Step 1: Understanding the Concept:
To check the properties of a piecewise function at a "breaking point" \( x=a \), we evaluate the Left Hand Limit (LHL), the Right Hand Limit (RHL), and the function value \( f(a) \).
If all three are equal, the function is continuous.
If it's continuous, we then check if the left-hand derivative and right-hand derivative exist and are equal to determine differentiability.
In this problem, the right-hand piece is defined by a definite integral whose lower limit is the variable \( x \).
Step 2: Key Formula or Approach:
1. Continuity check: \( \lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) = f(4) \).
2. Fundamental Theorem of Calculus: If \( F(x) = \int_x^a g(t) dt \), then \( F'(x) = -g(x) \).
Step 3: Detailed Explanation:
First, find \( f(4) \) using the second piece (\( x \leq 4 \)):
\[ f(4) = 2(4) + 8 = 8 + 8 = 16 \]
Since this is a linear function for \( x \leq 4 \), the LHL is also 16.
Now, calculate the RHL using the first piece (\( x>4 \)):
\[ \lim_{x \to 4^+} f(x) = \int_4^6 (|t - 2| + 3) dt \]
In the interval of integration \( [4, 6] \), the expression inside the absolute value \( (t - 2) \) is always positive because \( t \geq 4 \).
Thus, \( |t - 2| = t - 2 \).
The integral becomes:
\[ \int_4^6 (t - 2 + 3) dt = \int_4^6 (t + 1) dt \]
\[ = \left[ \frac{t^2}{2} + t \right]_4^6 \]
\[ = \left( \frac{36}{2} + 6 \right) - \left( \frac{16}{2} + 4 \right) \]
\[ = (18 + 6) - (8 + 4) = 24 - 12 = 12 \]
Comparing the results:
- LHL = 16
- RHL = 12
Since \( \text{LHL} \neq \text{RHL} \), the function is discontinuous at \( x = 4 \).
Step 4: Final Answer:
The function is discontinuous at \( x = 4 \).