Show that f is a Function and g is not a Function
Question:
The function \(f\) is defined by
$$
f(x)=
\begin{cases}
x^2, & 0\le x\le 3 \\
3x, & 3\le x\le 10
\end{cases}
$$
The relation \(g\) is defined by
$$
g(x)=
\begin{cases}
x^2, & 0\le x\le 2 \\
3x, & 2\le x\le 10
\end{cases}
$$
Show that \(f\) is a function and \(g\) is not a function.
Solution
For \(f(x)\)
At the common point \(x=3\),
$$ x^2=3^2=9 $$
and
$$ 3x=3(3)=9 $$
Both rules give the same value at \(x=3\).
Therefore, every input has exactly one output.
Hence, \(f\) is a function.
For \(g(x)\)
At the common point \(x=2\),
$$ x^2=2^2=4 $$
and
$$ 3x=3(2)=6 $$
The same input \(x=2\) gives two different outputs \(4\) and \(6\).
Therefore, \(g\) is not a function.
$$ \boxed{f \text{ is a function and } g \text{ is not a function}} $$