2.14 Odd, even and self-inverse functions · IB Maths AA | Augment Education
2.14 Odd, even and self-inverse functions
Test functions for odd and even symmetry, restrict domains so that inverses exist, find inverse functions, and recognise and prove self-inverse functions.
IB Mathematics: Analysis and Approaches
Part of 2 Functions
HL only
Key terms introduced here
Each is in the Maths AA glossary, with every other term in the subject.
A function for which f(−x)=−f(x) for every x in its domain. Its graph has rotational symmetry of order 2 about the origin, for example y=x3 and y=sinx.
A domain chosen as part of the largest possible one, usually so that a many-to-one function becomes one-to-one and has an inverse, for example f(x)=x2, x≥0.
Notes on 2.14 Odd, even and self-inverse functions
19 min readLast reviewed
Many functions have graphs with symmetry. The graph of y=x2 is unchanged by a reflection in the y-axis, and the graph of y=x3 is unchanged by a half-turn about the origin. Recognising these symmetries halves the work of sketching a graph, simplifies algebra and, later, simplifies integrals. A third kind of symmetry, in the line y=x, belongs to functions that are their own inverse.
This symmetry is closely linked to inverse functions. A function only has an inverse if it is one-to-one, so a function such as y=x2, whose graph is symmetric in the y-axis, must have its domain restricted before it can be inverted.
Even functions
The two statements in the definition say the same thing. The point (a,f(a)) reflects in the y-axis to (−a,f(a)), and this point is on the graph exactly when f(−a)=f(a)
Standard even functions include x2, x4, any power x2k with an even exponent, ∣x∣
Odd functions
A half-turn about the origin sends (a,f(a)) to (−a,−f(a)), which is the same as a reflection in the y-axis followed by a reflection in the x-axis. For an odd function, the output at −x
Standard odd functions include x, x3, any odd power x2k+1, x1
An odd function passes through the origin
If f is odd and 0 is in its domain, then putting x=0 into f(−x)=−f(x) gives f(0)=−f(0)
Functions that are neither odd nor even
Most functions are neither odd nor even. For example, f(x)=x2+x is a sum of an even and an odd term, and f(x)=ex has no symmetry in the y
For a function to be odd or even, its domain must be symmetric about 0: if a is in the domain, so is −a. The function f(x)=x2, x≥0
Testing whether a function is odd or even
The algebraic test is to replace x by −x throughout, simplify, and compare the result with f(x) and with −f(x).
Sums, products and composites
The parity of combined functions follows rules much like those for multiplying positive and negative numbers.
f
g
f×g
f+g
even
even
The product rules are easy to prove from the definitions, and proving one shows how all of them work.
For composite functions: if the inner function g is even, then f(g(−x))=f(g(x)), so f∘g is even whatever f
Finding unknown coefficients
The parity of a polynomial is fixed by which powers appear, so a condition such as "f is odd" sets the coefficients of all the even powers to zero.
Every function is a sum of an odd and an even function
Any function f with a domain symmetric about 0 can be split into an even part and an odd part:
f(x)=even2f(x)+f(−x)+odd2f(x)−f(−x).
The first part is even, since replacing x with −x swaps the two terms in the numerator and leaves the sum unchanged. The second is odd, since the swap changes the sign of the difference. For a polynomial this simply separates the even and odd powers: x3+2x2−x+5
Completing a graph using symmetry
If a function is known to be odd or even, its graph for x≥0 determines the whole graph.
Even: reflect the right-hand half in the y-axis.
Odd: rotate the right-hand half by 180° about the origin (equivalently, reflect it in the y-axis and then in the x-axis).
Inverse functions and domain restriction
An inverse function f−1 undoes f: if f(a)=b then f−1(b)=a
A many-to-one function fails this test. Every even function whose domain contains a non-zero value is many-to-one, since f(a)=f(−a) with a=−a; an inverse would have to send that one output back to two different inputs. Such a function has an inverse only after its domain is restricted.
For a quadratic, the natural choice is to restrict the domain to one side of the vertex: x≥h or x≤h, where (h,k) is the vertex. The largest such domain is the one that stops exactly at the vertex.
Finding the inverse on a restricted domain
The method is the one from 2.5: write y=f(x), rearrange to make x the subject, then swap x and y. With a restricted domain, one extra step decides which of two possible expressions is correct, usually the sign of a square root. The domain and range then swap:
domain of f−1=range of f,range of f−1=domain of f.
