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

Hint: To solve this question, we will first of all understand the definition of a function. A mapping $h:A\to B$ is called a function if $\forall a\in A\exists $ unique $b\in B$ such that $h\left( a \right)=b$.
To solve this further, we will determine all things of f and g and see whether both or one of them has a value of x having two or more images. If it has 2 or more images for a single value then, it is not a function otherwise it is.

Complete step-by-step answer:
We are given \[f\left( x \right)=\left\{ \begin{align}
  & {{x}^{2}},0\le x\le 3 \\
 & 3x,3\le x\le 10 \\
\end{align} \right.\]
First we will determine if f (x) is a function or not.
A mapping $h:A\to B$ where A and B are sets is called a function if $\forall a\in A\exists $ unique $b\in B$ such that $h\left( a \right)=b$.
A mapping is not a function if for a single pre image, we have two or more images.
Like here $h:A\to B$ is not a function if $h\left( a \right)=c\text{ and }h\left( a \right)=b$.
Now, consider \[f\left( x \right)=\left\{ \begin{align}
  & {{x}^{2}},0\le x\le 3 \\
 & 3x,3\le x\le 10 \\
\end{align} \right.\]
Compute f (0), f (1), f (2), f (3), f (4), f (5), f (6), f (7), f (8), f (9), f (10).
\[\begin{align}
  & f\left( 0 \right)=0 \\
 & f\left( 1 \right)={{1}^{2}}=1 \\
 & f\left( 2 \right)={{2}^{2}}=4 \\
 & f\left( 3 \right)={{3}^{2}}=3\times 3=9 \\
 & f\left( 4 \right)=12 \\
 & f\left( 5 \right)=15 \\
 & f\left( 6 \right)=18 \\
 & f\left( 7 \right)=21 \\
 & f\left( 8 \right)=24 \\
 & f\left( 9 \right)=27 \\
 & f\left( 10 \right)=30 \\
\end{align}\]
Similarly, for all x lying between $0\le x\le 10$ we have f (x) has a unique value.
So, we got that f (x) is a function.
Now, consider \[g\left( x \right)=\left\{ \begin{align}
  & {{x}^{2}},0\le x\le 2 \\
 & 3x,2\le x\le 10 \\
\end{align} \right.\]
Compute g (2) then \[g\left( 2 \right)={{2}^{2}}=4\text{ and }g\left( 2 \right)=3\times 2=6\]
So, for a single element 2 we have two images 4 and 6, therefore a unique element is not obtained.
Therefore, g (x) is not a function.
Hence, in case of f for each number in the domain, there exists a unique number in the co-domain. So f(x) is a function.
But, in g (x) number 2 had two images 4 and 6. So g (x) is not a function.

Note: To avoid confusion, students should understand what is co-domain and what is domain. If a function is given as $h:A\to B$ then all values of $a\in A$ are set of co-domain and all values $b\in B$ such that h (a) = b are in the set of domains.
A key point in this question is that even after taking same functional value that is ${{x}^{2}}\text{ and }3x$ then also by changing range of x we are able to define one mapping as a function and other not a function.