Step 1: Understanding the Concept:
Continuity of a function is a cornerstone concept in calculus that describes the behavior of a function without any interruptions, jumps, or holes.
For a piecewise function to be considered continuous at a specific transition point, say \( x = c \), the three primary requirements must be satisfied.
First, the function must actually exist at that point, meaning \( f(c) \) must be a well-defined real number.
Second, the limit of the function as it approaches the point from both the left and the right must exist and be equal.
This ensures that the two different mathematical rules or "pieces" of the function meet at the same point in the coordinate plane.
Third, the value of this mutual limit must exactly equal the functional value \( f(c) \).
In this specific problem, we are examining the boundary point \( x = 2 \), where the function switches from a quadratic form to a linear form.
Step 2: Key Formula or Approach:
The mathematical definition for continuity at the boundary \( x = 2 \) is expressed by the equality:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \]
We identify the pieces: for the Left-Hand Limit (LHL) and functional value \( f(2) \), we use the expression \( kx^2 \) because of the \( \le \) symbol.
For the Right-Hand Limit (RHL), we use the expression \( 3x - 2 \) because it applies to the region where \( x>2 \).
Step 3: Detailed Explanation:
We begin by evaluating the behavior of the function as \( x \) approaches 2 from the left side (values slightly smaller than 2).
Using the definition \( f(x) = kx^2 \):
\[ \text{LHL} = \lim_{x \to 2^-} (kx^2) = k(2)^2 = 4k \]
Because the first piece of the function includes the equality condition \( x \le 2 \), the functional value at 2 is also:
\[ f(2) = 4k \]
Now, we analyze the behavior of the function as \( x \) approaches 2 from the right side (values slightly larger than 2).
Using the definition \( f(x) = 3x - 2 \):
\[ \text{RHL} = \lim_{x \to 2^+} (3x - 2) = 3(2) - 2 \]
Performing the multiplication and subtraction steps:
\[ \text{RHL} = 6 - 2 = 4 \]
For the function to be continuous, there should be no "step" or "jump" at the boundary.
This means the LHL must match the RHL perfectly.
We set the two expressions equal to each other to form an algebraic equation:
\[ 4k = 4 \]
Dividing both sides of the equation by 4 to isolate the constant \( k \):
\[ k = \frac{4}{4} \]
\[ k = 1 \]
With \( k = 1 \), the parabola \( y = x^2 \) and the line \( y = 3x - 2 \) meet at the coordinate (2, 4), ensuring a continuous path.
Step 4: Final Answer:
The value of the constant \( k \) that ensures continuity at the transition point is 1.