Let `f(x)={:{(x^(2)+4x",",-3lexle0),(-sinx",",0ltxle(pi)/(2)),(-cosx-1",",(pi)/(2)ltxlepi):}` then

Options

A. x=-2 is the point of global minima
B. x=`pi` is the point of global maxima
C. f(x) is non-differentiable at `x=(pi)/(2)`
D. f(x) is dicontinuos at x=0

Answer

To find the global maxima and minima of the piecewise function

\[ f(x) = \begin{cases} x^2 + 4x & \text{for } -3 \leq x \leq 0 \\ -\sin x & \text{for } 0 < x \leq \frac{\pi}{2} \\ -\cos x - 1 & \text{for } \frac{\pi}{2} < x \leq \pi \end{cases} \]

we will analyze each piece of the function separately and then compare the values at the endpoints and critical points.

### Step 1: Analyze the first piece \( f(x) = x^2 + 4x \) for \( -3 \leq x \leq 0 \)

1. **Find the derivative**:
\[ f'(x) = 2x + 4 \]

2. **Set the derivative to zero to find critical points**:
\[ 2x + 4 = 0 \implies x = -2 \]

3. **Evaluate \( f(x) \) at the critical point and endpoints**:
- At \( x = -3 \):
\[ f(-3) = (-3)^2 + 4(-3) = 9 - 12 = -3 \]
- At \( x = -2 \):
\[ f(-2) = (-2)^2 + 4(-2) = 4 - 8 = -4 \]
- At \( x = 0 \):
\[ f(0) = 0^2 + 4(0) = 0 \]

### Step 2: Analyze the second piece \( f(x) = -\sin x \) for \( 0 < x \leq \frac{\pi}{2} \)

1. **Find the derivative**:
\[ f'(x) = -\cos x \]

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

3. **Evaluate \( f(x) \) at the critical point and endpoints**:
- At \( x = 0 \) (not included in this piece):
\[ f(0) = -\sin(0) = 0 \]
- At \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) = -1 \]

### Step 3: Analyze the third piece \( f(x) = -\cos x - 1 \) for \( \frac{\pi}{2} < x \leq \pi \)

1. **Find the derivative**:
\[ f'(x) = \sin x \]

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

3. **Evaluate \( f(x) \) at the critical point and endpoints**:
- At \( x = \frac{\pi}{2} \) (not included in this piece):
\[ f\left(\frac{\pi}{2}\right) = -\cos\left(\frac{\pi}{2}\right) - 1 = -0 - 1 = -1 \]
- At \( x = \pi \):
\[ f(\pi) = -\cos(\pi) - 1 = 1 - 1 = 0 \]

### Step 4: Compare all values

Now we have the following values:
- From the first piece:
- \( f(-3) = -3 \)
- \( f(-2) = -4 \)
- \( f(0) = 0 \)
- From the second piece:
- \( f\left(\frac{\pi}{2}\right) = -1 \)
- From the third piece:
- \( f(\pi) = 0 \)

### Conclusion

The minimum value occurs at \( f(-2) = -4 \), and the maximum value occurs at \( f(0) = 0 \) and \( f(\pi) = 0 \).

Thus, the global minimum is at \( x = -2 \) with \( f(-2) = -4 \), and the global maximum is at \( x = 0 \) and \( x = \pi \) with \( f(0) = 0 \) and \( f(\pi) = 0 \).