Understanding Function Continuity on Different Sets

The problem asks about the continuity of the function \({\rm{f}}\left( {\rm{x}} \right) = \frac{{{\rm{x}} + \left| {\rm{x}} \right|}}{{\rm{x}}}\) on different given sets. The domain of the function is specified as \({\rm{A}} = {\rm{R}}\backslash \{0\}\), which means all real numbers except zero.

Analyzing the Function f(x) based on Absolute Value

The function \({\rm{f}}\left( {\rm{x}} \right)\) involves the absolute value of \({\rm{x}}\), denoted as \({\left| {\rm{x}} \right|}\). The definition of the absolute value depends on whether \({\rm{x}}\) is positive or negative.

• If \({\rm{x}} > 0\), then \({\left| {\rm{x}} \right|} = {\rm{x}}\).
• If \({\rm{x}} < 0\), then \({\left| {\rm{x}} \right|} = -{\rm{x}}\).

Since the domain of \({\rm{f}}\left( {\rm{x}} \right)\) is \({\rm{A}} = {\rm{R}}\backslash \{0\}\), we only need to consider the cases where \({\rm{x}} > 0\) and \({\rm{x}} < 0\).

Defining f(x) Piece-wise on its Domain

Let's write the function \({\rm{f}}\left( {\rm{x}} \right)\) based on the cases for \({\rm{x}}\) within the domain \({\rm{A}}\).

• Case 1: \({\rm{x}} > 0\)
  In this case, \({\left| {\rm{x}} \right|} = {\rm{x}}\). So, the function becomes: $${\rm{f}}\left( {\rm{x}} \right) = \frac{{{\rm{x}} + {\rm{x}}}}{{\rm{x}}} = \frac{{2{\rm{x}}}}{{\rm{x}}} = 2$$ Thus, for all \({\rm{x}} > 0\) in the domain \({\rm{A}}\), \({\rm{f}}\left( {\rm{x}} \right) = 2\).
• Case 2: \({\rm{x}} < 0\)
  In this case, \({\left| {\rm{x}} \right|} = -{\rm{x}}\). So, the function becomes: $${\rm{f}}\left( {\rm{x}} \right) = \frac{{{\rm{x}} + \left( { - {\rm{x}}} \right)}}{{{\rm{x}}}} = \frac{{{\rm{x}} - {\rm{x}}}}{{\rm{x}}} = \frac{0}{{{\rm{x}}}} = 0$$ Thus, for all \({\rm{x}} < 0\) in the domain \({\rm{A}}\), \({\rm{f}}\left( {\rm{x}} \right) = 0\).

So, the function \({\rm{f}}\left( {\rm{x}} \right)\) on its domain \({\rm{A}} = {\rm{R}}\backslash \{0\}\) can be written as:

\[{\rm{f}}\left( {\rm{x}} \right) = \begin{cases} 2 & \text{if } {\rm{x}} > 0 \\ 0 & \text{if } {\rm{x}} < 0 \end{cases}\]

Checking Continuity on the Given Sets

A function is continuous on a set if it is continuous at every point in that set. For a function to be continuous at a point, it must be defined at that point.

• Set 1: \({\rm{A}} = {\rm{R}}\backslash \{0\}\)
  The domain \({\rm{A}}\) consists of two disjoint open intervals: \((-\infty, 0)\) and \((0, \infty)\). On the interval \((-\infty, 0)\), \({\rm{f}}\left( {\rm{x}} \right) = 0\). This is a constant function. Constant functions are continuous everywhere on their domain. So, \({\rm{f}}\left( {\rm{x}} \right)\) is continuous for all \({\rm{x}} < 0\). On the interval \((0, \infty)\), \({\rm{f}}\left( {\rm{x}} \right) = 2\). This is also a constant function and is continuous everywhere on its domain. So, \({\rm{f}}\left( {\rm{x}} \right)\) is continuous for all \({\rm{x}} > 0\). Since the function is continuous on each of the open intervals that make up the domain \({\rm{A}}\), the function \({\rm{f}}\left( {\rm{x}} \right)\) is continuous on the entire set \({\rm{A}}\). Note that we do not need to check continuity at \({\rm{x}} = 0\) because \(0\) is not in the domain \({\rm{A}}\).
• Set 2: \({\rm{B}} = \{{\rm{x}} \in {\rm{R}} : {\rm{x}} \ge 0\}\)
  This set includes the point \({\rm{x}} = 0\). However, the function \({\rm{f}}\left( {\rm{x}} \right)\) is not defined at \({\rm{x}} = 0\) as per its domain \({\rm{A}} = {\rm{R}}\backslash \{0\}\). A function cannot be continuous at a point where it is not defined. Therefore, \({\rm{f}}\left( {\rm{x}} \right)\) is not continuous on the set \({\rm{B}}\).
• Set 3: \({\rm{C}} = \{{\rm{x}} \in {\rm{R}} : {\rm{x}} \le 0\}\)
  This set also includes the point \({\rm{x}} = 0\). As explained for set B, the function \({\rm{f}}\left( {\rm{x}} \right)\) is not defined at \({\rm{x}} = 0\). Consequently, \({\rm{f}}\left( {\rm{x}} \right)\) cannot be continuous on the set \({\rm{C}}\).
• Set 4: \({\rm{D}} = {\rm{R}}\)
  The set \({\rm{D}}\) is the set of all real numbers, which includes \({\rm{x}} = 0\). Since the function \({\rm{f}}\left( {\rm{x}} \right)\) is not defined at \({\rm{x}} = 0\), it cannot be continuous on the set \({\rm{D}}\). The function has a discontinuity at \({\rm{x}}=0\).

Based on the analysis, the function \({\rm{f}}\left( {\rm{x}} \right)\) is continuous on its domain \({\rm{A}} = {\rm{R}}\backslash \{0\}\).

Conclusion on Function Continuity

The function \({\rm{f}}\left( {\rm{x}} \right) = \frac{{{\rm{x}} + \left| {\rm{x}} \right|}}{{\rm{x}}}\) with domain \({\rm{A}} = {\rm{R}}\backslash \{0\}\) is defined as \({\rm{f}}\left( {\rm{x}} \right) = 0\) for \({\rm{x}} < 0\) and \({\rm{f}}\left( {\rm{x}} \right) = 2\) for \({\rm{x}} > 0\). This function is continuous on the set \({\rm{A}}\).

Revision Table: Key Concepts for Continuity

Additional Information: Types of Discontinuities

Although the question focuses on where the function is continuous within its domain, understanding discontinuities is related. If the domain were extended to include 0, \({\rm{f}}\left( {\rm{x}} \right)\) would have a discontinuity at \({\rm{x}}=0\). This type of discontinuity is a jump discontinuity because the left-hand limit (as \({\rm{x}} \to 0^-\)) is 0, and the right-hand limit (as \({\rm{x}} \to 0^+\)) is 2. Since these limits exist but are not equal, there is a jump at \({\rm{x}}=0\).

However, in the context of the given problem, the point \({\rm{x}}=0\) is excluded from the domain \({\rm{A}}\). Thus, the function is continuous on its entire defined domain \({\rm{A}}\).