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Function
2. Function
Function notation
Graph of function
2.1. Function notation
Composition of functions
Identity function
The equation
Inverse function
f (x) = x2 + 3
defines a function f from the set R of real numbers to itself (written f : R → R). This
function accepts an input value x and returns an output value f (x), which, according to
the equation, is obtained by squaring the input value and adding 3. For instance, when the
input is 5, the output is f (5), which is 28.
One way to study a given function is to make a table of input values x versus output values
f (x), here illustrated for the function f (x) = x2 + 3:
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f (x)
7
4
3
4
7
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x
−2
−1
0
1
2
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Input values can be chosen to be anything, but they are usually chosen to be convenient
numbers (often integers) close to the region of interest (often not far from 0).
A Venn diagram provides a useful way to think about a function. Pictured below is a Venn
diagram representation of the function f (x) = x2 + 3. It shows a general input value x
being mapped to the corresponding output value f (x) as well as how this mapping works
for the chosen inputs.
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Function
Function notation
Graph of function
Composition of functions
Identity function
Inverse function
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The bubble on the left represents the domain of f (here, R), which is the set of input
values. The bubble on the right represents the codomain of f (here, R), which is a set
containing the output values.
The range of f is the set of all values that actually occur as outputs. Here, the range is
the set [3, ∞) of all numbers greater than or equal to 3 (the smallest x2 can be is 0).
If a function f has domain A and codomain B we write f : A → B and say that f is a
function from A to B. Strictly speaking, in order to define a function, one must specify the
domain and the codomain as we did above when we said that f was a function from the
set R of real numbers to itself. However, if a function is given by means of a formula, like
f (x) = x2 + 3, we assume that the domain is taken to be the set of all real numbers x for
which the expression on the right makes sense (usually meaning no division by 0 or even
roots of negative numbers), and the codomain is taken to be the set of all real numbers.
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Instead of saying something like “the function f given by f (x) = x2 + 3,” we will often say
“the function f (x) = x2 + 3,” or even just “the function x2 + 3” when there is no need to
give the function a name.
Function
Function notation
Graph of function
Composition of functions
2.1.1
Example
2x − 3
Let f (x) = √
.
x−1
Identity function
Inverse function
(a) Find f (5).
(b) Find f (x + h).
(c) Find the domain of f .
Solution
(a) We get f (5) by replacing every occurrence of x in the formula by 5:
2(5) − 3
7
f (5) = √
= .
2
5−1
(b) The expression f (x + h) will come up when we talk about the derivative of a function.
It is computed just as before: Replace every occurrence of x in the formula by x + h. If
this is found to be confusing, it might help to imagine the formula as being
2( ) − 3
f( ) = √
,
−1
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so that f (x + h) is obtained by putting x + h in every box:
2(x + h) − 3
f (x + h) = √
.
x+h−1
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The parentheses in 2(x + h) are essential.
(c) Since the square root of a negative number is undefined, the formula requires x − 1 ≥ 0.
But division by zero is also undefined, so there is the further requirement x−1 6= 0. Putting
these together, we get x − 1 > 0, which is the same as saying x > 1. Therefore, the domain
of f is (1, ∞), the set of all real numbers greater than 1.
Function
Function notation
Graph of function
Composition of functions
Identity function
Incidentally, the codomain of f in the example is taken to be R as usual. It is usually
not an easy task to find the range of a function, and that is the case in this example.
An accurate graph of the function (see 2.2) usually helps considerably; we will find that
calculus provides a means for producing accurate graphs.
Inverse function
2.2. Graph of function
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Let f be a function (by which we always mean a function from some subset of R to R).
The graph of f is the set of all points (x, f (x)) with x in the domain of f .
2.2.1
Example
Sketch the graph of the function f given by f (x) = x2 .
