Question

The relation f is defined by

$f\left(x\right)=\{\begin{array}{cc}{x}^2,& 0\le x\le 2\\ 3x,& 3\le x\le 10\end{array}$

The relation g is defined by

$g\left(x\right)=\{\begin{array}{cc}{x}^2,& 0\le x\le 3\\ 3x,& 2\le x\le 10\end{array}$

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

Answer 1

Here

$f\left(x\right)=x^2$ $0\le x\le 3$

$f\left(x\right)=3x$ $3\le x\le 10$

At x = 3,

$f\left(3\right)=(3{)}^2=9$ and $f\left(3\right)=3\times 3=9$

We observe that f(x) takes unique value at each point in its domain [0, 10]. So f is function.

Now $g\left(x\right)=x^2$ $0\le x\le 2$

$g\left(x\right)=3x$ $2\le x\le 10$

At x = 2,

$g\left(2\right)=(2{)}^2=4$ and $g\left(2\right)=3\times 2=6$

So g(x) does not have unique value at x = 2. Hence g(x) is not a function.