Range( f ′) =
Solving
m2 −1
2m
+
.
< 0 yields m < −1 or 0 < m < 1 . For instance, reflection of f about the line y = 12 x
(examined previously) is a function.
Corollary 2: If f :
→
is a C 1 non-increasing function (with f ′ ≤ 0 ), and either −1 < m < 0
or m > 1 , then reflection of f by R (m) yields a function.
Suppose the reflection of f by R (m) is a function y = y ( x) . Since ( x, y ) = ( x(t ), y (t ))
as before, a similar argument to the one preceding Result 1 shows that if y′(t ) ≠ 0 , then
x is a function of y . Since y is also a function of x , we see that y must be a
one-to-one function of x . Observing that y′(t ) = 0 whenever f ′(t ) = 1−2mm2 , we state Result 2.
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Result 2: Let f :
→
be a C 1 function. For every m such that
m2 − 1
2m
∉ Range( f ′) and
∉ Range( f ′) .
2m
1 − m2
then reflection of f by R (m) is a one-to-one function.
Corollary 3: If f is increasing, then its reflection by R (m) is a one-to-one function precisely
when m ∈ {0,1, −1} . In other words, the only possible one-to-one reflections are
− f , f −1 , and − (− f ) −1 , corresponding to R (m) acting on f for m = 0,1, and − 1 , respectively.
[For example, reflection of f ( x) = e x about y = 12 x yields a function, but not a one-to-one
function.]
Example: Let f ( x) =Arctan( x) . Then f ′( x) = 1+1x2 , and so Range( f ′) = (0,1] . Solving
m2 − 1
2m
∉ (0,1] together with
∉ (0,1] gives that reflection of f by R (m) is a one-to-one
2m
1 − m2
function whenever m ∈ [−1 − 2, −1] ∪ [1 − 2, 0] ∪ [−1 + 2,1] ∪ [1 + 2, ∞) . The algebra
involved here becomes less burdensome if one notices that
m2 − 1
2m
are negative
and
2m
1 − m2
reciprocals.
IV. Concluding Remarks
Results for reflections about lines of the form y = mx + b , b ≠ 0 , may be obtained from
the foregoing theory by vertical or horizontal translations of the plane. Finally, we state a pair of
open questions for investigation. Let f :
→
be a C 1 function which is invertible across the
line y = mx in the sense of Section III. Let f m−1 be the reflection of f about this line.
1. When is f o f m−1 = f m−1 o f ?
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2. How is
d
dx
(f
−1
m
( x) ) related to f ′( x) ?
Reference
[1] S. Beslin and D. Baney, A Different Type of Differentiability, PRIMUS, March 2000 (Vol.
X, No. 1), pp. 15-28.
Scott Beslin is currently department head of Mathematics and Computer Science at Nicholls
State University in Thibodaux, Louisiana. Although his administrative duties are often
demanding, he still enjoys working with students and colleagues on mathematical research and
exposition. Scott and the mathematics faculty are presently developing the Master of Science in
Community/Technical College Mathematics for online delivery
Brian K. Heck received his Ph. D. in mathematics from Louisiana State University in 1997. He
began teaching at Nicholls State University in 1998 and is currently a tenured associate
professor.
Jeremy J. Becnel was born in the small community of Kraemer, Louisiana on August 11, 1979.
He finished his undergraduate studies in mathematics and computer science at Nicholls State
University in May 2001. In August 2001 he went to Louisiana State University to pursue
graduate studies in mathematics. He earned a Master of Science degree in mathematics from
Louisiana State University in December 2002 and received his doctorate from Louisiana State
University in August 2006. Jeremy is currently an assistant professor at Stephen F. Austin State
University in Nacogdoches, Texas.
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