If `f(x)=2sinx-x^(2)`, then in `x in [0, pi]`

Options

A. `f(x)` has no local maximum
B. `f(x)` has one local minimum
C. `f(x)` has 2 local maxima
D. `f(x)` has one local maximum

Answer

To solve the problem, we need to find the local maxima and minima of the function \( f(x) = 2\sin x - x^2 \) in the interval \( x \in [0, \pi] \).

### Step-by-Step Solution:

1. **Find the derivative of the function**:
\[ f'(x) = \frac{d}{dx}(2\sin x - x^2) = 2\cos x - 2x \]

**Hint**: Use the basic differentiation rules for sine and power functions.

2. **Set the derivative to zero to find critical points**:
\[ 2\cos x - 2x = 0 \implies \cos x = x \]

**Hint**: This equation indicates where the slope of the function is zero, which corresponds to potential maxima or minima.

3. **Analyze the equation \( \cos x = x \)**:
- We need to find the intersection of the graphs of \( y = \cos x \) and \( y = x \) in the interval \( [0, \pi] \).
- The function \( y = \cos x \) decreases from 1 to -1, while \( y = x \) increases from 0 to \( \pi \).
- There will be one intersection point in the interval \( [0, \pi] \) since \( \cos x \) starts above \( x \) and ends below it.

**Hint**: Graphing both functions can help visualize the intersection point.

4. **Use numerical methods or graphical methods to find the critical point**:
- The critical point can be approximated using numerical methods (like the bisection method) or graphing calculators. Let's denote this point as \( x_0 \).

**Hint**: You can use numerical approximation techniques to find \( x_0 \) or use a calculator.

5. **Find the second derivative**:
\[ f''(x) = \frac{d}{dx}(2\cos x - 2x) = -2\sin x - 2 \]

**Hint**: Differentiate the first derivative to find the concavity of the function.

6. **Evaluate the second derivative at the critical point**:
- Since \( x_0 \) is in \( [0, \pi] \), \( \sin x \) is non-negative.
- Thus, \( f''(x_0) = -2\sin x_0 - 2 \) will be negative because \( -2 \) is always less than \( -2\sin x_0 \).

**Hint**: If \( f''(x_0) < 0 \), it indicates that \( f(x) \) is concave down at \( x_0 \), confirming a local maximum.

7. **Conclusion**:
- Since there is only one critical point in the interval and \( f''(x_0) < 0 \), we conclude that there is one local maximum at \( x_0 \).

**Hint**: The behavior of the second derivative helps confirm the nature of the critical point.

### Final Result:
The function \( f(x) = 2\sin x - x^2 \) has one local maximum in the interval \( [0, \pi] \).