The relation $f$ 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.
(N/A) For a relation to be a function,every element in the domain must have a unique image.
For $f(x)$:
At $x = 3$,the first part gives $f(3) = 3^2 = 9$.
The second part gives $f(3) = 3 \times 3 = 9$.
Since both parts yield the same value $9$ at $x = 3$,$f(x)$ is a function.
For $g(x)$:
At $x = 2$,the first part gives $g(2) = 2^2 = 4$.
The second part gives $g(2) = 3 \times 2 = 6$.
Since $x = 2$ has two different images ($4$ and $6$),$g(x)$ is not a function.
For $f(x)$:
At $x = 3$,the first part gives $f(3) = 3^2 = 9$.
The second part gives $f(3) = 3 \times 3 = 9$.
Since both parts yield the same value $9$ at $x = 3$,$f(x)$ is a function.
For $g(x)$:
At $x = 2$,the first part gives $g(2) = 2^2 = 4$.
The second part gives $g(2) = 3 \times 2 = 6$.
Since $x = 2$ has two different images ($4$ and $6$),$g(x)$ is not a function.
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