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A function $f(x)$ is differentiable at $x=a$ if the Left Hand Derivative (LHD) equals the Right Hand Derivative (RHD) at that point. That is, $LHD = RHD$.
For $x < -2$, $f(x) = 2x - 3$. The derivative $f'(x) = 2$. Thus, the LHD at $x = -2$ is: $$LHD = \lim_{h \to 0} \frac{f(-2-h) - f(-2)}{-h} = 2$$
For $x > -2$, $f(x) = x + 1$. The derivative $f'(x) = 1$. Thus, the RHD at $x = -2$ is: $$RHD = \lim_{h \to 0} \frac{f(-2+h) - f(-2)}{h} = 1$$
Since $LHD = 2$ and $RHD = 1$, we observe that $LHD \neq RHD$. Therefore, the function is not differentiable at $x = -2$.
Final Answer: The function is not differentiable at x = -2.
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