Question
The function f is defined by $\text{f(x)}=\begin{cases}\text{x}^2,& 0\leq\text{x}\leq3\\3\text{x},&3\leq\text{x}\leq10\end{cases}$
The relation g is defined by $\text{g(x)}=\begin{cases}\text{x}^2,& 0\leq\text{x}\leq2\\3\text{x},&2\leq\text{x}\leq10\end{cases}$
Show that f is a function and g is not a function.

Answer

We have,
$\text{f(x)}=\begin{cases}\text{x}^2,& 0\leq\text{x}\leq3\\3\text{x},&3\leq\text{x}\leq10\end{cases}$
and $\text{g(x)}=\begin{cases}\text{x}^2,& 0\leq\text{x}\leq2\\3\text{x},&2\leq\text{x}\leq10\end{cases}$
Now, $f(3) = (3)^2 = 9$ and $f(3) = 3 \times 3 = 9$
and $g(2) = (2)^2 = 4$ and $g(2) = 3 \times 2 = 6$
We observe $f(x)$ takes unique value at each point in its domain $[0, 10]$. However $g(x)$ does not takes unique value at each in its domain $[0, 10]$
Hence, $g(x)$ is not a function.

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