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Solution We are to depict all points in the plane of the form (x, f (x)) (that is, (x, x )),
with x in the domain of f , which is R. We make a table of inputs versus outputs for some
conveniently chosen input values, plot those points, and then connect them to form the
graph:
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Function
Function notation
x
f (x)
Graph of function
Composition of functions
−2
−1
0
1
2
4
1
0
1
4
Identity function
Inverse function
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This method of plotting points and connecting them to form the graph brings up a question:
How do we know that the graph proceeds smoothly between the points we plotted and does
not have, say, ripples? It turns out that calculus gives the answer; it shows that the graph
is in fact smooth. We will see why later, but for now we will take it on faith that this is
the case.
The graph of the function f (x) = x2 is the same as the graph of the equation y = x2 (so,
replace f (x) by y). The graph of y = x2 is the set of all points (x, y) that satisfy the
equation. More generally, the graph of a function f is the graph of the equation y = f (x).
2.2.2
Example
Sketch the graph of the function f given by
(
1 − x, x ≤ 0
f (x) =
2,
x > 0.
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Solution This is what is called a “piecewise-defined” function. For x ≤ 0, the function
is given by f (x) = 1 − x, and for x > 0, the function is given by f (x) = 2. The graph is
the same as that of the line y = 1 − x all the way up to and including x = 0, and then it
changes to the graph of the horizontal line y = 2 from that point on:
Function
Function notation
Graph of function
Composition of functions
Identity function
Inverse function
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It was mentioned earlier that having an accurate graph aids one in the determination of
the range of a function. Since the graph of f consists of all points (x, f (x)), the range of
f is the set of y-coordinates of the points on the graph. As a practical matter, one can
quickly visualize the range by projecting the graph horizontally (from both directions) onto
the y-axis. In the first example above, the range is [0, ∞) (which we could have determined
without looking at the graph). In the second example, the range is [1, ∞).
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Function
2.3. Composition of functions
Function notation
2
The function f (x) = (2x + 3) can be thought of as being built up from the two simpler
functions g(x) = 2x + 3 and h(x) = x2 :
Graph of function
f (x) = (2x + 3)2 = (g(x))2 = h(g(x)).
Identity function
Composition of functions
Inverse function
For a given input x, the function g produces the output g(x). With this output used as
input, the function h produces the output h(g(x)). According to the equation above, the
function f has the same effect as this two-step process.
The composition of the functions g and h is the function h ◦ g given by
(h ◦ g)(x) = h(g(x)).
Here is a Venn diagram depiction of the composition h ◦ g:
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For the particular functions defined above, we have the relationship f = h ◦ g (which means
that f (x) = (h ◦ g)(x) for all x).
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2.3.1
Example
Find (h ◦ g)(x), given that g(x) = x + 1 and h(x) =
√
x − 2x.
Function
Function notation
Solution
We have
Graph of function
(h ◦ g)(x) = h(g(x)) =
p
g(x) − 2g(x) =
√
x + 1 − 2(x + 1).
Composition of functions
Identity function
Inverse function
In the example, g has domain R, while h ◦ g has domain [−1, ∞). This shows that the
domain of a composition can be smaller than the domain of the first function applied.
2.4. Identity function
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The simplest function is the identity function ε defined by ε(x) = x. Given the input
value x, it returns the output value x, so, in effect, it does nothing. The graph of this
function is the graph of the equation y = x (the 45◦ line through the origin).
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For any function f , we have ε ◦ f = f and f ◦ ε = f , since
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(ε ◦ f )(x) = ε(f (x)) = f (x)
and
(f ◦ ε)(x) = f (ε(x)) = f (x).
With composition ◦ viewed as a sort of multiplication, these equations show that ε acts like
a multiplicative identity (like the number 1). This is why ε is called the identity function.
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2.5. Inverse function
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√
3
Let f (x) = x3 and g(x) = x. These functions satisfy g(f (x)) = x and f (g(x)) = x, since
√
p
√
3
g(f (x)) = 3 f (x) = x3 = x
and
f (g(x)) = (g(x))3 = ( 3 x)3 = x.