The same idea applies to other many-to-one functions. Squared denominators, even powers and ∣x∣ all need restricting, and so do the trigonometric functions: sinx restricted to −2π≤x≤2π
Where a function meets its inverse
The graphs of f and f−1 are reflections of each other in y=x. If f is increasing, any intersection point lies on the line y=x
Self-inverse functions
Some functions are their own inverse: applying them twice returns the starting value. The function f(x)=6−x is an example, since f(f(x))=6−(6−x)=x
The two conditions are equivalent because f∘f−1 is the identity function: if f∘f is also the identity, then f does what f−1
Standard self-inverse functions include
f(x)=x (the identity) and f(x)=c−x for any constant c;
f(x)=xk
Proving that a function is self-inverse
There are two ways to prove it: find f−1 and show it is the same as f, or show that f(f(x))=x. In both cases the domain of f
Finding a constant that makes a function self-inverse
Self-inverse functions need not be rational. The quarter circle f(x)=9−x2, 0≤x≤3
f(f(x))=9−(9−x2)=x2=∣x∣=xfor 0≤x≤3.
The last step uses x≥0; on the domain −3≤x≤3 the function would not be self-inverse, because its range would no longer equal its domain and it would not be one-to-one.
Exam-style questions on odd, even and self-inverse functions
The question that follows is typical of Paper 1 (no GDC).
Self-check
To do:
I can state the definitions of odd and even functions and describe the symmetry of each graph.
To do:
I can prove algebraically that a function is odd or even, and use a counterexample to show it is neither.
To do:
I can prove results about sums, products and composites of odd and even functions.
To do:
I can complete the graph of an odd or even function from its graph for x≥0.
To do:
I can restrict the domain of a many-to-one function so that it has an inverse.
To do:
I can find an inverse function on a restricted domain, choosing the correct sign of a square root and stating its domain and range.
To do:
I can find where a function meets its inverse, knowing when the intersection must lie on y=x
Questions on 2.14
25 exam-style questions on this subtopic, written in the style of the IB papers. The first has its markscheme open. Counted on
Question 1 · AA-001906 marks
Let g(x)=x2sinx, where x∈R.
(a)
Show that g is an odd function.
[2]2 marks
Markscheme (a)
g(−x)=(−x)2sin(−x)M1
=x2(−sinx)
(b)
Hence find the exact value of ∫−4π4π(x2sinx+cos2x)dx
Question 2 · AA-001925 marks
Let f(x)=x2cosx, where −2π≤x≤2π
Question 3 · AA-0019417 marks
Let f(x)=2sinx+sin2x, where 0≤x≤2π. The diagram below shows the curve y=f(x)
Try a quick-fire question
One of 30 short multiple-choice questions on 2.14. Pick an answer to see why it's right or wrong.
-axis is the mirror image of its graph on the right.
y = ¼x⁴ − 2x² + 3
An even function: every point (a, f(a)) has a partner (−a, f(a)) at the same height on the other side of the y-axis.
,
cosx
,
e−x2
and any constant function. A polynomial containing only even powers of
x
(and possibly a constant term) is even, because
(−x)2k=x2k
.
maps the graph onto itself.
is the negative of the output at
x
.
y = ¼x³ − 2x
An odd function: the point (2, −2) has a partner (−2, 2), its image under a half-turn about the origin.
,
sinx
and
tanx
. A polynomial containing only odd powers of
x
is odd, since
(−x)2k+1=−x2k+1
. This is the origin of the names.
, so
2f(0)=0
and
f(0)=0
. The graph of an odd function defined at
0
always passes through the origin. This gives a quick test: if
f(0)=0
, the function is not odd. The converse is false:
f(x)=x2
passes through the origin but is even, not odd.
-axis or about the origin. The only function that is both odd and even on a domain is the zero function
f(x)=0
: adding
f(−x)=f(x)
and
f(−x)=−f(x)
gives
f(x)=0
.
, is neither odd nor even, because
f(−1)
is not defined.
.
(c) h(x)=xsinx
(d) k(x)=x2+x
Show the solution
(a) f(−x)=(−x)3−2(−x)=−x3+2x=−(x3−2x)=−f(x). So f is odd.
(b) g(−x)=(−x)4+1(−x)2=x4+1x2=g(x)
(c) sin is odd, so sin(−x)=−sinx. Then h(−x)=(−x)sin(−x)=(−x)(−sinx)=xsinx=h(x)
(d) k(−x)=(−x)2+(−x)=x2−x. This is neither k(x)=x2+x
fully, then check whether it equals
f(x)
, equals
−f(x)
, or equals neither.
even
even
odd
odd
even
odd
odd
even
odd
usually neither
in the domain. Since
f
and
g
are odd,
f(−x)=−f(x)
and
g(−x)=−g(x)
. So
p(−x)=f(−x)g(−x)=(−f(x))(−g(x))=f(x)g(x)=p(x).
