Examine the differentiability of f, where f is defined by
`f(x) = {{:(x[x],if0lexlt2),((x-1)x,if2lexlt3):}` at `x = 2`

Explanation

To examine the differentiability of the function \( f \) defined as:

\[ f(x) = \begin{cases} x \lfloor x \rfloor & \text{if } 0 \leq x < 2 \\ (x - 1)x & \text{if } 2 \leq x < 3 \end{cases} \]

at \( x = 2 \), we need to find the left-hand derivative (LHD) and the right-hand derivative (RHD) at that point.

### Step 1: Calculate the Left-Hand Derivative (LHD) at \( x = 2 \)

The left-hand derivative at \( x = 2 \) is given by:

\[ f'_{-}(2) = \lim_{h \to 0} \frac{f(2 - h) - f(2)}{(2 - h) - 2} \]

For \( 2 - h < 2 \), we use the first case of the function:

\[ f(2 - h) = (2 - h) \lfloor 2 - h \rfloor = (2 - h) \cdot 1 = 2 - h \]

Now, we need to find \( f(2) \):

\[ f(2) = (2 - 1) \cdot 2 = 1 \cdot 2 = 2 \]

Substituting these into the LHD formula:

\[ f'_{-}(2) = \lim_{h \to 0} \frac{(2 - h) - 2}{(2 - h) - 2} = \lim_{h \to 0} \frac{-h}{-h} = \lim_{h \to 0} 1 = 1 \]

### Step 2: Calculate the Right-Hand Derivative (RHD) at \( x = 2 \)

The right-hand derivative at \( x = 2 \) is given by:

\[ f'_{+}(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{(2 + h) - 2} \]

For \( 2 + h \geq 2 \), we use the second case of the function:

\[ f(2 + h) = (2 + h - 1)(2 + h) = (1 + h)(2 + h) = 2 + 3h + h^2 \]

Now substituting into the RHD formula:

\[ f'_{+}(2) = \lim_{h \to 0} \frac{(2 + 3h + h^2) - 2}{(2 + h) - 2} = \lim_{h \to 0} \frac{3h + h^2}{h} = \lim_{h \to 0} (3 + h) = 3 \]

### Step 3: Compare LHD and RHD

We found:

\[ f'_{-}(2) = 1 \quad \text{and} \quad f'_{+}(2) = 3 \]

Since \( f'_{-}(2) \neq f'_{+}(2) \), the function \( f \) is not differentiable at \( x = 2 \).

### Conclusion

Thus, we conclude that the function \( f \) is not differentiable at \( x = 2 \).

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