If ` f(x) = {{:( 3x ^(2) + 12 x - 1",", - 1 le x le 2), (37- x",", 2 lt x le 3):}, `then

Options

A. f(x) is increasing on [-1,2]
B. f(x) is continuous on [-1,2]
C. f'(2) does not exist
D. f(x) has the minimum value at x=2

Answer

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

\[ f(x) = \begin{cases} 3x^2 + 12x - 1 & \text{for } -1 \leq x \leq 2 \\ 37 - x & \text{for } 2 < x \leq 3 \end{cases} \]

### Step 1: Check Continuity at \( x = 2 \)

To check if the function is continuous at \( x = 2 \), we need to find the left-hand limit and the right-hand limit at this point.

1. **Left-hand limit as \( x \) approaches 2:**
\[ f(2) = 3(2)^2 + 12(2) - 1 = 3(4) + 24 - 1 = 12 + 24 - 1 = 35 \]

2. **Right-hand limit as \( x \) approaches 2:**
\[ f(2^+) = 37 - 2 = 35 \]

Since both limits are equal, \( f(2^-) = f(2^+) = 35 \), the function is continuous at \( x = 2 \).

### Step 2: Determine if \( f(x) \) is Increasing or Decreasing on the Interval \( [-1, 2] \)

To determine if \( f(x) \) is increasing or decreasing, we need to find the derivative of \( f(x) \) in the interval \( [-1, 2] \).

1. **Differentiate \( f(x) \):**
\[ f'(x) = \frac{d}{dx}(3x^2 + 12x - 1) = 6x + 12 \]

2. **Evaluate the derivative in the interval \( [-1, 2] \):**
- At \( x = -1 \):
\[ f'(-1) = 6(-1) + 12 = -6 + 12 = 6 \quad (\text{positive}) \]
- At \( x = 2 \):
\[ f'(2) = 6(2) + 12 = 12 + 12 = 24 \quad (\text{positive}) \]

Since \( f'(x) > 0 \) for all \( x \) in \( [-1, 2] \), the function is increasing in this interval.

### Step 3: Determine if \( f(x) \) is Increasing or Decreasing on the Interval \( (2, 3] \)

For the interval \( (2, 3] \), we analyze the second part of the piecewise function.

1. **Differentiate \( f(x) \):**
\[ f'(x) = \frac{d}{dx}(37 - x) = -1 \]

Since \( f'(x) = -1 < 0 \) for \( x \) in \( (2, 3] \), the function is decreasing in this interval.

### Step 4: Find the Minimum Value of \( f(x) \)

Since \( f(x) \) is increasing on \( [-1, 2] \) and decreasing on \( (2, 3] \), the minimum value occurs at the endpoint of the increasing interval:

\[ \text{Minimum value at } x = 2: \quad f(2) = 35 \]

### Step 5: Check the Existence of the Derivative at \( x = 2 \)

1. **Left-hand derivative at \( x = 2 \):**
\[ f'(2^-) = 6(2) + 12 = 24 \]

2. **Right-hand derivative at \( x = 2 \):**
\[ f'(2^+) = -1 \]

Since \( f'(2^-) \neq f'(2^+) \), the derivative does not exist at \( x = 2 \).

### Conclusion

The function \( f(x) \) is continuous at \( x = 2 \), increasing on \( [-1, 2] \), decreasing on \( (2, 3] \), has a minimum value of 35 at \( x = 2 \), and the derivative does not exist at \( x = 2 \).