Analyzing Function Continuity and Differentiability

The question asks about the properties of a piecewise function defined as:

 \[ f(x) = \begin{cases} -\dfrac{x}{\sqrt{x^2}}, & x \ne 0, \\[6pt] 0, & x = 0. \end{cases} \] 

We need to determine if this function is continuous or differentiable at \({\rm{x}} = 0\).

Simplifying the Function Definition

Let's first simplify the expression for \({\rm{x}} \ne 0\). We know that \({\sqrt {{{\rm{x}}^2}} }\) is equal to the absolute value of \({\rm{x}}\), denoted as \(|{\rm{x}}|\). So, for \({\rm{x}} \ne 0\), the function is \({\rm{f}}({\rm{x}}) = - \frac{{\rm{x}}}{{|{\rm{x}}|}}\).

• If \({\rm{x}} > 0\), then \(|{\rm{x}}| = {\rm{x}}\). So, \({\rm{f}}({\rm{x}}) = - \frac{{\rm{x}}}{{\rm{x}}} = - 1\).
• If \({\rm{x}} < 0\), then \(|{\rm{x}}| = - {\rm{x}}\). So, \({\rm{f}}({\rm{x}}) = - \frac{{\rm{x}}}{{-{\rm{x}}}} = 1\).

Combining this with the definition for \({\rm{x}} = 0\), the function can be rewritten as:

\[ f(x) = \begin{cases} 1, & x < 0, \\[6pt] 0, & x = 0, \\[6pt] -1, & x > 0. \end{cases} \]

We can represent this piecewise function clearly:

Checking for Continuity at x = 0

For a function to be continuous at a point, say \({\rm{x}} = {\rm{a}}\), three conditions must be met:

1. \({\rm{f}}({\rm{a}})\) must be defined.
2. The limit of \({\rm{f}}({\rm{x}})\) as \({\rm{x}}\) approaches \({\rm{a}}\), \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}({\rm{x}})\), must exist. This requires the left-hand limit (LHL) and the right-hand limit (RHL) to be equal.
3. The limit must be equal to the function value: \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}({\rm{x}}) = {\rm{f}}({\rm{a}})\).

Let's check these conditions for our function at \({\rm{x}} = 0\):

• Is \({\rm{f}}(0)\) defined? Yes, the definition explicitly states \({\rm{f}}(0) = 0\).
• Does \(\mathop {\lim }\limits_{{\rm{x}} \to 0} {\rm{f}}({\rm{x}})\) exist? We need to check the LHL and RHL at \({\rm{x}} = 0\).
  • Left-Hand Limit (LHL): As \({\rm{x}}\) approaches 0 from the left (i.e., \({\rm{x}} < 0\)), \({\rm{f}}({\rm{x}}) = 1\). \[\mathop {\lim }\limits_{{\rm{x}} \to {0^-}} {\rm{f}}({\rm{x}}) = \mathop {\lim }\limits_{{\rm{x}} \to {0^-}} 1 = 1\]
  • Right-Hand Limit (RHL): As \({\rm{x}}\) approaches 0 from the right (i.e., \({\rm{x}} > 0\)), \({\rm{f}}({\rm{x}}) = - 1\). \[\mathop {\lim }\limits_{{\rm{x}} \to {0^+}} {\rm{f}}({\rm{x}}) = \mathop {\lim }\limits_{{\rm{x}} \to {0^+}} (-1) = - 1\]

Because the second condition for continuity is not met (the limit does not exist), the function \({\rm{f}}({\rm{x}})\) is discontinuous at \({\rm{x}} = 0\).

Checking for Differentiability at x = 0

For a function to be differentiable at a point, it must first be continuous at that point. Since we have already determined that \({\rm{f}}({\rm{x}})\) is discontinuous at \({\rm{x}} = 0\), it cannot be differentiable at \({\rm{x}} = 0\).

Evaluating the Options

Let's look at the given options based on our analysis:

• Option 1: \({\rm{f}}({\rm{x}})\) is continuous at \({\rm{x}} = 0\) but not differentiable at \({\rm{x}} = 0\). This is incorrect because \({\rm{f}}({\rm{x}})\) is discontinuous at \({\rm{x}} = 0\).
• Option 2: \({\rm{f}}({\rm{x}})\) is continuous as well as differentiable at \({\rm{x}} = 0\). This is incorrect because \({\rm{f}}({\rm{x}})\) is discontinuous at \({\rm{x}} = 0\).
• Option 3: \({\rm{f}}({\rm{x}})\) is discontinuous at \({\rm{x}} = 0\). This is correct based on our analysis of the limits.
• Option 4: None of the above. This is incorrect because Option 3 is correct.

Therefore, the function \({\rm{f}}({\rm{x}})\) is discontinuous at \({\rm{x}} = 0\).

Revision Table: Function Properties at x = 0

Additional Information: Continuity and Differentiability

Continuity: A function is continuous at a point if its graph can be drawn through that point without lifting the pen. Mathematically, it means the function value at the point is equal to the limit of the function as the input approaches that point. If there's a break, jump, or hole in the graph at that point, the function is discontinuous.

Types of Discontinuity:

• Removable Discontinuity: The limit exists, but either the function is not defined at the point or the function value does not equal the limit. Can often be 'removed' by redefining the function at that single point.
• Jump Discontinuity: The left-hand limit and the right-hand limit exist but are not equal. Our function \({\rm{f}}({\rm{x}})\) at \({\rm{x}}=0\) is an example of a jump discontinuity.
• Infinite Discontinuity: One or both of the one-sided limits are infinite.

Differentiability: A function is differentiable at a point if the derivative exists at that point. The derivative represents the instantaneous rate of change or the slope of the tangent line at that point. For the derivative to exist at a point, the function must be continuous at that point, and the graph must not have a sharp corner or a vertical tangent line at that point. Differentiability is a stronger condition than continuity; if a function is differentiable at a point, it must be continuous at that point, but the reverse is not always true (e.g., \({\rm{f}}({\rm{x}}) = |{\rm{x}}|\) is continuous at \({\rm{x}}=0\) but not differentiable).

In this problem, the function \({\rm{f}}({\rm{x}})\) has a clear jump at \({\rm{x}}=0\), jumping from 1 (as \({\rm{x}}\) approaches from the left) to -1 (as \({\rm{x}}\) approaches from the right), while the function value at \({\rm{x}}=0\) is 0. This jump confirms the discontinuity.