If `f(x)={:{(3x^2+12x-1"," -1le x le2),(37-x ","2 lt x le 3):}` then

Options

A. f(X) increasing on [-1,2]
B. f(x) is continuos on [-1,3]
C. f'(2) does not exist
D. all of these

Explanation

To solve the given problem, we need to analyze the piecewise function \( f(x) \) defined as follows:

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

### Step 1: Differentiate \( f(x) \)

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

2. For the second piece \( f(x) = 37 - x \):
\[ f'(x) = \frac{d}{dx}(37 - x) = -1 \]

### Step 2: Analyze the derivative in the intervals

1. For \( -1 \leq x \leq 2 \):
\[ f'(x) = 6x + 12 \]
- At \( x = -1 \):
\[ f'(-1) = 6(-1) + 12 = 6 \quad (\text{positive}) \]
- At \( x = 2 \):
\[ f'(2) = 6(2) + 12 = 24 \quad (\text{positive}) \]
- Since \( f'(x) \) is a linear function that increases from 6 to 24, \( f'(x) > 0 \) for all \( x \) in the interval \( -1 \leq x \leq 2 \). Thus, \( f(x) \) is increasing in this interval.

2. For \( 2 < x \leq 3 \):
\[ f'(x) = -1 \quad (\text{constant and negative}) \]
- This means \( f(x) \) is decreasing in the interval \( 2 < x \leq 3 \).

### Step 3: Check continuity at the boundary \( x = 2 \)

1. Calculate \( f(2) \):
\[ f(2) = 3(2)^2 + 12(2) - 1 = 12 + 24 - 1 = 35 \]

2. Calculate \( f(2^+) \):
\[ f(2^+) = 37 - 2 = 35 \]

3. Since \( f(2) = f(2^+) \), the function is continuous at \( x = 2 \).

### Step 4: Check differentiability at \( x = 2 \)

1. Calculate the left-hand derivative \( f'(2^-) \):
\[ f'(2^-) = 6(2) + 12 = 24 \]

2. Calculate the right-hand derivative \( f'(2^+) \):
\[ f'(2^+) = -1 \]

3. Since \( f'(2^-) \neq f'(2^+) \), the function is not differentiable at \( x = 2 \).

### Conclusion

- \( f(x) \) is increasing in the interval \( -1 \leq x \leq 2 \).
- \( f(x) \) is continuous in the interval \( -1 \leq x \leq 3 \).
- \( f(x) \) is not differentiable at \( x = 2 \).