If a function f(x) is defined as `f(x) = {{:(-x",",x lt 0),(x^(2)",",0 le x le 1),(x^(2)-x + 1",",x gt 1):}` then

Options

A. f(x) is differentiable at x = 0 and x = 1
B. f(x) is differentiable at x = 0 but not at x = 1
C. f(x) is not differentiable at x = 1 but not at x = 0
D. f(x) is not differentiable at x = 0 and x = 1

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined piecewise as follows:

\[ f(x) = \begin{cases} -x & \text{if } x < 0 \\ x^2 & \text{if } 0 \leq x \leq 1 \\ x^2 - x + 1 & \text{if } x > 1 \end{cases} \]

### Step 1: Check Continuity

To check continuity at the points where the definition of the function changes, we will evaluate the limits and the function values at \( x = 0 \) and \( x = 1 \).

#### At \( x = 0 \):
- \( f(0^-) = f(0) = 0 \) (using the first case, \( -x \))
- \( f(0^+) = f(0) = 0 \) (using the second case, \( x^2 \))

Since \( f(0^-) = f(0^+) = f(0) \), the function is continuous at \( x = 0 \).

#### At \( x = 1 \):
- \( f(1^-) = f(1) = 1 \) (using the second case, \( x^2 \))
- \( f(1^+) = f(1) = 1 \) (using the third case, \( x^2 - x + 1 \))

Since \( f(1^-) = f(1^+) = f(1) \), the function is continuous at \( x = 1 \).

### Conclusion for Continuity:
The function \( f(x) \) is continuous at both \( x = 0 \) and \( x = 1 \).

---

### Step 2: Check Differentiability

Next, we need to check the differentiability of the function at the points \( x = 0 \) and \( x = 1 \).

#### At \( x = 0 \):
- For \( x < 0 \), \( f(x) = -x \) so \( f'(x) = -1 \).
- Thus, \( f'(0^-) = -1 \).

- For \( 0 < x < 1 \), \( f(x) = x^2 \) so \( f'(x) = 2x \).
- Thus, \( f'(0^+) = 2(0) = 0 \).

Since \( f'(0^-) \neq f'(0^+) \), the function is **not differentiable** at \( x = 0 \).

#### At \( x = 1 \):
- For \( 0 < x < 1 \), \( f(x) = x^2 \) so \( f'(x) = 2x \).
- Thus, \( f'(1^-) = 2(1) = 2 \).

- For \( x > 1 \), \( f(x) = x^2 - x + 1 \) so \( f'(x) = 2x - 1 \).
- Thus, \( f'(1^+) = 2(1) - 1 = 1 \).

Since \( f'(1^-) \neq f'(1^+) \), the function is **not differentiable** at \( x = 1 \).

### Conclusion for Differentiability:
The function \( f(x) \) is not differentiable at both \( x = 0 \) and \( x = 1 \).

### Final Answer:
The function \( f(x) \) is continuous at \( x = 0 \) and \( x = 1 \), but it is not differentiable at these points.

---