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 (i.e., $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$).
  3. $\lim_{x \to c} f(x) = f(c)$.
• Greatest Integer Function $[x]$: This function gives the greatest integer less than or equal to $x$. It has jump discontinuities at integer values.
• Absolute Value Function $|x|$: This function is defined as $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$. It is continuous everywhere but has a sharp corner at $x=0$, making it non-differentiable there.

Step-by-Step Solution

Step 1: Rewrite the function by simplifying the absolute value and greatest integer functions in each interval.

The function is defined piecewise: f(x) = \left\{ {\matrix{ {\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \cr {x + \left| x \right|} & , & {1 \le x < 2} \cr {x + \left[ x \right]} & , & {2 \le x \le 3} \cr } } \right.

We need to analyze each interval separately:

• Interval 1: $-1 \le x < 1$

  • For $-1 \le x < 0$: $|x| = -x$ and $[x] = -1$. So, $f(x) = -x + (-1) = -x - 1$.
  • For $0 \le x < 1$: $|x| = x$ and $[x] = 0$. So, $f(x) = x + 0 = x$. Combining these, for $-1 \le x < 1$, we have: f(x) = \left\{ {\matrix{ {-x - 1} & , & {-1 \le x < 0} \cr {x} & , & {0 \le x < 1} \cr } } \right.

• Interval 2: $1 \le x < 2$

  • For $1 \le x < 2$: $|x| = x$. So, $f(x) = x + x = 2x$.

• Interval 3: $2 \le x \le 3$

  • For $2 \le x < 3$: $[x] = 2$. So, $f(x) = x + 2$.
  • For $x = 3$: $[x] = 3$. So, $f(x) = 3 + 3 = 6$. Combining these, for $2 \le x \le 3$, we have: f(x) = \left\{ {\matrix{ {x + 2} & , & {2 \le x < 3} \cr {6} & , & {x = 3} \cr } } \right.

The rewritten function is: f(x) = \left\{ {\matrix{ {-x - 1} & , & {-1 \le x < 0} \cr {x} & , & {0 \le x < 1} \cr {2x} & , & {1 \le x < 2} \cr {x + 2} & , & {2 \le x < 3} \cr {6} & , & {x = 3} \cr } } \right.

Step 2: Identify potential points of discontinuity.

Discontinuities can occur at:

• The endpoints of the domain: $x = -1$ and $x = 3$.
• The points where the definition of the function changes: $x = 0$, $x = 1$, and $x = 2$.
• Points where the component functions themselves are discontinuous (e.g., the greatest integer function at integers). In our rewritten piecewise function, the component functions ($-x-1, x, 2x, x+2$) are all linear and thus continuous within their respective open intervals. Therefore, we only need to check the points where the definition changes or at the domain endpoints. The points to check are $x = -1, 0, 1, 2, 3$.

Step 3: Check for continuity at $x = -1$.

• $f(-1) = -(-1) - 1 = 1 - 1 = 0$.
• $\lim_{x \to -1^-} f(x)$: This limit is not applicable as the domain starts at $x=-1$. However, we consider the limit from the right.
• $\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (-x - 1) = -(-1) - 1 = 1 - 1 = 0$. Since $f(-1)$ is defined and the limit from the right exists and equals $f(-1)$, the function is continuous at $x = -1$.

Step 4: Check for continuity at $x = 0$.

• $f(0) = 0$ (from the $0 \le x < 1$ case).
• $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x - 1) = -(0) - 1 = -1$.
• $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x) = 0$. Since $\lim_{x \to 0^-} f(x) \ne \lim_{x \to 0^+} f(x)$, the limit $\lim_{x \to 0} f(x)$ does not exist. Therefore, $f(x)$ is discontinuous at $x = 0$.

Step 5: Check for continuity at $x = 1$.

• $f(1) = 2(1) = 2$ (from the $1 \le x < 2$ case).
• $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x) = 1$.
• $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x) = 2(1) = 2$. Since $\lim_{x \to 1^-} f(x) \ne \lim_{x \to 1^+} f(x)$, the limit $\lim_{x \to 1} f(x)$ does not exist. Therefore, $f(x)$ is discontinuous at $x = 1$.

Step 6: Check for continuity at $x = 2$.

• $f(2) = 2 + 2 = 4$ (from the $2 \le x < 3$ case).
• $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x + 2) = 2 + 2 = 4$.
• $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x + 2) = 2 + 2 = 4$. Since $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = 4$, the limit $\lim_{x \to 2} f(x)$ exists and is equal to 4. Also, $f(2) = 4$. Thus, $\lim_{x \to 2} f(x) = f(2)$. Therefore, $f(x)$ is continuous at $x = 2$.

Step 7: Check for continuity at $x = 3$.

• $f(3) = 6$ (as defined for $x=3$).
• $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (x + 2) = 3 + 2 = 5$.
• $\lim_{x \to 3^+} f(x)$: This limit is not applicable as the domain ends at $x=3$. Since $\lim_{x \to 3^-} f(x) = 5$ and $f(3) = 6$, we have $\lim_{x \to 3^-} f(x) \ne f(3)$. Therefore, $f(x)$ is discontinuous at $x = 3$.

Step 8: Summarize the points of discontinuity.

Based on the analysis, the function $f(x)$ is discontinuous at $x = 0$, $x = 1$, and $x = 3$. These are exactly three points.

Common Mistakes & Tips

• Incorrectly simplifying $|x|$ or $[x]$: Ensure you correctly evaluate $|x|$ and $[x]$ for each specific sub-interval. For example, $[x]$ is $-1$ for $-1 \le x < 0$, not $0$.
• Forgetting to check both left and right limits: Continuity requires the limit to exist, which means the left-hand limit must equal the right-hand limit.
• Not checking the function value at the point: Even if the left and right limits are equal, the function must also be defined at that point and equal to the limit for continuity.
• Assuming continuity within intervals: While linear functions are continuous, the piecewise definition means you must check the transition points between these linear segments.

Summary

To determine the points of discontinuity, we first simplified the piecewise function by correctly evaluating the absolute value and greatest integer functions in each interval. We then identified the potential points of discontinuity, which are the points where the function definition changes and the endpoints of the domain. For each of these points ($x = -1, 0, 1, 2, 3$), we checked the three conditions for continuity: the existence of the function value, the existence of the limit, and whether the function value equals the limit. We found that the function is continuous at $x = -1$ and $x = 2$, but discontinuous at $x = 0$, $x = 1$, and $x = 3$. Therefore, there are exactly three points of discontinuity.

The final answer is \boxed{A}.