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 differentiable at x = 0 2. The function f(x) is differentiable at \({\rm{x}} = \frac{{\rm{\pi }}}{2}\) .
Neither 1 nor 2
Understanding Differentiability of Piecewise Functions
This problem asks us to analyze the differentiability of a given piecewise function at two specific points: \({\rm{x}} = 0\) and \({\rm{x}} = \frac{{\rm{\pi}}}{2}\). For a function to be differentiable at a point, two conditions must be met:
- The function must be continuous at that point.
- The left-hand derivative must be equal to the right-hand derivative at that point.
Let's evaluate each statement separately.
The function is defined as:
\({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {{\rm{x}} + {\rm{\pi \;for\;x}} \in \left[ { - {\rm{\pi }},{\rm{\;}}0} \right)}\\ {{\rm{\pi }}\cos {\rm{x\;for\;x}} \in \left[ {0,\frac{{\rm{\pi }}}{2}} \right]}\\ {{{\left( {{\rm{x}} - \frac{{\rm{\pi }}}{2}} \right)}^2}{\rm{\;for\;x}} \in \left( {\frac{{\rm{\pi }}}{2},{\rm{\;\pi }}} \right]} \end{array}} \right.\)
Analyzing Statement 1: Differentiability at x = 0
First, let's check for continuity at \({\rm{x}} = 0\). We need to compare the left-hand limit, the right-hand limit, and the function value at \({\rm{x}} = 0\).
- Left-hand limit at \({\rm{x}} = 0\): As \({\rm{x}}\) approaches 0 from the left (\({\rm{x}} < 0\)), we use the first part of the function definition: \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}} + {\rm{\pi }}\).
\(\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \left( {{\rm{x}} + {\rm{\pi }}} \right) = 0 + {\rm{\pi }} = {\rm{\pi }}\) - Right-hand limit at \({\rm{x}} = 0\): As \({\rm{x}}\) approaches 0 from the right (\({\rm{x}} > 0\)), we use the second part of the function definition: \({\rm{f}}\left( {\rm{x}} \right) = {\rm{\pi }}\cos {\rm{x}}\).
\(\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \left( {{\rm{\pi }}\cos {\rm{x}}} \right) = {\rm{\pi }}\cos \left( 0 \right) = {\rm{\pi }} \times 1 = {\rm{\pi }}\) - Function value at \({\rm{x}} = 0\): At \({\rm{x}} = 0\), the function is defined by the second part: \({\rm{f}}\left( 0 \right) = {\rm{\pi }}\cos \left( 0 \right) = {\rm{\pi }} \times 1 = {\rm{\pi }}\).
Since \(\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( 0 \right) = {\rm{\pi }}\), the function \({\rm{f}}\left( {\rm{x}} \right)\) is continuous at \({\rm{x}} = 0\).
Next, let's check for differentiability by calculating the left-hand derivative (LHD) and the right-hand derivative (RHD) at \({\rm{x}} = 0\).
- Left-hand derivative at \({\rm{x}} = 0\): For \({\rm{x}} \in \left[ { - {\rm{\pi }},{\rm{\;}}0} \right)\), \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}} + {\rm{\pi }}\). The derivative is \({\rm{f}}'\left( {\rm{x}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{x}} + {\rm{\pi }}} \right) = 1\).
LHD at \({\rm{x}} = 0\) is 1. - Right-hand derivative at \({\rm{x}} = 0\): For \({\rm{x}} \in \left[ {0,\frac{{\rm{\pi}}}{2}} \right]\), \({\rm{f}}\left( {\rm{x}} \right) = {\rm{\pi }}\cos {\rm{x}}\). The derivative is \({\rm{f}}'\left( {\rm{x}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{\pi }}\cos {\rm{x}}} \right) = -{\rm{\pi }}\sin {\rm{x}}\).
RHD at \({\rm{x}} = 0\) is \(-{\rm{\pi }}\sin \left( 0 \right) = -{\rm{\pi }} \times 0 = 0\).
