Q.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.

Concept: A relation is a function if every input in its domain maps to exactly one output. Piecewise definitions fail this test when an input receives conflicting values from overlapping pieces.

For $f$:

The pieces meet at $x = 3$. Check consistency:

• First piece: $f(3) = 3^2 = 9$
• Second piece: $f(3) = 3 \cdot 3 = 9$

Both pieces assign the same value at the boundary, so every $x \in [0, 10]$ has a unique image. $f$ is a function.

For $g$:

The pieces meet at $x = 2$. Check consistency:

• First piece: $g(2) = 2^2 = 4$
• Second piece: $g(2) = 3 \cdot 2 = 6$

The input $x = 2$ receives two different outputs ($4$ and $6$), violating the definition of a function. $g$ is not a function.

The heart of this problem lies in understanding what makes a relation a function. A function must satisfy one non-negotiable rule: every input in the domain must map to exactly one output. When we define a relation piecewise—using different formulas on different intervals—we need to be especially careful at the boundary points where the pieces meet. If the two pieces disagree at a shared boundary, the relation fails to be a function.

Let me show you why $f$ passes this test while $g$ fails.

Checking $f(x)$

The relation $f$ is defined by two pieces that meet at $x = 3$. Both intervals $[0, 3]$ and $[3, 10]$ include this point, so we must verify that both formulas give the same output.

1. Using the first piece (valid for $0 \le x \le 3$):

2. Using the second piece (valid for $3 \le x \le 10$):

Both formulas agree at $x = 3$. The point $(3, 9)$ appears in both pieces, but since the $y$-value is identical, there's no conflict. Every input from $0$ to $10$ maps to exactly one output.

3. For all other points, there's no overlap issue:
   • When $0 \le x < 3$, only the first formula applies
   • When $3 < x \le 10$, only the second formula applies

Therefore, $f$ satisfies the definition of a function.

Checking $g(x)$

The relation $g$ has its pieces meeting at $x = 2$. Let's apply the same test.

1. Using the first piece (valid for $0 \le x \le 2$):

2. Using the second piece (valid for $2 \le x \le 10$):

Here's the problem: the input $x = 2$ produces two different outputs, $4$ and $6$. According to the definition, $g$ would have to map $2$ to both values simultaneously. This violates the fundamental requirement that each input correspond to exactly one output.

3. The contradiction is fatal: We cannot say "$g(2) = 4$" and "$g(2) = 6$" at the same time. The relation $g$ is therefore not a function.

Visual intuition

If you were to graph these relations, $f$ would be a smooth curve (parabola transitioning to a line) with no break at $x = 3$—the point $(3, 9)$ lies on both pieces. But $g$ would have a "split" at $x = 2$: the parabola reaches $(2, 4)$ while the line starts at $(2, 6)$. A vertical line at $x = 2$ would intersect the graph at two points, the classic signature of "not a function."