Download Real Functions

Workshop: Calculus
Real Functions
Topics Covered:
• Definition of function
• The domain and the range
• Composite function
• Inverse function
by Dr.I.Namestnikova
1
Definition
A fundamental concept in calculus is that of a function. We will consider only
real functions.
To define a real function f we need to specify:
• a set of real numbers X , called the domain of f .
• a set of real numbers Y , called the codomain of f .
• a rule that associates with each real number x from set X a unique real
number f (x) in the set Y .
Functions are denoted as e.g.
f (x), g(x) or F (x)
name of a function
f(x)
argument (independent variable)
no multiplication
An expression like f (2) means the value of the function f at point 2.
Example
Let f (u)
= 3u − 4. Evaluate
1.
f (−2)
2.
f (t − 2)
3.
f (t2 )
Solution:
1.
f (−2) = 3 × (−2) − 4 = −10
2.
f (t − 2) = 3(t − 2) − 4 = 3t − 10
3.
f (t2 ) = 3t2 − 4
2
Diagrammatical representation
Y
X
f Hx2L
x2
The number f (x) is
f Hx1L
x1
the value of f at x.
f Hx3L
x3
Codomain
Domain
The natural domain of a real function is the largest subset of the real numbers
for which the function (the rule) has meaning.
For example, the function
f (x) =
√
x − 3 is undefined for x < 3 since
the number under the square root is negative for these values of the argument
x. So the natural domain is x > 3.
To each x ∈ X , f (x) assigns one and only one value f (x) ∈ Y
(single valued function).
Impossible
Possible
Y
X
Y
X
x2
a = f(x)
x
x1
f Hx1L = f Hx2L
b = f(x)
Y
The range of a function f is the set
X
of all ”output” values produced by f .
x2
The range of a function is a subset
Range
x1
x3
Domain
of Y .
Codomain
3
Example
For each of the following functions determine the natural domain and the range
1.
f (x) = x2 − 3x + 4
2.
f (x) = ln(x)
3.
f (x) = ex
f (x) = sin(x)
1
5. f (x) =
x−1
1
6. f (x) =
x2 − 1
1
7. f (x) = √
1 − x2
4.
Solution:
1. The domain is a set of all real numbers
also R.
R or −∞ < x < ∞. The range is
ln(x) is not defined for negative x. Hence the domain is 0 <
x < ∞ but the range is −∞ < f (x) < ∞, i.e. R.
2. The function
3. The function ex is defined for all real x. Hence the domain is −∞
but the range is 0
< f (x) < ∞.
4. The domain is −∞
−1 6 f (x) 6 1.
<x<∞
< x < ∞ but the range is
5. We cannot divide by zero, so we must exclude 1 from the domain. Therefore the
domain is x
6= 1 and the range −∞ < f (x) < ∞.
= ±1. Therefore the domain is x 6= ±1 and
the range −∞ < f (x) < ∞.
6. This function is not defined for x
2.5
2.0
7.
1.5
1.0
0.5
-1.0
-0.5
0.0
R because the
square root is a real number only when 1−x2 > 0,
that is, when x2 6 1. We also cannot divide by
zero, so we must exclude ±1 from the domain. Thus
the domain is −1 < x < 1.
The range is 1 6 f (x) < ∞.
The function does not have domain
f HxL
3.0
0.5
1.0
x
4
Composite functions
Let us consider two functions
y = f (x) and z = g(y) and define a
composite function g(f (x))
z
x
f HxL
y
gHyL
Z
X
Y
Note that (f
Example
◦ g)(x) = f (g(x))
Can the following functions be considered as composite function?
1.
ln(3x2 )
2.
(3x + 4)5
3.
sin(x)
(x − 3)2
2x2 tan(5x)
p
5. 3 cos(x2 − 3x)
4.
Solution:
1. Yes.
y = (3x2 ),
2. Yes.
y = 3x + 4,
z = ln(y).
z = y5.
3. No, this a quotient of two composite functions.
4. No, this is a product of two composite functions.
5. Yes.
y = x2 − 3x,
z = cos(y),
5
w=
√
3
z.
Inverse function
f −1 (x) : f ◦ f −1 (x) = f −1 ◦ f (x) = x
f(x)
f(x)
x
f -1 IxM
Y
X
Note that the inverse function does not always exist (e.g. for f (x)
Example
Find f −1 (x) if f (x)
Solution:
1. Let y
= 2x − 1.
= f (x) ≡ 2x − 1
2. Solve for x
x=
3. Rename x as f
−1
(y) =
y+1
2
y+1
2
or rename y
f −1 (x) =
as required.
6
x+1
2
= x2 ).
Example If f (x)
1.
f (−1)
2.
f (y 2 )
3.
g(y 2 )
4.
f ◦ g(x)
5.
g ◦ f (x)
g(x) = x2 find
= x − 3,
Solution:
1.
f (−1) = −1 − 3 = −4
2.
f (y 2 ) = y 2 − 3
3.
g(y 2 ) = (y 2 )2 = y 4
x2 - 3
x2
x
4.
f
g
gH L =
fH
2
L=
-3
f ◦ g(x) = x2 − 3
x
Hx - 3L2
x-3
5.
g
f
fH
L=
-3
gH L =
g ◦ f (x) = (x − 3)2
7
2
Example
Given that
x2
f (x) = 3x − 2, g(x) =
4
1
h(x) =
and
• f (3)
• (f ◦ g)(x)
• g(1)
• (g ◦ f )(x)
• f (2) − h(2)
• (f ◦ h)(x)
• f −1 (x)
• (g ◦ h ◦ f )(x)
• h−1 (x)
• (f ◦ g ◦ h)(x)
x
find the following
Solution:
• f (3) = 3 × 3 − 2 = 7
• g(1) =
12
4
1
4
=
• f (2) − h(2) = 3 × 2 − 2 −
• f −1 (x) =
• h−1 (x) =
1
2
= 3 12
2+x
3
1
x
• (f ◦ g)(x) =
3x3
−2
4
1
• (g ◦ f )(x) = (3x − 2)2
4
3
• (f ◦ h)(x) = − 2
x
1
• (g ◦ h ◦ f )(x) =
4(3x − 2)2
x
3x-2
f
fH
L = 3
-2
• (f ◦ g ◦ h)(x) =
h
hH
3
4x2
1  H3 x - 2L
L = 1
g
gH
−2
8
L =
1 ‘ I4 H3 x - 2L2 M
2
‘4
Example
Given that
f (x) = x3 − 2x, g(x) = sin x and h(x) = arcsin x find the following
1.
(f ◦ h)(x)
2.
(g ◦ h)(x)
3.
(h ◦ g)(x)
4.
(f ◦ g)(x)
5.
(g ◦ f )(x)
6.
(g ◦ f ◦ h)(x)
7.
(g ◦ h ◦ f )(x)
Solution:
1.
(f ◦ h)(x) = (arcsin x)3 − 2 arcsin x
2.
(g ◦ h)(x) = x
3.
(h ◦ g)(x) = x
4.
(f ◦ g)(x) = (sin x)3 − 2(sin x)
5.
(g ◦ f )(x) = sin(x3 − 2x)
6.
(g ◦ f ◦ h)(x) = sin ((arcsin x)3 − 2 arcsin x)
7.
(g ◦ h ◦ f )(x) = x3 − 2x
9