Question

The function f is defined by$f\left(x\right)=\left\{\begin{array}{cc}{x}^2,& 0\le x\le 3\\ 3x,& 3\le x\le 10\end{array}\right.$
The relation g is defined by $g\left(x\right)=\left\{\begin{array}{ll}{x}^2,& 0\le x\le 2\\ 3x,& 2\le x\le 10\end{array}\right.$
Show that f is a function and g is not a function.

Answer 1

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