Cosider the function `f(x)=|x-1|+x^(2)" where " x""inR`.
Which one of the following statements is correct?

Options

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

Explanation

To analyze the function \( f(x) = |x - 1| + x^2 \), we need to determine its continuity and differentiability.

### Step 1: Identify the components of the function
The function can be broken down into two parts:
1. \( f_1(x) = |x - 1| \)
2. \( f_2(x) = x^2 \)

### Step 2: Check the continuity of each component
Both \( f_1(x) \) and \( f_2(x) \) are continuous functions:
- The absolute value function \( |x - 1| \) is continuous everywhere.
- The polynomial function \( x^2 \) is also continuous everywhere.

Since both components are continuous, their sum \( f(x) = |x - 1| + x^2 \) is continuous for all \( x \in \mathbb{R} \).

### Step 3: Check differentiability of \( f(x) \)
To check differentiability, we need to analyze the derivative of \( f(x) \):
- For \( x < 1 \), \( |x - 1| = 1 - x \), so:
\[ f(x) = (1 - x) + x^2 = -x + x^2 + 1 \]
The derivative is:
\[ f'(x) = -1 + 2x \]

- For \( x > 1 \), \( |x - 1| = x - 1 \), so:
\[ f(x) = (x - 1) + x^2 = x + x^2 - 1 \]
The derivative is:
\[ f'(x) = 1 + 2x \]

- At \( x = 1 \), we need to check the left-hand and right-hand derivatives:
- Left-hand derivative at \( x = 1 \):
\[ f'(1^-) = -1 + 2(1) = 1 \]
- Right-hand derivative at \( x = 1 \):
\[ f'(1^+) = 1 + 2(1) = 3 \]

Since the left-hand derivative \( f'(1^-) = 1 \) and the right-hand derivative \( f'(1^+) = 3 \) are not equal, \( f(x) \) is not differentiable at \( x = 1 \).

### Step 4: Summary of findings
- The function \( f(x) \) is continuous everywhere.
- The function \( f(x) \) is differentiable everywhere except at \( x = 1 \).

### Conclusion
The correct statement about the function \( f(x) = |x - 1| + x^2 \) is:
- **It is continuous everywhere but not differentiable at \( x = 1 \)**.