For the next two (2) items that follow: A function f(x) is defined as follows: \({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {{\rm{x}} + {\rm{\pi \;for\;x}} \in \left[ { - {\rm{\pi }},{\rm{\;}}0} \right)}\\ {{\rm{\pi }}\cos {\rm{x\;for\;x}} \in \left[ {0,\frac{{\rm{\pi }}}{2}} \right]}\\ {{{\left( {{\rm{x}} - \frac{{\rm{\pi }}}{2}} \right)}^2}{\rm{\;for\;x}} \in \left( {\frac{{\rm{\pi }}}{2},{\rm{\;\pi }}} \right]} \end{array}} \right.\)
Consider the following statements: 1. The function f(x) is continuous at x = 0 2. The function f(x) is continuous at \({\rm{x}} = \frac{{\rm{\pi }}}{2}\)
Both 1 and 2
Understanding Continuity of a Piecewise Function
The question asks us to examine the continuity of a given piecewise function \(f(x)\) at two specific points: \(x = 0\) and \(x = \frac{\pi}{2}\). A function is considered continuous at a point 'a' if the function is defined at 'a', the limit of the function as x approaches 'a' exists, and the limit is equal to the function's value at 'a'. Mathematically, this means:
- \(f(a)\) is defined.
- \({\lim_{x \to a} f(x)}\) exists (i.e., \({\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)}\)).
- \({\lim_{x \to a} f(x) = f(a)}\).
Let's analyze the function definition:
\[{\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {{\rm{x}} + {\rm{\pi \;for\;x}} \in \left[ { - {\rm{\pi }},{\rm{\;}}0} \right)}\\ {{\rm{\pi }}\cos {\rm{x\;for\;x}} \in \left[ {0,\frac{{\rm{\pi }}}{2}} \right]}\\ {{{\left( {{\rm{x}} - \frac{{\rm{\pi }}}{2}} \right)}^2}{\rm{\;for\;x}} \in \left( {\frac{{\rm{\pi }}}{2},{\rm{\;\pi }}} \right]} \end{array}} \right.\]We need to check the continuity at \(x = 0\) and \(x = \frac{\pi}{2}\).
Analyzing Continuity at x = 0
For the function \(f(x)\) to be continuous at \(x = 0\), we must check the three conditions mentioned above.
- Value of the function at x = 0:
At \(x = 0\), the function definition is \(f(x) = \pi \cos x\) (from the second interval \(x \in [0, \frac{\pi}{2}]\)).
So, \(f(0) = \pi \cos(0) = \pi \times 1 = \pi\).
\(f(0)\) is defined and equals \(\pi\). - Left-Hand Limit (LHL) at x = 0:
As \(x\) approaches 0 from the left (\(x < 0\)), the function definition is \(f(x) = x + \pi\) (from the first interval \(x \in [-\pi, 0)\)).
\({\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x + \pi) = 0 + \pi = \pi}\). - Right-Hand Limit (RHL) at x = 0:
As \(x\) approaches 0 from the right (\(x > 0\)), the function definition is \(f(x) = \pi \cos x\) (from the second interval \(x \in [0, \frac{\pi}{2}]\)).
\({\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (\pi \cos x) = \pi \cos(0) = \pi \times 1 = \pi}\).
Comparing the values: LHL = \(\pi\), RHL = \(\pi\), and \(f(0) = \pi\). Since LHL = RHL = \(f(0)\), the function \(f(x)\) is continuous at \(x = 0\). Therefore, Statement 1 is correct.
Analyzing Continuity at \(x = \frac{\pi}{2}\)
For the function \(f(x)\) to be continuous at \(x = \frac{\pi}{2}\), we again check the continuity conditions.
- Value of the function at \(x = \frac{\pi}{2}\):
At \(x = \frac{\pi}{2}\), the function definition is \(f(x) = \pi \cos x\) (from the second interval \(x \in [0, \frac{\pi}{2}]\)).
