Question

The relation f is defined by

$f(x) = \begin{cases} x^2, & \quad 0≤x≤3\\\ 3x, & \quad 3≤x≤10 \end{cases}$
The relation g is defined by
$g(x) = \begin{cases} x^2, & \quad 0≤x≤2\\\ 3x, & \quad 2≤x≤10 \end{cases}$
Show that f is a function and g is not a function.

Answer 1

The relation f is defined as $f(x) = \begin{cases} x^2, & \quad 0≤x≤3\\\ 3x, & \quad 3≤x≤10 \end{cases}$
It is observed that for
0 ≤ x < 3, f(x) = x2
3 < x ≤10, f(x) = 3x
Also, at x = 3, f(x) = 32= 9 or f(x) = 3 × 3 = 9
i.e., at x = 3, f(x) = 9
Therefore, for 0 ≤ x ≤ 10, the images of f(x) are unique.

Thus, the given relation is a function.

The relation g is defined as $g(x) = \begin{cases} x^2, & \quad 0≤x≤2\\\ 3x, & \quad 2≤x≤10 \end{cases}$

It can be observed that for x=2, g(x) = 22 = 4 and g(x) = 3 × 2 = 6
Hence, element 2 of the domain of the relation g corresponds to two different images i.e., 4 and 6.

Hence, this relation is not a function.