Since the LHD at \({\rm{x}} = 0\) (which is 1) is not equal to the RHD at \({\rm{x}} = 0\) (which is 0), the function \({\rm{f}}\left( {\rm{x}} \right)\) is not differentiable at \({\rm{x}} = 0\).
Therefore, Statement 1 is incorrect.
Analyzing Statement 2: Differentiability at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\)
First, let's check for continuity at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\). We need to compare the left-hand limit, the right-hand limit, and the function value at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\).
- Left-hand limit at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\): As \({\rm{x}}\) approaches \(\frac{{\rm{\pi}}}{2}\) from the left (\({\rm{x}} < \frac{{\rm{\pi}}}{2}\)), we use the second part of the function definition: \({\rm{f}}\left( {\rm{x}} \right) = {\rm{\pi }}\cos {\rm{x}}\).
\(\mathop {\lim }\limits_{{\rm{x}} \to {{\frac{{\rm{\pi}}}{2}}^-}} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\frac{{\rm{\pi}}}{2}}^-}} \left( {{\rm{\pi }}\cos {\rm{x}}} \right) = {\rm{\pi }}\cos \left( \frac{{\rm{\pi}}}{2} \right) = {\rm{\pi }} \times 0 = 0\) - Right-hand limit at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\): As \({\rm{x}}\) approaches \(\frac{{\rm{\pi}}}{2}\) from the right (\({\rm{x}} > \frac{{\rm{\pi}}}{2}\)), we use the third part of the function definition: \({\rm{f}}\left( {\rm{x}} \right) = {{\left( {{\rm{x}} - \frac{{\rm{\pi}}}{2}} \right)}^2}\).
\(\mathop {\lim }\limits_{{\rm{x}} \to {{\frac{{\rm{\pi}}}{2}}^+}} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\frac{{\rm{\pi}}}{2}}^+}} {{\left( {{\rm{x}} - \frac{{\rm{\pi}}}{2}} \right)}^2} = {{\left( {\frac{{\rm{\pi}}}{2}} - \frac{{\rm{\pi}}}{2}} \right)}^2} = 0^2 = 0\) - Function value at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\): At \({\rm{x}} = \frac{{\rm{\pi}}}{2}\), the function is defined by the second part: \({\rm{f}}\left( \frac{{\rm{\pi}}}{2} \right) = {\rm{\pi }}\cos \left( \frac{{\rm{\pi}}}{2} \right) = {\rm{\pi }} \times 0 = 0\).
Since \(\mathop {\lim }\limits_{{\rm{x}} \to {{\frac{{\rm{\pi}}}{2}}^-}} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\frac{{\rm{\pi}}}{2}}^+}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( \frac{{\rm{\pi}}}{2} \right) = 0\), the function \({\rm{f}}\left( {\rm{x}} \right)\) is continuous at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\).
Next, let's check for differentiability by calculating the left-hand derivative (LHD) and the right-hand derivative (RHD) at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\).
- Left-hand derivative at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\): For \({\rm{x}} \in \left[ {0,\frac{{\rm{\pi}}}{2}} \right]\), \({\rm{f}}\left( {\rm{x}} \right) = {\rm{\pi }}\cos {\rm{x}}\). The derivative is \({\rm{f}}'\left( {\rm{x}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{\pi }}\cos {\rm{x}}} \right) = -{\rm{\pi }}\sin {\rm{x}}\).
LHD at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\) is \(-{\rm{\pi }}\sin \left( \frac{{\rm{\pi}}}{2} \right) = -{\rm{\pi }} \times 1 = -{\rm{\pi }}\). - Right-hand derivative at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\): For \({\rm{x}} \in \left( {\frac{{\rm{\pi}}}{2},{\rm{\;\pi }}} \right]\), \({\rm{f}}\left( {\rm{x}} \right) = {{\left( {{\rm{x}} - \frac{{\rm{\pi}}}{2}} \right)}^2}\). The derivative is \({\rm{f}}'\left( {\rm{x}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}{{\left( {{\rm{x}} - \frac{{\rm{\pi}}}{2}} \right)}^2} = 2\left( {{\rm{x}} - \frac{{\rm{\pi}}}{2}} \right)\).
