If \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}}\left( {\sqrt {\rm{x}} - \sqrt {{\rm{x}} + 1} } \right)\) , then f(x) is
differentiable at x = 0
Analyzing Differentiability of f(x) at x = 0
We are given the function \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}}\left( {\sqrt {\rm{x}} - \sqrt {{\rm{x}} + 1} } \right)\) and asked about its differentiability at \({\rm{x}} = 0\).
Understanding the Function's Domain
The function involves square roots. For \({\sqrt {\rm{x}}} \) to be defined, we need \({\rm{x}} \ge 0\). For \({\sqrt {{\rm{x}} + 1} }\) to be defined, we need \({\rm{x}} + 1 \ge 0\), which means \({\rm{x}} \ge -1\).
Combining these conditions, the domain of \({\rm{f}}\left( {\rm{x}} \right)\) is \({\rm{x}} \ge 0\). This means we only need to consider the behavior of the function for \({\rm{x}} \ge 0\).
Checking for Continuity at x = 0
A function is continuous at a point if the limit of the function as \({\rm{x}}\) approaches the point equals the function's value at that point. Since the domain is \({\rm{x}} \ge 0\), we will check the right-hand continuity at \({\rm{x}} = 0\).
First, evaluate the function at \({\rm{x}} = 0\):
\[ {\rm{f}}\left( 0 \right) = 0 \left( {\sqrt {0} - \sqrt {0 + 1} } \right) = 0 \left( {0 - 1} \right) = 0 \]Next, evaluate the right-hand limit as \({\rm{x}}\) approaches 0:
\[ \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 0^+} {\rm{f}}\left( {\rm{x}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 0^+} {\rm{x}}\left( {\sqrt {\rm{x}} - \sqrt {{\rm{x}} + 1} } \right) \]Substituting \({\rm{x}} = 0\) into the expression:
\[ \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 0^+} {\rm{x}}\left( {\sqrt {\rm{x}} - \sqrt {{\rm{x}} + 1} } \right) = 0 \left( {\sqrt {0} - \sqrt {0 + 1} } \right) = 0 \left( {0 - 1} \right) = 0 \]Since \( \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 0^+} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( 0 \right) = 0 \), the function \({\rm{f}}\left( {\rm{x}} \right)\) is continuous at \({\rm{x}} = 0\).
Checking for Differentiability at x = 0
A function is differentiable at a point if the limit of the difference quotient exists at that point. Since the domain is \({\rm{x}} \ge 0\), we will evaluate the right-hand derivative at \({\rm{x}} = 0\).
The derivative of \({\rm{f}}\left( {\rm{x}} \right)\) at \({\rm{x}} = 0\) is given by the limit:
\[ {\rm{f}}'\left( 0 \right) = \mathop {{\rm{lim}}}\limits_{{\rm{h}} \to 0} \frac{{{\rm{f}}\left( {0 + {\rm{h}}} \right) - {\rm{f}}\left( 0 \right)}}{{\rm{h}}} \]Considering the domain, we use the right-hand limit (\({\rm{h}} \to 0^+\)):
\[ {\rm{f}}'\left( 0 \right) = \mathop {{\rm{lim}}}\limits_{{\rm{h}} \to 0^+} \frac{{{\rm{f}}\left( {\rm{h}} \right) - {\rm{f}}\left( 0 \right)}}{{\rm{h}}} \]We know \({\rm{f}}\left( {\rm{h}} \right) = {\rm{h}}\left( {\sqrt {\rm{h}} - \sqrt {{\rm{h}} + 1} } \right)\) and \({\rm{f}}\left( 0 \right) = 0\). Substitute these into the expression:
\[ {\rm{f}}'\left( 0 \right) = \mathop {{\rm{lim}}}\limits_{{\rm{h}} \to 0^+} \frac{{{\rm{h}}\left( {\sqrt {\rm{h}} - \sqrt {{\rm{h}} + 1} } \right) - 0}}{{\rm{h}}} \]For \({\rm{h}} > 0\), we can cancel the \({\rm{h}}\) terms:
\[ {\rm{f}}'\left( 0 \right) = \mathop {{\rm{lim}}}\limits_{{\rm{h}} \to 0^+} \left( {\sqrt {\rm{h}} - \sqrt {{\rm{h}} + 1} } \right) \]Now, substitute \({\rm{h}} = 0\):
\[ {\rm{f}}'\left( 0 \right) = \sqrt {0} - \sqrt {0 + 1} = 0 - \sqrt {1} = 0 - 1 = -1 \]Since the limit of the difference quotient exists and is a finite value (\(-1\)), the function \({\rm{f}}\left( {\rm{x}} \right)\) is differentiable at \({\rm{x}} = 0\).
Conclusion on Differentiability at x = 0
Based on our analysis, the function \({\rm{f}}\left( {\rm{x}} \right)\) is continuous at \({\rm{x}} = 0\), and its derivative at \({\rm{x}} = 0\) exists and is equal to \(-1\).
Therefore, \({\rm{f}}\left( {\rm{x}} \right)\) is differentiable at \({\rm{x}} = 0\).
Revision Table: Continuity and Differentiability Analysis
| Concept | Definition at a Point 'a' | Requirement for Differentiability |
|---|---|---|
| Continuity | \( \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} \right) \) | Necessary condition (A differentiable function must be continuous) |
| Differentiability | The limit of the difference quotient \( \mathop {{\rm{lim}}}\limits_{{\rm{h}} \to 0} \frac{{{\rm{f}}\left( {{\rm{a}} + {\rm{h}}} \right) - {\rm{f}}\left( {\rm{a}} \right)}}{{\rm{h}}} \) exists and is finite. | Sufficient condition (If differentiable, it is continuous) |
Additional Information on Differentiability Concepts
Differentiability implies continuity. If a function is differentiable at a point, it must also be continuous at that point. However, the converse is not true; a continuous function may not be differentiable (e.g., \({\rm{f}}\left( {\rm{x}} \right) = \left| {\rm{x}} \right|\) at \({\rm{x}} = 0\)).
For a function defined only on one side of a boundary point (like our function at \({\rm{x}} = 0\) where the domain is \({\rm{x}} \ge 0\)), we check the corresponding one-sided limit of the difference quotient to determine differentiability at that boundary point.
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\({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {3{{\rm{x}}^2} + 12{\rm{x}} - 1,{\rm{\;\;}} - 1 \le {\rm{x}} \le 2}\\ {37 - {\rm{x}},{\rm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}2 < {\rm{x}} \le 3} \end{array}} \right.\)
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Select the correct answer using the code given below:
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Which of the above statements is/are correct?What is f’(4) equal to?
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?
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