91Ó°ÊÓ

Piecewise functions are mathematical expressions defined by different rules or formulas for different intervals of the domain. These functions are "pieces" of other functions, each applied to a specific portion of the domain. In the context of the exercise, we have the function:

• For part (a): \( f(x) = 9-x^2 \) for \( x \geq -3 \) and \( f(x) = \frac{k}{x^2} \) for \( x < -3 \).
• For part (b): \( f(x) = 9-x^2 \) for \( x \geq 0 \) and \( f(x) = \frac{k}{x^2} \) for \( x < 0 \).

It is crucial for piecewise functions to be continuous at the points where the definition changes. To ensure continuity, these function segments must meet smoothly at their boundaries. This often involves determining constants—which in this case is the constant \( k \)—to make the transition seamless, maintaining the conceptual integrity of the function.

Limits and continuity are foundational concepts in calculus, defining how functions behave near certain points. A function is continuous at a point if the following three conditions are met:

• The function is defined at the point.
• The limit of the function as it approaches the point from the left and the right exists.
• The left-hand limit, right-hand limit, and the function value at the point are equal.

In part (a) of our exercise, the function must be continuous at \( x = -3 \). We set the right-hand limit and the left-hand limit equal to the function's value at that point to find \( k \).
For part (b), the function should be continuous at \( x = 0 \). However, as we approach 0 from the negative side, the expression \( \frac{k}{x^2} \) becomes undefined unless \( k = 0 \), not allowing for a smooth transition at \( x=0 \) without leading to infinite values. Thus, making the function in part (b) continuous is impossible.

Solving equations in the context of continuity involves ensuring that the values of limits and function outputs align perfectly. In part (a), we are tasked with finding a suitable \( k \) that equalizes the two limits:

• Right-hand limit: As \( x \to -3^+ \), the limit equals \( 0 \) (from the formula \( 9 - x^2 \)).
• Left-hand limit: As \( x \to -3^- \), the expression \( \frac{k}{x^2} \) transforms into \( \frac{k}{9} \).

The equation \( 0 = \frac{k}{9} \) is directly solved for \( k \), meaning \( k = 0 \) balances the limits. For part (b), no solution for \( k \) exists since \( \frac{k}{x^2} \) diverges as \( x \to 0^- \).
Equation solving in calculus often requires careful limit examination and boundary adjustment, testing understanding of both algebraic manipulation and function behavior at crucial points.