So, \(f(\frac{\pi}{2}) = \pi \cos(\frac{\pi}{2}) = \pi \times 0 = 0\).
\(f(\frac{\pi}{2})\) is defined and equals 0. - Left-Hand Limit (LHL) at \(x = \frac{\pi}{2}\):
As \(x\) approaches \(\frac{\pi}{2}\) from the left (\(x < \frac{\pi}{2}\)), the function definition is \(f(x) = \pi \cos x\) (from the second interval \(x \in [0, \frac{\pi}{2}]\)).
\({\lim_{x \to (\frac{\pi}{2})^-} f(x) = \lim_{x \to (\frac{\pi}{2})^-} (\pi \cos x) = \pi \cos(\frac{\pi}{2}) = \pi \times 0 = 0}\). - Right-Hand Limit (RHL) at \(x = \frac{\pi}{2}\):
As \(x\) approaches \(\frac{\pi}{2}\) from the right (\(x > \frac{\pi}{2}\)), the function definition is \(f(x) = (x - \frac{\pi}{2})^2\) (from the third interval \(x \in (\frac{\pi}{2}, \pi]\)).
\({\lim_{x \to (\frac{\pi}{2})^+} f(x) = \lim_{x \to (\frac{\pi}{2})^+} (x - \frac{\pi}{2})^2 = (\frac{\pi}{2} - \frac{\pi}{2})^2 = 0^2 = 0}\).
Comparing the values: LHL = 0, RHL = 0, and \(f(\frac{\pi}{2}) = 0\). Since LHL = RHL = \(f(\frac{\pi}{2})\), the function \(f(x)\) is continuous at \(x = \frac{\pi}{2}\). Therefore, Statement 2 is correct.
Conclusion on Statements
Based on our analysis:
- Statement 1: The function f(x) is continuous at x = 0. (Correct)
- Statement 2: The function f(x) is continuous at \(x = \frac{\pi}{2}\). (Correct)
Both statements are correct.
| Point | Function Value | Left Limit | Right Limit | Continuity |
|---|---|---|---|---|
| \(x = 0\) | \(f(0) = \pi\) | \({\lim_{x \to 0^-} f(x) = \pi}\) | \({\lim_{x \to 0^+} f(x) = \pi}\) | Continuous |
| \(x = \frac{\pi}{2}\) | \(f(\frac{\pi}{2}) = 0\) | \({\lim_{x \to (\frac{\pi}{2})^-} f(x) = 0}\) | \({\lim_{x \to (\frac{\pi}{2})^+} f(x) = 0}\) | Continuous |
Revision Table: Key Concepts for Continuity
| Concept | Description |
|---|---|
| Continuity at a Point | A function \(f(x)\) is continuous at \(x=a\) if \({\lim_{x \to a} f(x)}\) exists and equals \(f(a)\). |
| Limit Existence | \({\lim_{x \to a} f(x)}\) exists if and only if the left-hand limit \({\lim_{x \to a^-} f(x)}\) and the right-hand limit \({\lim_{x \to a^+} f(x)}\) both exist and are equal. |
| Piecewise Function | A function defined by multiple sub-functions, each applying to a certain interval of the domain. Continuity needs to be checked at the points where the definition changes. |
Additional Information: Types of Discontinuity
If a function is not continuous at a point, it is said to be discontinuous. There are different types of discontinuity:
- Removable Discontinuity: This occurs if the limit \({\lim_{x \to a} f(x)}\) exists but is not equal to \(f(a)\), or \(f(a)\) is undefined. It can be 'removed' by redefining the function at that point.
- Jump Discontinuity: This occurs if the left-hand limit and the right-hand limit both exist but are not equal. There is a 'jump' in the function's value at that point.
- Infinite Discontinuity: This occurs if the function approaches infinity or negative infinity as x approaches 'a'.
In this question, we found that the function is continuous at the points where the definition changes, so there are no discontinuities at \(x=0\) or \(x=\frac{\pi}{2}\).
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