The function f is defined by
Question:

The function $f$ is defined by $f(x)=\left\{\begin{array}{lc}x^{2}, & 0 \leq x \leq 3 \\ 3 x, & 3 \leq x \leq 10\end{array}\right.$

The relation $g$ is defined by $g(x)= \begin{cases}x^{2}, & 0 \leq x \leq 2 \\ 3 x, & 2<x<10\end{cases}$

Show that f is a function and g is not a function.

Solution:

The function $f$ is defined by $f(x)= \begin{cases}x^{2} & 0 \leqslant x \leqslant 3 \\ 3 x & 3 \leqslant x \leqslant 10\end{cases}$

It is observed that for 0 ≤ x < 3, (x) = x2 .

3 <  x ≤ 10, f (x) = 3x

Also, at x = 3, f(x) = 32 = 9. And

(x) = 3 × 3 = 9.

That is, 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.

Again,

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

It can be observed that for x = 2, g(x) = 22 = 4 and also, 

g(x) = 3 × 2 = 6.

Hence, 2 in the domain of the relation g corresponds to two different images, i.e. 4 and 6.

Hence, this relation is not a function.

Hence proved.

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