The question asks to discuss the differentiability of the function $f(x) = |x-1| + |x-2|$.

Steps to Analyze Differentiability:

1. Break Down Absolute Values: Recall that the absolute value function $|a|$ can be expressed as:

   \begin{cases} a & \text{if } a \geq 0, \\ -a & \text{if } a < 0. \end{cases}$$ Using this property, analyze $$f(x) = |x-1| + |x-2|$$ by splitting it into intervals where $$x-1$$ and $$x-2$$ change sign. - $$|x-1|$$: $$|x-1| = \begin{cases} x-1 & \text{if } x \geq 1, \\ -(x-1) = 1-x & \text{if } x < 1. \end{cases}$$ - $$|x-2|$$: $$|x-2| = \begin{cases} x-2 & \text{if } x \geq 2, \\ -(x-2) = 2-x & \text{if } x < 2. \end{cases}$$ Combine these into cases based on the intervals $$x < 1$$, $$1 \leq x < 2$$, and $$x \geq 2$$.

2. Piecewise Definition: For $f(x) = |x-1| + |x-2|$, the function can be expressed as:

   • For $x < 1$: $f(x) = (1-x) + (2-x) = 3 - 2x.$

   • For $1 \leq x < 2$: $f(x) = (x-1) + (2-x) = 1.$

   • For $x \geq 2$: $f(x) = (x-1) + (x-2) = 2x - 3.$

   Thus, the piecewise definition of $f(x)$ is:

   \begin{cases} 3 - 2x & \text{if } x < 1, \\ 1 & \text{if } 1 \leq x < 2, \\ 2x - 3 & \text{if } x \geq 2. \end{cases}$$

3. Check Continuity: $f(x)$ is continuous if the left-hand limit (LHL) equals the right-hand limit (RHL) at $x = 1$ and $x = 2$.

   • At $x = 1$: $\text{LHL} = \lim_{x \to 1^-} f(x) = 3 - 2(1) = 1, \quad \text{RHL} = \lim_{x \to 1^+} f(x) = 1.$ Hence, $f(x)$ is continuous at $x = 1$.

   • At $x = 2$: $\text{LHL} = \lim_{x \to 2^-} f(x) = 1, \quad \text{RHL} = \lim_{x \to 2^+} f(x) = 2(2) - 3 = 1.$ Hence, $f(x)$ is continuous at $x = 2$.

   Therefore, $f(x)$ is continuous everywhere.

4. Check Differentiability: $f(x)$ is differentiable if the left-hand derivative (LHD) equals the right-hand derivative (RHD) at $x = 1$ and $x = 2$.

   • At $x = 1$: For $x < 1$, $f(x) = 3 - 2x$, so $f'(x) = -2$. For $1 \leq x < 2$, $f(x) = 1$, so $f'(x) = 0$. $\text{LHD} = -2, \quad \text{RHD} = 0.$ Since $\text{LHD} \neq \text{RHD}$, $f(x)$ is not differentiable at $x = 1$.

   • At $x = 2$: For $1 \leq x < 2$, $f(x) = 1$, so $f'(x) = 0$. For $x \geq 2$, $f(x) = 2x - 3$, so $f'(x) = 2$. $\text{LHD} = 0, \quad \text{RHD} = 2.$ Since $\text{LHD} \neq \text{RHD}$, $f(x)$ is not differentiable at $x = 2$.

Conclusion:

• $f(x)$ is continuous everywhere.
• $f(x)$ is not differentiable at $x = 1$ and $x = 2$.

Would you like further details or have any questions? Here are five related questions to deepen your understanding:

1. What is the general method to test differentiability at a point for any function?
2. How is the concept of absolute value used in defining piecewise functions?
3. Can you think of another function involving absolute values that behaves similarly?
4. What happens to differentiability when there are sharp corners in a graph?
5. How would the second derivative of $f(x)$ behave, and what does it tell us about the function?

Tip: Always verify both continuity and differentiability at critical points separately. Continuity does not guarantee differentiability!