Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if the following three conditions are met:
  1. $f(c)$ is defined.
  2. $\lim_{x \to c} f(x)$ exists.
  3. $\lim_{x \to c} f(x) = f(c)$. For a piecewise function, this implies that the left-hand limit, the right-hand limit, and the function value at the point must all be equal. $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$
• Absolute Value Function: The absolute value of a real number $a$, denoted by $|a|$, is its distance from zero.
  • $|a| = a$ if $a \ge 0$
  • $|a| = -a$ if $a < 0$
• Limits using Substitution: For continuous functions, the limit as $x$ approaches a point $c$ can often be found by direct substitution, i.e., $\lim_{x \to c} f(x) = f(c)$.

Step-by-Step Solution

Step 1: Understand the Condition for Continuity The problem states that the function $f(x)$ is continuous at $x=5$. For a function to be continuous at a point, the left-hand limit, the right-hand limit, and the function's value at that point must be equal. $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5)$

Step 2: Evaluate the Left-Hand Limit and the Function Value at x = 5 For $x \le 5$, the function is defined as $f(x) = a|\pi - x| + 1$. So, $f(5) = a|\pi - 5| + 1$. The left-hand limit is: $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (a|\pi - x| + 1)$ Since the function is continuous for $x \le 5$ near $x=5$, we can substitute $x=5$ into the expression: $\lim_{x \to 5^-} f(x) = a|\pi - 5| + 1$ Thus, $f(5) = \lim_{x \to 5^-} f(x) = a|\pi - 5| + 1$.

Step 3: Evaluate the Right-Hand Limit at x = 5 For $x > 5$, the function is defined as $f(x) = b|x - \pi| + 3$. The right-hand limit is: $\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (b|x - \pi| + 3)$ Since the function is continuous for $x > 5$ near $x=5$, we can substitute $x=5$ into the expression: $\lim_{x \to 5^+} f(x) = b|5 - \pi| + 3$

Step 4: Equate the Limits and Function Value According to the continuity condition: $a|\pi - 5| + 1 = b|5 - \pi| + 3$

Step 5: Simplify the Absolute Value Expressions We know that $\pi \approx 3.14159$. Therefore, $\pi - 5$ is a negative number: $\pi - 5 < 0$. Using the definition of absolute value: $|\pi - 5| = -(\pi - 5) = 5 - \pi$. Also, $5 - \pi$ is a positive number. $|5 - \pi| = 5 - \pi$. Notice that $|\pi - 5| = |-(5 - \pi)| = |5 - \pi|$. So, $|\pi - 5| = |5 - \pi| = 5 - \pi$.

Step 6: Substitute the Simplified Absolute Values into the Equation Substitute $|5 - \pi| = 5 - \pi$ and $|\pi - 5| = 5 - \pi$ into the equation from Step 4: $a(5 - \pi) + 1 = b(5 - \pi) + 3$

Step 7: Rearrange the Equation to Solve for a - b Move all terms involving $a$ and $b$ to one side and constant terms to the other: $a(5 - \pi) - b(5 - \pi) = 3 - 1$ Factor out $(5 - \pi)$ from the terms on the left side: $(a - b)(5 - \pi) = 2$

Step 8: Isolate a - b Divide both sides by $(5 - \pi)$ to find the value of $a - b$: $a - b = \frac{2}{5 - \pi}$

Common Mistakes & Tips

• Incorrectly handling absolute values: Always determine the sign of the expression inside the absolute value before removing the absolute value bars. In this case, $\pi - 5$ is negative, so $|\pi - 5| = -(\pi - 5)$. Similarly, $|5 - \pi| = 5 - \pi$.
• Confusing left-hand and right-hand limits: Ensure you use the correct definition of $f(x)$ for the left-hand limit (where $x \le 5$) and the right-hand limit (where $x > 5$).
• Algebraic errors: Be careful when rearranging terms and factoring. Double-check each step of your algebraic manipulation.

Summary

The problem requires us to find the value of $a - b$ given that the piecewise function $f(x)$ is continuous at $x = 5$. We used the definition of continuity, which states that the left-hand limit, the right-hand limit, and the function value at $x=5$ must be equal. By evaluating these three quantities using the given piecewise definitions of $f(x)$ and simplifying the absolute value expressions, we arrived at an equation relating $a$, $b$, and constants. Rearranging this equation allowed us to solve for $a - b$.

The final answer is $\boxed{\frac{2}{\pi - 5 }}$.