Let f(x) = `[{:( sin (pi)/(2) x "," , 0 le x le 1) , (3 - 2x "," , x ge 1):}` then

Options

A. f(x) has local maxima at x = 1
B. f(x) has local minima at x = 1
C. f(x) does not have any local extrema at x = 1
D. f(x) has a global minima at x = 1

Answer

To solve the problem, we need to analyze the piecewise function given:

\[ f(x) = \begin{cases} \sin\left(\frac{\pi}{2} x\right) & \text{for } 0 \leq x \leq 1 \\ 3 - 2x & \text{for } x > 1 \end{cases} \]

### Step 1: Analyze the function on the interval \(0 \leq x \leq 1\)

1. **Find the derivative of \(f(x)\)** for \(0 \leq x \leq 1\):
\[ f'(x) = \frac{d}{dx}\left(\sin\left(\frac{\pi}{2} x\right)\right) = \frac{\pi}{2} \cos\left(\frac{\pi}{2} x\right) \]

2. **Set the derivative to zero to find critical points**:
\[ \frac{\pi}{2} \cos\left(\frac{\pi}{2} x\right) = 0 \]
This occurs when:
\[ \cos\left(\frac{\pi}{2} x\right) = 0 \implies \frac{\pi}{2} x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
The only solution in the interval \(0 \leq x \leq 1\) is:
\[ x = 1 \]

### Step 2: Analyze the function at \(x = 1\)

1. **Evaluate \(f(1)\)**:
\[ f(1) = \sin\left(\frac{\pi}{2} \cdot 1\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]

### Step 3: Analyze the function for \(x > 1\)

1. **Evaluate \(f(x)\)** for \(x > 1\):
\[ f(x) = 3 - 2x \]
This is a linear function with a negative slope, which means it is decreasing for \(x > 1\).

### Step 4: Check the behavior around \(x = 1\)

1. **Evaluate \(f(1)\)** from both sides:
- As \(x\) approaches \(1\) from the left (\(x \to 1^-\)):
\[ f(1^-) = \sin\left(\frac{\pi}{2}\right) = 1 \]
- As \(x\) approaches \(1\) from the right (\(x \to 1^+\)):
\[ f(1^+) = 3 - 2(1) = 1 \]

2. **Determine if it is a local maximum or minimum**:
- Since \(f(x)\) is increasing on \([0, 1]\) and then decreases for \(x > 1\), we conclude that \(f(x)\) has a local maximum at \(x = 1\).

### Conclusion

The function \(f(x)\) has a local maximum at \(x = 1\) with a value of \(f(1) = 1\).