Understanding Function Differentiability at a Point

Differentiability of a function at a specific point is a fundamental concept in calculus. A function \(f(x)\) is said to be differentiable at a point \(x=c\) if the limit of the difference quotient exists at that point. This limit is the derivative of the function at \(x=c\), denoted as \(f'(c)\).

Mathematically, the definition of the derivative at \(x=c\) is:

\(f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}\)

For piecewise functions, like the ones given, to be differentiable at the point where the definition changes (in this case, \(x=0\)), two conditions must be met:

1. The function must be continuous at the point.
2. The left-hand derivative and the right-hand derivative must exist and be equal at the point.

Let's analyze each function given in the question for differentiability at \(x=0\).

Analysis of Function 1: \(f(x) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{{\rm{x}}}{\rm{\;\;if\;\;x}} \ne 0}\\ {0{\rm{\;\;if\;\;x}} = 0} \end{array}} \right.\)

We need to check if Function 1 is differentiable at \(x=0\).

Step 1: Check Continuity at x=0

For continuity at \(x=0\), we need \(\lim_{x \to 0} f(x) = f(0)\).

• \(f(0) = 0\), as given by the definition.
• The limit as \(x \to 0\): \(\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1}{x}\).

Let's check the one-sided limits:

• Left-hand limit: \(\lim_{x \to 0^-} \frac{1}{x} = -\infty\).
• Right-hand limit: \(\lim_{x \to 0^+} \frac{1}{x} = +\infty\).

Since the left-hand limit and the right-hand limit are not finite and not equal, the limit \(\lim_{x \to 0} \frac{1}{x}\) does not exist. Therefore, Function 1 is not continuous at \(x=0\).

Conclusion for Function 1: Since Function 1 is not continuous at \(x=0\), it cannot be differentiable at \(x=0\). Continuity is a necessary condition for differentiability.

Analysis of Function 2: \(f(x) = \left\{ {\begin{array}{*{20}{c}} {2{\rm{x}} + 5{\rm{\;\;if\;\;x}} > 0}\\ {{{\rm{x}}^2} + 2{\rm{x}} + 5{\rm{\;\;if\;\;x}} \le 0} \end{array}} \right.\)

We need to check if Function 2 is differentiable at \(x=0\).

Step 1: Check Continuity at x=0

For continuity at \(x=0\), we need \(\lim_{x \to 0} f(x) = f(0)\).

• Function value at \(x=0\): \(f(0) = (0)^2 + 2(0) + 5 = 5\) (using the \(x \le 0\) part of the definition).
• Left-hand limit at \(x=0\): \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x^2 + 2x + 5) = (0)^2 + 2(0) + 5 = 5\).
• Right-hand limit at \(x=0\): \(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2x + 5) = 2(0) + 5 = 5\).

Since the left-hand limit, the right-hand limit, and the function value at \(x=0\) are all equal to 5, Function 2 is continuous at \(x=0\).

Step 2: Check Differentiability at x=0 (Using Left and Right Derivatives)

We need to calculate the left-hand derivative (\(f'(0^-)\)) and the right-hand derivative (\(f'(0^+)\)) at \(x=0\) using the limit definition:

\(f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}\)

Here, \(c=0\) and \(f(0)=5\).

• Left-hand derivative at \(x=0\): We consider \(h \to 0^-\), which means \(h < 0\). For \(x=0+h=h\), we use the part of the function definition where \(x \le 0\).

  \(f'(0^-) = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{(h^2 + 2h + 5) - 5}{h}\)

  \(f'(0^-) = \lim_{h \to 0^-} \frac{h^2 + 2h}{h} = \lim_{h \to 0^-} \frac{h(h + 2)}{h}\)

  For \(h \ne 0\), we can cancel \(h\):

  \(f'(0^-) = \lim_{h \to 0^-} (h + 2) = 0 + 2 = 2\)

• Right-hand derivative at \(x=0\): We consider \(h \to 0^+\), which means \(h > 0\). For \(x=0+h=h\), we use the part of the function definition where \(x > 0\).

  \(f'(0^+) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{(2h + 5) - 5}{h}\)

  \(f'(0^+) = \lim_{h \to 0^+} \frac{2h}{h}\)

  For \(h \ne 0\), we can cancel \(h\):

  \(f'(0^+) = \lim_{h \to 0^+} 2 = 2\)

Since the left-hand derivative (\(f'(0^-) = 2\)) is equal to the right-hand derivative (\(f'(0^+) = 2\)), the derivative of Function 2 at \(x=0\) exists and is equal to 2.

Conclusion for Function 2: Function 2 is differentiable at \(x=0\).

Summary of Findings

Based on the analysis, only Function 2 is differentiable at \(x=0\).

Revision Table: Differentiability at a Point

Additional Information: Smoothness and Derivatives

Differentiability at a point implies that the function is 'smooth' at that point, meaning there is no sharp corner, cusp, or break in the graph. If a function is differentiable at every point in an interval, it is differentiable on that interval.

For polynomial functions, they are differentiable everywhere. Rational functions are differentiable everywhere in their domain. Piecewise functions require checking continuity and the equality of one-sided derivatives at the points where the definition changes.

The derivative of a function gives the instantaneous rate of change of the function at a point, which can be interpreted geometrically as the slope of the tangent line to the graph at that point.

The differentiability of a function is a stronger condition than continuity. 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 (e.g., \(f(x) = |x|\) at \(x=0\)).