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

Options

A. Not differentiable at `x = 2`
B. Differentiable at `x = 2`
C. None of these
D. Can not say

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 check the left-hand derivative and the right-hand derivative at that point.

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

The left-hand derivative is given by the limit:

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

Since \( 2 - h < 2 \) for small \( h \), 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 find \( f(2) \):

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

Now substituting these values into the limit:

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

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

The right-hand derivative is given by the limit:

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

Since \( 2 + h \geq 2 \) for small \( h \), 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 limit:

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

### Step 3: Compare the left-hand and right-hand derivatives

We have:

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

Since the left-hand derivative does not equal the right-hand derivative:

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

### Conclusion

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

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