Real-Valued Functions

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connecting reals

Real-Valued Functions

For the most part, our attention in this course is focused on real-valued functions.

Real-Valued Function

A real-valued function is one whose range is a subset of the real numbers.

The values of a real-valued function are real numbers.

Squaring Function

Let the function $SQ$ be defined as follows.

Domain of $SQ$ is $(-3, 5]$ .
Codomain of $SQ$ is $[-30, 30)$ .
$SQ$ pairs a domain number with its square.

First, this function is well-defined, since the square of any number in $(-3, 5]$ will be in $[-30, 30)$ and each domain number has exactly one square.

Shorthand Notation: $SQ: (-3, 5] \mapsto [-30, 30)$ .

Evaluate the following:

$SQ(-2) = \answer {4}$
$SQ(0) = \answer {0}$
$SQ(1.1) = \answer {1.21}$
$SQ\left (\frac {7}{5}\right ) = \answer {\frac {49}{25}}$

Range

The range of $SQ$ is $\left [ \answer {0}, \answer {25} \right ]$ .

$SQ$ is not an onto function. For instance, $-1 \in [-30,30)$ , yet $-1$ is not in the range, since squares of real numbers cannot be negative.

$SQ$ is not a one-to-one function, since $SQ(-1)=SQ(1)$ .

Collatz

Let the function $C$ be defined as follows.

Domain of $C$ is the positive integers: $\mathbb {N}$ .
Codomain of $C$ is the positive integers: $\mathbb {N}$ .
pairs a domain number with a range number according to the following rule:
- If the domain number is even, then $C$ pairs it with half the domain number:
- If the domain number is odd, then $C$ pairs it with one more than three times the domain number.

First, this function is well-defined, since the calculations can only produce one result.

Shorthand Notation: $C: \mathbb {N} \mapsto \mathbb {N}$ .

Evaluate the following:

$C(8) = \answer {4}$
$C(7) = \answer {22}$
$C(1) = \answer {4}$
$C(28) = \answer {14}$

$C$ is not a one-to-one function since $C(8) = C(1)$ .

Identity Function

Let the function $Id$ be defined as follows.

Domain of $Id$ is all real numbers: $\mathbb {R}$ .
Codomain of $Id$ is all real numbers: $\mathbb {R}$ .
$Id$ pairs a domain number with itself.

First, this function is well-defined.

Shorthand Notation: $Id: \mathbb {R} \mapsto \mathbb {R}$ .

Evaluate the following:

$Id(\pi ) = \answer {\pi }$
$Id(\sqrt {5}) = \answer {\sqrt {5}}$
$Id\left (\frac {13}{27}\right ) = \answer {\frac {13}{27}}$
$Id(0) = \answer {0}$

The Identity function is an onto function. If $r \in \mathbb {R}$ , the range, then $Id(r) = r$ .

The Identity function is a one-to-one function, since if $Id(r)=Id(s)$ , then $r = s$ . If two $Id$ values are equal, then the domain numbers are equal. They were not different domain numbers.

Remainder

Let the function $Remainder$ be defined as follows.

Domain of $Remainder$ is all natural numbers: $\mathbb {N}$ .
Codomain of $Remainder$ is $\{ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 \}$ .
$Remainder$ pairs a natural number with the remainder when divided by $10$ .

First, this function is well-defined. Dividing by $10$ can only have one remainder.

Shorthand Notation: $Remainder: \mathbb {N} \mapsto \{ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 \}$ .

$Remainder(12) = 2$
$Remainder(25) = 5$
$Remainder(77) = 7$
$Remainder(12 + 11) = 3$
$Remainder(7 \cdot 7) = 9$

Evaluate the following:

$Remainder(101) = \answer {1}$
$Remainder(3) = \answer {3}$
$Remainder(1652435) = \answer {5}$
$Remainder(94) = \answer {4}$

Linear

Let the function $L$ be defined as follows.

Domain of $L$ is $[-3, 4)$ .
Codomain of $L$ is $\mathbb {R}$ .
$L$ pairs a number with twice the number.

First, this function is well-defined.

Shorthand Notation: $L: [-3, 4) \mapsto \mathbb {R}$ .

The codomain of $L$ is $\mathbb {R}$ but the range is not.

The range is [( -8-6,68 , -8-6,68 ]) .

...more communication.

Communication Summary

The range is also called the image of the function.

Sometimes the range partner of a domain number is called the image of the domain number.

$f(a)$ is called the “value of $f$ at $a$ ” or “the image of $a$ under $f$ ”.

$f(a)$ is pronounced “ $f$ of $a$ ”.

And, we have the reverse direction.

The preimage of a subset of the codomain consists of the domain members whose function values are inside the given subset.

The preimage of $S$ is the set

$f^{-1}(S) = \{ d \in D \, | \, f(d) \in S \}$

Notation: The $-1$ exponent does not mean reciprocal. Instead, it is conveying an “opposite” direction.

The preimage of the range is the domain.

Sometimes when the preimage is a single domain member, then we drop the idea of a set and just quote that one domain member.

Note: The preimage of a codomain number, which is not in the range, is the empty set, $\emptyset$ .

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more examples can be found by following this link
More Examples of Real-Valued Functions