If `f(x)={{:(3x^(2)+12x-1",",-1lexle2),(37-x",",2ltxle3):}`, Then

Options

A. f(x) is increasing in `[-1, 2]`
B. f(x) is continuous in `[-1, 3]`
C. f'(2) does not exist
D. f(x) has minimum at x = 2

Answer

To solve the problem step by step, we need to analyze the given piecewise function and determine the properties of \( f(x) \).

### Given Function
The function is defined as:
\[ f(x) = \begin{cases} 3x^2 + 12x - 1 & \text{for } -1 \leq x \leq 2 \\ 37 - x & \text{for } 2 < x \leq 3 \end{cases} \]

### Step 1: Determine if \( f(x) \) is increasing on the interval \((-1, 2)\)
To check if \( f(x) \) is increasing, we need to find the derivative \( f'(x) \) and check if it is greater than 0.

#### Calculation of \( f'(x) \):
For the first part of the function \( f(x) = 3x^2 + 12x - 1 \):
\[ f'(x) = \frac{d}{dx}(3x^2 + 12x - 1) = 6x + 12 \]

#### Check if \( f'(x) > 0 \):
We need to find when \( 6x + 12 > 0 \):
\[ 6x + 12 > 0 \implies 6x > -12 \implies x > -2 \]
Since \( -2 < -1 \), \( f'(x) > 0 \) for all \( x \) in the interval \((-1, 2)\). Thus, \( f(x) \) is increasing in this interval.