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The equation g(f (x)) = x says that an input x, changed by f to f (x), gets changed back
f
g
to x by g, that is, g undoes what f does. For example, 2 7→ 8 7→ 2. Similarly, the equation
f (g(x)) = x says that f undoes what g does. We say that g is an “inverse” of f .
Function notation
Graph of function
Composition of functions
Identity function
Existence of inverse function.
Not every function has an inverse. For instance, the
squaring function f (x) = x2 has no inverse. One might suppose that the square root
√
f
g
function g(x) = x undoes what f does, since, for instance, 2 7→ 4 7→ 2. However, it does
f
Inverse function
g
not work for every input value. For example, −2 7→ 4 7→ 2.
The problem is that the squaring function f sends two inputs to the same output (both 2
and −2 get sent to 4). Put another way, the graph of f fails the horizontal line test. (A
graph is said to pass the horizontal line test if each horizontal line intersects the graph
in at most one point.) Indeed, the horizontal line y = 4 intersects the graph of f at both
(−2, 4) and (2, 4).
The remedy is to use a smaller domain for f , that is, to allow fewer input values. If we
declare the domain of f to be just the set [0, ∞) of nonnegative real numbers, then it is no
longer the case that f sends two inputs to the same output (for example, −2 is no longer
an allowable input). In this new situation, the left half of the graph of f disappears and we
are left with a graph that
√ passes the horizontal line test. With this adjustment, the square
root function g(x) = x does indeed serve as an inverse for f .
2.5.1
Example
(a) Find
√
4.
(b) Solve the equation x2 = 4.
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Solution
Function notation
(a) The square√root function is the inverse of the squaring function with reduced domain
[0,
√ ∞), so 4 is the number in the interval [0, ∞) that you square to get 4. Therefore,
4 = 2.
(b) Here, we are searching for all real numbers x satisfying the equation. The answer is
x = ±2.
This example is intended to serve as an illustration in a familiar setting of an analogous
example given later in the less-familiar setting of inverse trigonometric functions
√ (see Exa is ± the
ample 4.4.1). Also, it is intended to correct
the
common
misconception
that
√
number you square to get a. Rather, a is the number ≥ 0 that you square to get a. The
issue of ± comes up only in solving equations like in (b).
A function f is injective if it never sends two inputs to the same output (i.e., its graph
passes the horizontal line test). If f is injective, then it has an inverse function that sends
an output of f back to the (unique) input that produced the output.
Graph of function
Composition of functions
Identity function
Inverse function
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Definition of inverse function. Let f : A → B be an injective function with range
B. The function f −1 : B → A defined by
f −1 (y) = x
⇐⇒
f (x) = y
(1)
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is called the inverse of the function f .
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Inverse function identities. With notation as above
(i) f −1 (f (x)) = x for all x in A,
(ii) f (f −1 (x)) = x for all x in B.
Function
Function notation
Graph of function
Composition of functions
If a function is not injective, then it does not have an inverse, but we can reduce its domain
in order to achieve an injective function just as we did for the squaring function. This
process is illustrated here using Venn diagrams:
Identity function
Inverse function
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2.5.2
Example
|x + 2|
Let f (x) =
.
3
(a) Explain why f has no inverse as is.
(b) From here on, assume that the domain of f has been reduced to [−2, ∞). Then the
function f : [−2, ∞) → [0, ∞) is injective and its range is [0, ∞) so that it has an
inverse f −1 : [0, ∞) → [−2, ∞). Find a formula for the function f −1 .
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(c) Verify that f and f −1 satisfy the inverse function identities (i) and (ii) (see 2.5).
Function
Function notation
Solution
Graph of function
Composition of functions
(a) f (−3) = 1/3 = f (−1) so f is not injective (the two inputs −3 and −1 produce the
same output 1/3).
(b) For x in the restricted domain [−2, ∞), the quantity x + 2 is always greater than or
equal to 0, so the absolute value sign in the formula can be dropped giving f (x) =
(x + 2)/3. By (1) in the definition of inverse function, f −1 (y) is the x satisfying
f (x) = y, that is, the x satisfying (x + 2)/3 = y. Solving for x, we get x = 3y − 2,
so f −1 (y) = 3y − 2. Replacing y with the more traditional letter x, we get f −1 (x) =
3x − 2.