This holds for every x in the domain, so p is even.
is. For example,
cos(x2)
and
e−x2
are even. If both
f
and
g
are odd, then
f(g(−x))=f(−g(x))=−f(g(x))
, so
f∘g
is odd; for example,
sin(x3)
is odd.
is even, and
x3+1
is neither.
. Find the values of
p
,
q
and
r
.
Show the solution
Since f is odd, f(−x)=−f(x):
−x3+px2−qx+r=−x3−px2−qx−r.
Comparing coefficients of x2: p=−p, so p=0. Comparing constants: r=−r
Then f(x)=x3+qx, and f(1)=1+q=4
has even part
2x2+5
and odd part
x3−x
.
.
(a) Given that f is even, find an expression for f(x) when x<0, and sketch the graph of y=f(x).
(b) Given instead that f is odd, find an expression for f(x) when x<0, and sketch the graph.
Show the solution
(a) If x<0, then −x>0, so the given formula applies to −x: f(−x)=(−x)2−2(−x)=x2+2x. Since f is even, f(x)=f(−x)=x2+2x for x<0. (Both pieces can be written together as f(x)=x2−2∣x∣.)
(b) Since f is odd, f(x)=−f(−x)=−(x2+2x)=−x2−2x
. For
f−1
to be a function, each output of
f
must come from exactly one input, so
f
must be a
one-to-one function
. Graphically, every horizontal line meets the graph of
f
at most once (the
horizontal line test
).
for which
f
has an inverse function.
(b) For this value of a, find f−1(x), and state the domain and range of f−1.
(c) Sketch the graphs of y=f(x) and y=f−1(x) on the same axes.
Show the solution
(a) Complete the square: f(x)=(x−3)2−4. The vertex is (3,−4). To the left of the vertex f is decreasing, so it is one-to-one for x≤3. Any larger a would include values on both sides of x=3 with equal outputs. So a=3.
(b) Let y=(x−3)2−4 with x≤3. Then
(x−3)2=y+4⇒x−3=±y+4.
Since x≤3, x−3≤0, so take the negative root: x=3−y+4
f−1(x)=3−x+4.
The range of f for x≤3 is f(x)≥−4, so the domain of f−1
(c) The graph of f−1 is the reflection of the graph of f in the line y=x.
gives values greater than
3
, which are outside the domain of
f
and so cannot be outputs of
f−1
. Use the restricted domain to choose the sign, and check:
f−1(5)
should be
0
, since
f(0)=5
.
is one-to-one, and its inverse is
arcsinx
(3.9).
. Find
g−1(x)
and state its domain.
Show the solution
Without the restriction g would be many-to-one, since its graph is symmetric in the line x=1. For x>1 it is decreasing, so it is one-to-one.
Let y=(x−1)24. Then (x−1)2=y4, so x−1=±y2. Since x>1, x−1>0: take the positive root, x=1+y2. Swapping,
g−1(x)=1+x2.
For x>1 the values of g are all positive real numbers (as x→1+, g→∞
, so the equation
f(x)=f−1(x)
can be replaced by the much simpler
f(x)=x
.
and
y=f−1(x)
intersect.
Show the solution
For x≥0, f is increasing, so the graphs meet on the line y=x. Solve f(x)=x:
x2−2=x⇒x2−x−2=0⇒(x−2)(x+1)=0.
Since x≥0, reject x=−1. The graphs meet at (2,2).
Check with the inverse, f−1(x)=x+2, x≥−2
, the inverse is
f−1(x)=−3x
, and the graphs meet at
(0,0)
but also at
(1,−1)
and
(−1,1)
, which are not on
y=x
.
.
for every
x
in the domain. The graph of a self-inverse function is symmetric in the line
y=x
.
does. Since the graph of
f−1
is the reflection of the graph of
f
in
y=x
, a
self-inverse function
is one whose graph is its own reflection in
y=x
.
,
x=0
, for any
k=0
(in particular the reciprocal function, 2.8);
f(x)=cx−aax+b, whenever a2+bc=0 (so that f is not constant);
f(x)=r2−x2, 0≤x≤r, a quarter of a circle.
must equal its range, since the domain of
f−1
is the range of
f
.
.
(a) Show that (f∘f)(x)=x.
(b) Hence write down f−1(x).
(c) Sketch the graph of y=f(x), and describe its symmetry.
Show the solution
(a) Substitute f(x) into f:
f(f(x))=x−22x+3−22(x−22x+3)+3.
Multiply the numerator and the denominator by (x−2):