Question

If \( f(x) = \begin{cases 2x - 3, & -3 \leq x \leq -2
x + 1, & -2<x \leq 0 \end{cases} \), check the differentiability of \( f(x) \) at \( x = -2 \).}

Hint:

A function must be continuous at a point to be differentiable there. If discontinuity exists, differentiability fails.

Answer 1

To check differentiability at \( x = -2 \), we must first check continuity and then differentiability. Step 1: Check continuity at \( x = -2 \).
A function is continuous at \( x = -2 \) if: \[ \lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x) = f(-2). \] Left-hand limit: \[ \lim_{x \to -2^-} f(x) = 2(-2) - 3 = -4 - 3 = -7. \] Right-hand limit: \[ \lim_{x \to -2^+} f(x) = (-2) + 1 = -1. \] Since \( -7 \neq -1 \), \( f(x) \) is not continuous at \( x = -2 \). If a function is not continuous at a point, it is not differentiable there. Conclusion: Since \( f(x) \) is not continuous at \( x = -2 \), it is not differentiable at \( x = -2 \).