Question

The points of discontinuity of the function \[ f(x) = \begin{cases} \frac{1}{x - 1} & \text{if } 0 \leq x \leq 2 \\ \frac{x + 5}{x + 3} & \text{if } 2<x \leq 4 \end{cases} \] in its domain are

• A. \( x = 2 \) only
• B. \( x = 1, x = 2 \)
• C. \( x = 4 \) only
• D. \( x = 0, x = 2 \)

Hint:

To identify points of discontinuity, check where the denominators of rational functions are zero and where the function changes pieces.

Answer 1

The Correct Option is B

Step 1: Checking for discontinuities in the first piece of the function.
The function \( f(x) = \frac{1}{x - 1} \) is discontinuous at \( x = 1 \), since the denominator becomes zero at this point.

Step 2: Checking for discontinuities in the second piece of the function.
The function \( f(x) = \frac{x + 5}{x + 3} \) is discontinuous at \( x = -3 \), but since \( -3 \) is not in the domain \( 2<x \leq 4 \), we only need to check \( x = 2 \), where the piecewise function transitions. The function is continuous at \( x = 2 \) as the limit from both sides exists and is finite.

Step 3: Conclusion.
Thus, the points of discontinuity in the domain are \( x = 1 \) and \( x = 2 \), making option (B) the correct answer.