RHD at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\) is \(2\left( {\frac{{\rm{\pi}}}{2}} - \frac{{\rm{\pi}}}{2}} \right) = 2 \times 0 = 0\).
Since the LHD at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\) (which is \(-\pi\)) is not equal to the RHD at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\) (which is 0), the function \({\rm{f}}\left( {\rm{x}} \right)\) is not differentiable at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\).
Therefore, Statement 2 is incorrect.
Conclusion
Based on our analysis, neither Statement 1 nor Statement 2 is correct. The function \({\rm{f}}\left( {\rm{x}} \right)\) is not differentiable at \({\rm{x}} = 0\) because the left-hand derivative (1) is not equal to the right-hand derivative (0). Similarly, the function \({\rm{f}}\left( {\rm{x}} \right)\) is not differentiable at \({\rm{x}} = \frac{{\rm{\pi}}}{2}\) because the left-hand derivative (\(-\pi\)) is not equal to the right-hand derivative (0).
| Point | Continuity Check | LHD | RHD | Differentiable? |
|---|---|---|---|---|
| \({\rm{x}} = 0\) | Continuous (Limit = Function value = \(\pi\)) | 1 | 0 | No (1 \(\neq\) 0) |
| \({\rm{x}} = \frac{{\rm{\pi}}}{2}\) | Continuous (Limit = Function value = 0) | \(-\pi\) | 0 | No (\(-\pi\) \(\neq\) 0) |
Revision Table: Differentiability and Continuity
It is important to remember the relationship between continuity and differentiability. Differentiability is a stronger condition than continuity.
| Property | Description | Condition |
|---|---|---|
| Continuity at a point 'a' | The function has no breaks or jumps at 'a'. | \(\mathop {\lim }\limits_{x \to a^ - } f(x) = \mathop {\lim }\limits_{x \to a^ + } f(x) = f(a)\) |
| Differentiability at a point 'a' | The function has a well-defined tangent line at 'a'. The rate of change is the same from both sides. | Function is continuous at 'a', AND LHD at 'a' = RHD at 'a'. |
Note: If a function is differentiable at a point, it must be continuous at that point. However, the converse is not true; a function can be continuous at a point but not differentiable there (as seen in this problem).
Additional Information: Calculating Derivatives of Piecewise Functions
When dealing with piecewise functions, calculating derivatives at the points where the definition changes (the boundary points) requires special attention. You must use the definition of the derivative or calculate the derivatives of the pieces and then evaluate the left-hand and right-hand limits of these derivatives at the boundary point.
- For a point 'a' where the function definition changes:
- Left-hand derivative at 'a' is the derivative of the left piece evaluated at 'a' (or the limit of the difference quotient from the left).
- Right-hand derivative at 'a' is the derivative of the right piece evaluated at 'a' (or the limit of the difference quotient from the right).
- The function is differentiable at 'a' if and only if it is continuous at 'a' and the left-hand derivative equals the right-hand derivative.
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Select the correct answer using the code given below:
\({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {3{{\rm{x}}^2} + 12{\rm{x}} - 1,{\rm{\;\;}} - 1 \le {\rm{x}} \le 2}\\ {37 - {\rm{x}},{\rm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}2 < {\rm{x}} \le 3} \end{array}} \right.\)
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Select the correct answer using the code given below:
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1. The function f(x) is continuous at x = 0
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What is f’’(2.5) equal to?
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2. \({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {2{\rm{x}} + 5{\rm{\;\;if\;\;x}} > 0}\\ {{{\rm{x}}^2} + 2{\rm{x}} + 5{\rm{\;\;if\;\;x}} \le 0} \end{array}} \right.\)
Which of the above functions is/are derivable at x = 0?
Important Questions from Differentiability
What is the value of f'(x) at x = 4 from the following table of values?
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The set of all point where the function f(x) = 2x|x| is differentiable, is:
