Question

The relation f is defined by

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

The relation g is defined by

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

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

Answer 1

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

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

Now, check given relation is function or not.

Put x = 3

If $f\left(x\right)=x^2$

$f\left(3\right)=9$

and if $f\left(x\right)=3x$

f(3) = 9

i.e., at x = 3, f (x) = 9 (unique image)

So, relation has unique image in [0,10]

Hence, the given relation is a function.

g(x) also has two relations at x = 2

Now, check given raltion is function or not.

at x = 2

If $f\left(x\right)=x^2$

If f (2) = 4

and if f (x) = 3x

$f\left(2\right)=3\times 2=6$

So, relation g (x) has two values at

x = 2 i.e., 4 and 6

Hence the given relation is not a function.

Therefore, it is proved that f is a function and g is not a function.