Function `{{:(-x,","if,,x<0),(x^(2),","if,,0lexle1),(x^(3)-x+1,","if,,x>1):}`
is differentiable at-

Options

A. `x = 0` but not at `x = 1`
B. `x = 1` but not at `x = 0`
C. `x = 0` and `x = 1` both
D. neither `x = 0` nor `x = 1`

Answer

To determine the differentiability of the given piecewise function at the points \( x = 0 \) and \( x = 1 \), we will follow these steps:

### Step 1: Define the Function
The function \( f(x) \) is defined as follows:
- \( f(x) = -x \) for \( x < 0 \)
- \( f(x) = x^2 \) for \( 0 \leq x \leq 1 \)
- \( f(x) = x^3 - x + 1 \) for \( x > 1 \)

### Step 2: Check Differentiability at \( x = 0 \)
To check if \( f(x) \) is differentiable at \( x = 0 \), we need to find the left-hand derivative (LHD) and the right-hand derivative (RHD) at this point.

#### Left-Hand Derivative at \( x = 0 \)
For \( x < 0 \), \( f(x) = -x \).
- The derivative \( f'(x) = -1 \).
- Therefore, the left-hand derivative at \( x = 0 \) is:
\[ \text{LHD} = \lim_{x \to 0^-} f'(x) = -1 \]

#### Right-Hand Derivative at \( x = 0 \)
For \( 0 \leq x \leq 1 \), \( f(x) = x^2 \).
- The derivative \( f'(x) = 2x \).
- Therefore, the right-hand derivative at \( x = 0 \) is:
\[ \text{RHD} = \lim_{x \to 0^+} f'(x) = 2(0) = 0 \]

#### Conclusion at \( x = 0 \)
Since \( \text{LHD} \neq \text{RHD} \) (i.e., \(-1 \neq 0\)), the function \( f(x) \) is **not differentiable at \( x = 0 \)**.

### Step 3: Check Differentiability at \( x = 1 \)
Now, we will check if \( f(x) \) is differentiable at \( x = 1 \).

#### Left-Hand Derivative at \( x = 1 \)
For \( 0 \leq x \leq 1 \), \( f(x) = x^2 \).
- The derivative \( f'(x) = 2x \).
- Therefore, the left-hand derivative at \( x = 1 \) is:
\[ \text{LHD} = \lim_{x \to 1^-} f'(x) = 2(1) = 2 \]

#### Right-Hand Derivative at \( x = 1 \)
For \( x > 1 \), \( f(x) = x^3 - x + 1 \).
- The derivative \( f'(x) = 3x^2 - 1 \).
- Therefore, the right-hand derivative at \( x = 1 \) is:
\[ \text{RHD} = \lim_{x \to 1^+} f'(x) = 3(1)^2 - 1 = 3 - 1 = 2 \]

#### Conclusion at \( x = 1 \)
Since \( \text{LHD} = \text{RHD} \) (i.e., \( 2 = 2 \)), the function \( f(x) \) is **differentiable at \( x = 1 \)**.

### Final Conclusion
The function \( f(x) \) is differentiable at \( x = 1 \) but not at \( x = 0 \).