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 increasing in `(-oo,(1)/(2))` and decreasing in `((1)/(2),oo)`
B. f(x) is decreasing in `(-oo,(1)/(2))` and increasing in `((1)/(2),oo)`
C. f(x) is increasing in `(-oo,1)` and increasing in `(1,oo)`
D. f(x) is decreasing in `(-oo,1)` and increasing in `(1,oo)`

Explanation

To analyze the function \( f(x) = |x - 1| + x^2 \), we will first break down the absolute value function and then determine the intervals where the function is increasing or decreasing.

### Step 1: Define the function based on the absolute value
The absolute value function \( |x - 1| \) can be expressed as:
- \( x - 1 \) when \( x \geq 1 \)
- \( -(x - 1) = -x + 1 \) when \( x < 1 \)

Thus, we can define \( f(x) \) in two cases:
1. For \( x \geq 1 \):
\[ f(x) = (x - 1) + x^2 = x^2 + x - 1 \]
2. For \( x < 1 \):
\[ f(x) = (-x + 1) + x^2 = x^2 - x + 1 \]

### Step 2: Find the derivative of each piece
To determine where the function is increasing or decreasing, we will find the derivative \( f'(x) \) for each case.

1. For \( x \geq 1 \):
\[ f(x) = x^2 + x - 1 \]
\[ f'(x) = 2x + 1 \]

2. For \( x < 1 \):
\[ f(x) = x^2 - x + 1 \]
\[ f'(x) = 2x - 1 \]

### Step 3: Analyze the derivative
Now we will analyze the sign of the derivatives in their respective intervals.

1. For \( x \geq 1 \):
\[ f'(x) = 2x + 1 \]
Since \( 2x + 1 > 0 \) for all \( x \geq 1 \), the function is **increasing** in this interval.

2. For \( x < 1 \):
\[ f'(x) = 2x - 1 \]
Setting \( f'(x) = 0 \) gives:
\[ 2x - 1 = 0 \implies x = \frac{1}{2} \]
- For \( x < \frac{1}{2} \), \( f'(x) < 0 \) (decreasing).
- For \( \frac{1}{2} < x < 1 \), \( f'(x) > 0 \) (increasing).

### Step 4: Summary of intervals
- The function is **decreasing** on the interval \( (-\infty, \frac{1}{2}) \).
- The function is **increasing** on the intervals \( (\frac{1}{2}, 1) \) and \( [1, \infty) \).

### Conclusion
Based on the analysis, the correct statement regarding the function \( f(x) = |x - 1| + x^2 \) is:
- The function is decreasing on \( (-\infty, \frac{1}{2}) \) and increasing on \( (\frac{1}{2}, \infty) \).