To determine whether the given relations are functions, we need to check if each element in the domain has exactly one image.

For relation f:

The relation f is defined by $f(x) = \begin{cases} x^2, & 0 \leq x \leq 3 \\ 3x, & 3 \leq x \leq 10 \end{cases}$

The domain of f is $[0, 10]$. We need to verify that at the point where both definitions overlap ($x = 3$), the values are equal.

At $x = 3$:

• From first definition: $f(3) = 3^2 = 9$
• From second definition: $f(3) = 3(3) = 9$

Since both definitions give the same value at $x = 3$, every element in the domain $[0, 10]$ has exactly one image. Therefore, f is a function.

For relation g:

The relation g is defined by $g(x) = \begin{cases} x^2, & 0 \leq x \leq 2 \\ 3x, & 2 \leq x \leq 10 \end{cases}$

The domain of g is $[0, 10]$. We need to check the point where both definitions overlap ($x = 2$).

At $x = 2$:

• From first definition: $g(2) = 2^2 = 4$
• From second definition: $g(2) = 3(2) = 6$

Since both definitions give different values at $x = 2$, the element 2 has two images (4 and 6). This violates the definition of a function. Therefore, g is not a function.