91Ó°ÊÓ

Continuity is a fundamental concept in mathematics, ensuring that there are no abrupt leaps or gaps in a function's graph. For a function to be continuous at a specific point, three criteria must be satisfied:

• The function must be defined at that point.
• The left-hand limit and the right-hand limit at that point must exist.
• The left-hand limit, right-hand limit, and the actual function value must all be equal.

In the case of piecewise functions, such as the one in the exercise, checking continuity at the boundary where the function's definition changes is crucial. Here, the boundary is at \( x = 2 \), where we need to ensure that the value of the function from both sides matches. For instance, from the left, the function value is given as a constant 12. From the right, it is determined by the expression \( a^2x - 2a \). When both are equal at \( x = 2 \), the function is continuous there.

Limits are essential for analyzing and defining how functions behave as inputs approach a certain value. A limit essentially tells us the value that a function approaches as the input gets arbitrarily close to a certain point. It's especially useful when dealing with functions that are not explicitly defined at a point.
In the given exercise, we face a piecewise function. The continuity requirement at \( x = 2 \) heavily relies on limits. We evaluate both the left-hand limit and the right-hand limit as \( x \) approaches 2:

• The left-hand limit involves the function taking the constant value, 12, since \( x < 2 \).
• The right-hand limit is assessed using the expression \( a^2x - 2a \), specifically evaluating it at \( x = 2 \). This results in the expression \( 2a^2 - 2a \).

Setting these two limits equal ensures that the function has no sudden jumps at the transition point, thus maintaining continuity.

Quadratic equations appear frequently in algebra and are identified by their characteristic form \( ax^2 + bx + c = 0 \). Solving them requires finding the values of \( x \) for which the equation holds true. This process often involves factoring, completing the square, or using the quadratic formula.
In this exercise, deriving the condition for continuity led us to a quadratic equation in terms of \( a \):
\[ 2a^2 - 2a - 12 = 0. \]
Simplifying this by dividing through by 2, we obtain:
\[ a^2 - a - 6 = 0. \]
The next step is to factor the quadratic. Observe that:

• The two numbers that multiply to \(-6\) (the constant term) and add to \(-1\) (the linear coefficient of \(a\)) are \(-3\) and \(2\).

This yields us the factors \((a - 3)(a + 2) = 0\). When set equal to zero, each leads to solutions for \( a \):
\( a = 3 \) and \( a = -2 \).
These values ensure continuity for the piecewise function at the transition point.