Let `f(x){:{(1+sinx", "xlt0),(x^(2)-x+1ge0):}` Then

Options

A. f has a local maximum at x=0
B. f has a local minimum at x=0
C. f is increasing everywhere
D. f is decreasing everywhere

Answer

To solve the problem, we need to analyze the function defined piecewise as follows:

1. **Function Definition**:
- \( f(x) = 1 + \sin x \) for \( x < 0 \)
- \( f(x) = x^2 - x + 1 \) for \( x \geq 0 \)

2. **Finding the Value at \( x = 0 \)**:
- For \( x < 0 \): \( f(0) = 1 + \sin(0) = 1 + 0 = 1 \)
- For \( x \geq 0 \): \( f(0) = 0^2 - 0 + 1 = 1 \)
- Thus, \( f(0) = 1 \) from both sides.

3. **Finding the Derivative**:
- For \( x < 0 \):
\[ f'(x) = \cos x \]
- For \( x \geq 0 \):
\[ f'(x) = 2x - 1 \]

4. **Analyzing the Derivative for \( x < 0 \)**:
- Since \( \cos x \) oscillates between -1 and 1, we need to evaluate the behavior as \( x \) approaches 0 from the left.
- For \( x \) close to 0 (but negative), \( \cos x \) is positive, indicating that \( f(x) \) is increasing as \( x \) approaches 0 from the left.

5. **Analyzing the Derivative for \( x \geq 0 \)**:
- Set \( f'(x) = 0 \):
\[ 2x - 1 = 0 \implies x = \frac{1}{2} \]
- To determine if this is a local minimum or maximum, we can check the second derivative or the behavior of \( f'(x) \):
- For \( x < \frac{1}{2} \), \( f'(x) < 0 \) (decreasing).
- For \( x > \frac{1}{2} \), \( f'(x) > 0 \) (increasing).
- Thus, \( x = \frac{1}{2} \) is a local minimum.

6. **Behavior at \( x = 0 \)**:
- Since \( f'(x) \) changes from positive (increasing) to negative (decreasing) at \( x = 0 \), we can conclude that \( x = 0 \) is a local maximum.

7. **Conclusion**:
- The function has a local maximum at \( x = 0 \).

### Final Answer:
The correct option is that \( f \) has a local maxima at \( x = 0 \).