91Ó°ÊÓ

Piecewise functions are special types of functions where different rules apply to different parts of the domain. In simple terms, it's like having a function that behaves one way in a certain range of values and another way in a different range.
For example, consider the function provided in the exercise:

• When the value of \(x\) is less than or equal to 3, the function \(f(x, y)\) behaves as \(c + y\).
• When \(x\) is greater than 3, the function behaves as \(5 - y\).

The important thing about piecewise functions is to know at which points the behavior changes. This point or value is called the boundary, and we need to pay extra attention to this boundary to make sure the function does not "jump" from one piece to the next without a smooth transition.

To guarantee that a function is continuous, its values must align perfectly at each point, particularly across boundaries. For a piecewise function, this means ensuring the different pieces match up seamlessly at the boundary.
Continuity at a boundary, like \(x = 3\) in this exercise, is crucial. We want the function's value approaching from both sides of this boundary to be equal. Mathematically, this is expressed by making the output from both function definitions equal at the boundary line.
For example, the function \(f(x, y) = \begin{cases} c+y, & \text{if}\ x \leq 3 \ 5-y, & \text{if}\ x > 3 \end{cases}\) requires that \(c + y = 5 - y\) when \(x = 3\). Thus, continuity is all about equality of function outputs at critical points.

Boundaries in piecewise functions, like the point \(x = 3\) in our problem, are pivotal because they are where the rules change, potentially causing discontinuities.
To tackle these function boundaries, first identify the expressions that apply at the boundary. Ensure that these expressions give the same result at the boundary for the function to remain continuous.
In our example, evaluating both expressions at \(x = 3\) leads to our key condition: the values \(c + y\) (from \(x \leq 3\)) and \(5 - y\) (from \(x > 3\)) must be equal to ensure continuity. Solving \(c + y = 5 - y\) lets us explore if a constant \(c\) could achieve this. Unfortunately, as worked through in the exercise, no single value of \(c\) can uniformly satisfy this across all \(y\) values, thereby highlighting a challenge at the boundary that prevents full continuity.