Question

Check the differentiability of the function \( f(x) = \lfloor x \rfloor \) at \( x = -3 \), where \( \lfloor \cdot \rfloor \) denotes the greatest integer function.

Hint:

The greatest integer function is not differentiable at integer values of \( x \) because of the discontinuity in the derivative at those points.

Answer 1

The greatest integer function \( f(x) = \lfloor x \rfloor \) gives the greatest integer less than or equal to \( x \). For \( x = -3 \), we have:

\[ f(x) = \lfloor x \rfloor = \begin{cases} -4, & -4 \leq x < -3, \\ -3, & x = -3. \end{cases} \]
To check differentiability, we need to verify the left-hand derivative (LHD) and right-hand derivative (RHD) at \( x = -3 \). 1. Left-hand derivative (LHD): \[ \text{LHD} = \lim_{h \to 0^-} \frac{f(-3 + h) - f(-3)}{h}. \] For \( h \to 0^- \), \( -3 + h \in (-4, -3) \), so \( f(-3 + h) = -4 \). Substituting: \[ \text{LHD} = \lim_{h \to 0^-} \frac{-4 - (-3)}{h} = \lim_{h \to 0^-} \frac{-1}{h}. \] Since \( h \to 0^- \), the denominator is negative, and \( \text{LHD} \to \infty \). 2. Right-hand derivative (RHD): \[ \text{RHD} = \lim_{h \to 0^+} \frac{f(-3 + h) - f(-3)}{h}. \] For \( h \to 0^+ \), \( -3 + h \in (-3, -2) \), so \( f(-3 + h) = -3 \). Substituting: \[ \text{RHD} = \lim_{h \to 0^+} \frac{-3 - (-3)}{h} = \lim_{h \to 0^+} \frac{0}{h} = 0. \] Since \( \text{LHD} \neq \text{RHD} \), the function \( f(x) = \lfloor x \rfloor \) is not differentiable at \( x = -3 \).