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 solve the problem, we need to check the differentiability of the piecewise function:
\[ f(x) = \begin{cases} 2x - 3, & -3 \leq x \leq -2 \\ x + 1, & -2 < x \leq 0 \end{cases} \] at the point \( x = -2 \).

1. Step 1: Check Continuity at \( x = -2 \):
We must first check if the function is continuous at \( x = -2 \).

Left-hand limit (LHL):
As \( x \to -2^- \), we use the first definition \( f(x) = 2x - 3 \):
\[ \lim_{x \to -2^-} f(x) = 2(-2) - 3 = -4 - 3 = -7 \]

Right-hand limit (RHL):
As \( x \to -2^+ \), we use the second definition \( f(x) = x + 1 \):
\[ \lim_{x \to -2^+} f(x) = -2 + 1 = -1 \]

Since LHL \( \neq \) RHL, the function is not continuous at \( x = -2 \).

2. Conclusion:
Since the function is not even continuous at \( x = -2 \), it is not differentiable at that point.

Final Answer:
The function \( f(x) \) is not differentiable at \( x = -2 \).