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 by

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

at \( x = 2 \), we need to find the left-hand derivative and the right-hand derivative at that point and check if they are equal.

### Step 1: Find the left-hand derivative at \( x = 2 \)

The left-hand derivative is defined as:

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

Since we are approaching from the left, we will use the piece of the function valid for \( 0 \leq x < 2 \):

\[ f(x) = x \lfloor x \rfloor \]

At \( x = 2 \):

\[ f(2) = 2 \lfloor 2 \rfloor = 2 \cdot 2 = 4 \]

For \( 2 - h \) (where \( h \) is a small positive number), we have:

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

Now substituting into the left-hand derivative limit:

\[ f'_-(2) = \lim_{h \to 0^-} \frac{(2-h) - 4}{-h} = \lim_{h \to 0^-} \frac{-h - 2}{-h} = \lim_{h \to 0^-} \frac{2 + h}{h} \]

As \( h \) approaches \( 0 \), this simplifies to:

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

### Step 2: Find the right-hand derivative at \( x = 2 \)

The right-hand derivative is defined as:

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

For \( 2 + h \) (where \( h \) is a small positive number), we will use the piece of the function valid for \( 2 \leq x < 3 \):

\[ f(x) = (x-1)x \]

At \( x = 2 \):

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

Now substituting into the right-hand derivative limit:

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

Expanding this:

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

### Step 3: Compare the derivatives

Now we have:

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

Since the left-hand derivative is not equal to the right-hand derivative:

\[ f'_-(2) \neq f'_+(2) \]

### Conclusion

The function \( f \) is not differentiable at \( x = 2 \).

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