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A function is considered continuous if there are no abrupt changes or jumps in its graph. This means you can draw the graph of the function without lifting your pen off the paper. Mathematically, a function is continuous at a point if

• The limit of the function as it approaches the point from the left equals the limit as it approaches from the right.
• Both of these limits are equal to the function's actual value at that point.

This rule ensures that not only does the function behave nicely at most points, but also at any point you choose to test. In the case of our exercise with the piecewise function, we checked each piece separately. We found that the segments of the function defined by linear and polynomial equations are continuously smooth across their domains, as expected from linear and polynomial characteristics. The only trouble arises at points where different pieces meet, which we'll delve into next.

Piecewise functions are defined by different expressions based on different intervals of the domain. In the exercise we had two parts:

• For values of \(x < 0\), the function is given by \(-x + 1\).
• For \(x \geq 0\), the function is defined as \(x^2\).

This type of function is especially useful when dealing with situations where the rule changes at certain points or intervals. Evaluating the continuity of a piecewise function means verifying each part separately and assessing how they connect, as in steps 2 and 3 from the solution. Because each part separately is continuous in their respective domains, discontinuity usually arises at the borders where the expressions meet. Understanding the behavior at these junctions is critical in piecewise continuity analysis.

Discontinuity points occur when a function is not continuous at a certain point in its domain. For piecewise functions, these points usually occur where the expressions switch over.
In our exercise, we examined the function at \(x = 0\), where it shifts from \(-x+1\) to \(x^2\). To confirm continuity, both pieces should agree at this transition point. However, we found:

• The limit of \(-x+1\) as \(x\) approaches 0 from the left is 1.
• The right-hand limit of \(x^2\) as \(x\) approaches 0 is 0.
• The function value at \(x = 0\) is also 0.

Since the limits from both sides did not equate, a discontinuity is present at \(x = 0\). This example illustrates how just a single mismatch at the border can create a point of discontinuity in an otherwise smooth function.