Understanding Function Continuity

The question asks us to determine the points of continuity and discontinuity for the given piecewise function \({\rm{f}}\left( {\rm{x}} \right)\). A function is continuous at a point \({\rm{c}}\) if the following three conditions are met:

1. The function \({\rm{f}}\left( {\rm{x}} \right)\) is defined at \({\rm{c}}\) (i.e., \({\rm{f}}\left( {\rm{c}} \right)\) exists).
2. The limit of \({\rm{f}}\left( {\rm{x}}\right)\) as \({\rm{x}}\) approaches \({\rm{c}}\) exists (i.e., \(\lim_{x \to c} {\rm{f}}\left( {\rm{x}} \right)\) exists). This means the left-hand limit and the right-hand limit are equal: \(\lim_{x \to c^-} {\rm{f}}\left( {\rm{x}} \right) = \lim_{x \to c^+} {\rm{f}}\left( {\rm{x}} \right)\).
3. The limit of \({\rm{f}}\left( {\rm{x}} \right)\) as \({\rm{x}}\) approaches \({\rm{c}}\) is equal to the function value at \({\rm{c}}\): \(\lim_{x \to c} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{c}} \right)\).

For a piecewise function, we need to check for continuity at the points where the definition of the function changes. In this case, the points are \({\rm{x}} = -2\) and \({\rm{x}} = 0\). We also need to consider the continuity within the open intervals defined by the pieces, but polynomial functions are continuous everywhere, so the function is continuous within the open intervals \((-3, -2)\), \((-2, 0)\), and \((0, 1)\).

Checking Continuity at x = -2

At \({\rm{x}} = -2\), the function changes from \(2{\rm{x}} + 1\) to \({\rm{x}} - 1\).

• Left-hand limit (LHL) as \({\rm{x}} \to -2^-\): Using the definition for \(-3 < {\rm{x}} < -2\), which is \({\rm{f}}\left( {\rm{x}} \right) = 2{\rm{x}} + 1\).

  \(\lim_{x \to -2^-} {\rm{f}}\left( {\rm{x}} \right) = \lim_{x \to -2^-} \left( {2{\rm{x}} + 1} \right) = 2\left( { - 2} \right) + 1 = -4 + 1 = -3\)

• Right-hand limit (RHL) as \({\rm{x}} \to -2^+\): Using the definition for \(-2 \le {\rm{x}} < 0\), which is \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}} - 1\).

  \(\lim_{x \to -2^+} {\rm{f}}\left( {\rm{x}} \right) = \lim_{x \to -2^+} \left( {{\rm{x}} - 1} \right) = \left( { - 2} \right) - 1 = -3\)

• Function value at \({\rm{x}} = -2\): Using the definition for \(-2 \le {\rm{x}} < 0\), which is \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}} - 1\).

  \({\rm{f}}\left( { - 2} \right) = \left( { - 2} \right) - 1 = -3\)

Since \(\lim_{x \to -2^-} {\rm{f}}\left( {\rm{x}} \right) = \lim_{x \to -2^+} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( { - 2} \right) = -3\), the function \({\rm{f}}\left( {\rm{x}} \right)\) is continuous at \({\rm{x}} = -2\).

Checking Continuity at x = 0

At \({\rm{x}} = 0\), the function changes from \({\rm{x}} - 1\) to \({\rm{x}} + 2\).

• Left-hand limit (LHL) as \({\rm{x}} \to 0^-\): Using the definition for \(-2 \le {\rm{x}} < 0\), which is \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}} - 1\).

  \(\lim_{x \to 0^-} {\rm{f}}\left( {\rm{x}} \right) = \lim_{x \to 0^-} \left( {{\rm{x}} - 1} \right) = 0 - 1 = -1\)

• Right-hand limit (RHL) as \({\rm{x}} \to 0^+\): Using the definition for \(0 \le {\rm{x}} < 1\), which is \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}} + 2\).

  \(\lim_{x \to 0^+} {\rm{f}}\left( {\rm{x}} \right) = \lim_{x \to 0^+} \left( {{\rm{x}} + 2} \right) = 0 + 2 = 2\)

• Function value at \({\rm{x}} = 0\): Using the definition for \(0 \le {\rm{x}} < 1\), which is \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}} + 2\).

  \({\rm{f}}\left( 0 \right) = 0 + 2 = 2\)

Since \(\lim_{x \to 0^-} {\rm{f}}\left( {\rm{x}} \right) = -1\) and \(\lim_{x \to 0^+} {\rm{f}}\left( {\rm{x}} \right) = 2\), the left-hand limit is not equal to the right-hand limit (\(-1 \neq 2\)). Therefore, the limit as \({\rm{x}} \to 0\) does not exist. This means the function \({\rm{f}}\left( {\rm{x}} \right)\) is discontinuous at \({\rm{x}} = 0\).

Summary of Continuity Analysis

Based on the analysis, the function \({\rm{f}}\left( {\rm{x}} \right)\) is continuous everywhere in its domain except at the point \({\rm{x}} = 0\).

Matching with Given Statements

• Statement 1: It is discontinuous at x = -2 but continuous at every other point. This is incorrect because the function is continuous at \({\rm{x}} = -2\).
• Statement 2: It is continuous only in the interval (-3, -2). This is incorrect because the function is continuous at \({\rm{x}} = -2\) and also in the intervals \((-2, 0)\) and \((0, 1)\), except at \({\rm{x}} = 0\).
• Statement 3: It is discontinuous at x = 0 but continuous at every other point. This matches our finding that the function is discontinuous only at \({\rm{x}} = 0\) within its domain.
• Statement 4: It is discontinuous at every point. This is incorrect as the function is continuous within the defined intervals and at \({\rm{x}} = -2\).

Therefore, the correct statement is that the function is discontinuous at \({\rm{x}} = 0\) but continuous at every other point in its domain.

Revision Table: Function Continuity Basics

Additional Information: Types of Discontinuity

Understanding different types of discontinuity can provide deeper insight into the behavior of functions.

• Removable Discontinuity: Occurs when the limit of the function exists at a point, but either the function is not defined at that point or the function's value at the point does not equal the limit. This often looks like a "hole" in the graph.
• Jump Discontinuity: Occurs when the left-hand limit and the right-hand limit at a point both exist but are not equal. This results in a "jump" in the graph at that point, as seen in piecewise functions like the one in this question at \({\rm{x}} = 0\).
• Infinite Discontinuity: Occurs when the function value or the limit approaches infinity at a point. This is typically associated with vertical asymptotes.

In this problem, the discontinuity at \({\rm{x}} = 0\) is a jump discontinuity because the left-hand limit is -1 and the right-hand limit is 2.