Identity function
Inverse function
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(c) For x ∈ [−2, ∞), we have
f −1 (f (x)) = f −1 (
x+2
x+2
) = 3(
) − 2 = (x + 2) − 2 = x,
3
3
and for x ∈ [0, ∞), we have
f (f −1 (x)) = f (3x − 2) =
|(3x − 2) + 2|
3|x|
=
= |x| = x,
3
3
as desired.
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The inverse function identity (i) can be written (f −1 ◦ f )(x) = ε(x), which says that
f −1 ◦ f = ε. Similarly, (ii) can be written f ◦ f −1 = ε. With composition ◦ viewed as
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Function
multiplication and ε viewed as a multiplicative identity (as in 2.4), these equations say that
f −1 is a multiplicative inverse of f . This is the reason for the notation f −1 as well as the
terminology “inverse function.”
The trigonometric functions (see 4.2) are not injective, but we will reduce their domains in
order to achieve injective functions, just as we did for the squaring function, and this will
enable us to define the inverse trigonometric functions (see 4.4).
Function notation
Graph of function
Composition of functions
Identity function
Inverse function
Graph of inverse function. The graph of the function f (x) = x2 , with domain reduced
√
to [0, ∞), is shown below together with the graph of its inverse function f −1 (x) = x.
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We see that the graph of f −1 is the reflection of the graph of f across the 45◦ line y = x. If
the graph of f is drawn, and the paper is folded along the dotted line before the ink dries,
then the graph of f −1 is produced (albeit, in a different color here).
This is a general property relating the graph of a function f to the graph of its inverse f −1
(if such exists). It amounts to saying that the point (x, y) is on the graph of f if and only
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if the point (y, x) is on the graph of f −1 , and this is precisely what (1) says.
Function
Function notation
2.5.3 Example
Let f be as in Example 2.5.2, with domain [−2, ∞). Sketch the graph
of f together with the graph of f −1 .
Graph of function
Composition of functions
Identity function
Solution As noted in the earlier example, for x in the domain [−2, ∞), the function f
is given by f (x) = (x + 2)/3. The graph of f is the same as the graph of the equation
y = (x + 2)/3 = 31 x + 23 , which is a line with slope 13 and y-intercept 32 . The graph of the
inverse f −1 is obtained by reflecting across the 45◦ line y = x:
Inverse function
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(By the earlier example, f −1 (x) = 3x − 2, the graph of which is the same as that of
y = 3x − 2, a line with slope 3 and y-intercept −2. This agrees with the graph of f −1 as
shown above.)
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Function
2 – Exercises
Function notation
Graph of function
√
2–1
Let f be the function given by f (x) = x2 −
1 + 3x
.
x−1
Composition of functions
Identity function
Inverse function
(a) Find f (5).
(b) Find f (x + h).
(c) Find the domain of f .
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2–2
2–3
Sketch the graph of the function f given by

2

1 − x , x < 0,
f (x) = 1 − 2x, 0 ≤ x ≤ 1,


1/x,
x > 1.
Let f (x) = 2x + x2 and g(x) =
(a) Find (f ◦ g)(x).
(b) Find (g ◦ f )(3).
√
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x − 1.
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2–4
Let f (x) = (x − 1)2 − 1.
Function
Function notation
(a) Explain why f has no inverse as is.
(b) From here on, assume that the domain of f has been reduced to [1, ∞). Then the
function f : [1, ∞) → [−1, ∞) is injective and its range is [−1, ∞) so that it has an
inverse f −1 : [−1, ∞) → [1, ∞). Sketch the graph of f together with the graph of
f −1 .
Graph of function
Composition of functions
Identity function
Inverse function
(c) Find a formula for the function f −1 .
(d) Verify that f and f −1 satisfy the inverse function identities (i) and (ii) (see 2.5).
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