Chapter 2 - Relations and Functions

The Cartesian product $A \times B$ is the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$.
Given $A = \{1, 2, 3, 4\}$ and $B = \{5, 7, 9\}$.
$A \times B = \{(1,5), (1,7), (1,9), (2,5), (2,7), (2,9), (3,5), (3,7), (3,9), (4,5), (4,7), (4,9)\}$.

Given $x \in \{0, 1, 2, 3, 4, 5\}$ and $y = x + 5$.
When $x=0, y=5$
When $x=1, y=6$
When $x=2, y=7$
When $x=3, y=8$
When $x=4, y=9$
When $x=5, y=10$
$\therefore R = \{(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)\}$
Domain = Set of first elements = $\{0, 1, 2, 3, 4, 5\}$
Range = Set of second elements = $\{5, 6, 7, 8, 9, 10\}$.

A relation is a function if every element in the domain has a unique image.
Here, the domain elements are $\{2, 5, 8, 11, 14, 17\}$.
Each element in the domain has a unique image (which is 1). No two distinct ordered pairs have the same first component.
Therefore, $R$ is a function.

Since $f$ is a linear function, let $f(x) = mx + c$.
Using $(1, 1)$: $1 = m(1) + c \implies m + c = 1$ ...(i)
Using $(0, -1)$: $-1 = m(0) + c \implies c = -1$
Substitute $c = -1$ in (i):
$m - 1 = 1 \implies m = 2$
Therefore, the function is $f(x) = 2x - 1$.

The function is defined for all real numbers except where the denominator is zero.
Denominator: $x^2 - 8x + 12 = 0$
$(x-6)(x-2) = 0$
$x = 6$ or $x = 2$
Therefore, Domain = $\mathbb{R} - \{2, 6\}$.

We know that the absolute value function $|x|$ is always non-negative for all real $x$.
$\therefore |x-1| \ge 0$ for all $x \in \mathbb{R}$.
Hence, the range of $f(x)$ is $[0, \infty)$.

Given $f(x) = x^2$.
$f(1.1) = (1.1)^2 = 1.21$
$f(1) = (1)^2 = 1$
$\frac{f(1.1) - f(1)}{1.1 - 1} = \frac{1.21 - 1}{0.1} = \frac{0.21}{0.1} = 2.1$.