Finding Parameter 'a' for Piecewise Function Continuity

We are given a piecewise function \({\rm{f}}\left( {\rm{x}} \right)\) defined as:

\[ {\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {{\rm{ax}} - 2{\rm{\;\;for\;}} - 2 < {\rm{x}} < - 1}\\ { - 1{\rm{\;\;for\;\;}} - 1 \le {\rm{x}} \le 1}\\ {{\rm{a}} + 2{{\left( {{\rm{x}} - 1} \right)}^2}{\rm{\;\;for\;\;\;}}1 < {\rm{x}} < 2} \end{array}} \right. \]

For a function to be continuous at a point \({\rm{c}}\), the following conditions must be met:

• The function must be defined at \({\rm{x}} = {\rm{c}}\), i.e., \({\rm{f}}({\rm{c}})\) exists.
• The limit of the function as \({\rm{x}}\) approaches \({\rm{c}}\) must exist, i.e., \(\mathop {\lim }\limits_{x \to c} {\rm{f}}(x)\) exists. This means the left-hand limit and the right-hand limit must be equal: \(\mathop {\lim }\limits_{x \to c^-} {\rm{f}}(x) = \mathop {\lim }\limits_{x \to c^+} {\rm{f}}(x)\).
• The limit of the function as \({\rm{x}}\) approaches \({\rm{c}}\) must be equal to the function value at \({\rm{c}}\), i.e., \(\mathop {\lim }\limits_{x \to c} {\rm{f}}(x) = {\rm{f}}({\rm{c}})\).

We need to find the value of 'a' for which \({\rm{f}}({\rm{x}})\) is continuous at \({\rm{x}} = -1\) and \({\rm{x}} = 1\).

Analyzing Continuity at x = -1

To check for continuity at \({\rm{x}} = -1\), we examine the function value and the limits around this point.

Step 1: Evaluate the function at x = -1

At \({\rm{x}} = -1\), the function is defined by the second piece \({\rm{f}}({\rm{x}}) = -1\) for \(-1 \le {\rm{x}} \le 1\).

So, \({\rm{f}}(-1) = -1\).

Step 2: Evaluate the left-hand limit at x = -1

The left-hand limit as \({\rm{x}}\) approaches -1 comes from the first piece, where \({\rm{x}} < -1\):

\[ \mathop {\lim }\limits_{x \to -1^-} {\rm{f}}(x) = \mathop {\lim }\limits_{x \to -1^-} ({\rm{ax}} - 2) \]

Substituting \({\rm{x}} = -1\) into the expression:

\[ {\rm{a}}(-1) - 2 = -{\rm{a}} - 2 \]

Step 3: Evaluate the right-hand limit at x = -1

The right-hand limit as \({\rm{x}}\) approaches -1 comes from the second piece, where \({\rm{x}} > -1\):

\[ \mathop {\lim }\limits_{x \to -1^+} {\rm{f}}(x) = \mathop {\lim }\limits_{x \to -1^+} (-1) = -1 \]

Step 4: Apply the continuity condition at x = -1

For continuity at \({\rm{x}} = -1\), the left-hand limit, right-hand limit, and function value must be equal:

\[ \mathop {\lim }\limits_{x \to -1^-} {\rm{f}}(x) = \mathop {\lim }\limits_{x \to -1^+} {\rm{f}}(x) = {\rm{f}}(-1) \]

\[ -{\rm{a}} - 2 = -1 = -1 \]

From this equality, we get the equation:

\[ -{\rm{a}} - 2 = -1 \]

Solving for 'a':

\[ -{\rm{a}} = -1 + 2 \]

\[ -{\rm{a}} = 1 \]

\[ {\rm{a}} = -1 \]

So, for the function to be continuous at \({\rm{x}} = -1\), the value of 'a' must be -1.

Analyzing Continuity at x = 1

Next, we check the conditions for continuity at \({\rm{x}} = 1\).

Step 1: Evaluate the function at x = 1

At \({\rm{x}} = 1\), the function is defined by the second piece \({\rm{f}}({\rm{x}}) = -1\) for \(-1 \le {\rm{x}} \le 1\).

So, \({\rm{f}}(1) = -1\).

Step 2: Evaluate the left-hand limit at x = 1

The left-hand limit as \({\rm{x}}\) approaches 1 comes from the second piece, where \({\rm{x}} < 1\):

\[ \mathop {\lim }\limits_{x \to 1^-} {\rm{f}}(x) = \mathop {\lim }\limits_{x \to 1^-} (-1) = -1 \]

Step 3: Evaluate the right-hand limit at x = 1

The right-hand limit as \({\rm{x}}\) approaches 1 comes from the third piece, where \({\rm{x}} > 1\):

\[ \mathop {\lim }\limits_{x \to 1^+} {\rm{f}}(x) = \mathop {\lim }\limits_{x \to 1^+} ({\rm{a}} + 2{{\left( {{\rm{x}} - 1} \right)}^2}) \]

Substituting \({\rm{x}} = 1\) into the expression:

\[ {\rm{a}} + 2{{\left( {1 - 1} \right)}^2} = {\rm{a}} + 2(0)^2 = {\rm{a}} + 0 = {\rm{a}} \]

Step 4: Apply the continuity condition at x = 1

For continuity at \({\rm{x}} = 1\), the left-hand limit, right-hand limit, and function value must be equal:

\[ \mathop {\lim }\limits_{x \to 1^-} {\rm{f}}(x) = \mathop {\lim }\limits_{x \to 1^+} {\rm{f}}(x) = {\rm{f}}(1) \]

\[ -1 = {\rm{a}} = -1 \]

From this equality, we get the equation:

\[ {\rm{a}} = -1 \]

So, for the function to be continuous at \({\rm{x}} = 1\), the value of 'a' must be -1.

Conclusion for Continuity

Both continuity conditions, at \({\rm{x}} = -1\) and \({\rm{x}} = 1\), require the value of 'a' to be -1. Therefore, for \({\rm{f}}({\rm{x}})\) to be continuous at both points, the value of 'a' must be -1.

The value of 'a' that satisfies both conditions is -1.

Revision Table: Key Continuity Concepts

Additional Information: Piecewise Functions and Limits

A piecewise function is a function defined by multiple sub-functions, each applying to a different interval of the domain. To analyze the continuity of a piecewise function, we need to pay special attention to the points where the definition of the function changes. These points are called junction points or breakpoints.

For a function to be continuous at a breakpoint \({\rm{c}}\), the pieces of the function must "meet" seamlessly at \({\rm{c}}\). This requires the limit of the function as \({\rm{x}}\) approaches \({\rm{c}}\) from the left (using the function definition for \({\rm{x}} < {\rm{c}}\)), the limit as \({\rm{x}}\) approaches \({\rm{c}}\) from the right (using the function definition for \({\rm{x}} > {\rm{c}}\)), and the value of the function exactly at \({\rm{x}} = {\rm{c}}\) to all be equal.

In this problem, the breakpoints are \({\rm{x}} = -1\) and \({\rm{x}} = 1\). We evaluated the left-hand limit, right-hand limit, and function value at each of these points and set them equal to find the required value of the parameter 'a'. If the question asked for continuity over the entire domain, we would also need to check if each individual piece is continuous within its defined interval, but polynomials (like \({\rm{ax}} - 2\) and \({\rm{a}} + 2{{\left( {{\rm{x}} - 1} \right)}^2}\)) and constant functions (like -1) are continuous everywhere within their domains, so continuity only needs to be checked at the breakpoints.