`f(x)=x^(3)-9x^(2)+36x+2` is increasing in

Options

A. R
B. `phi`
C. `R^(-)`
D. `R^(+)`

Answer

To determine where the function \( f(x) = x^3 - 9x^2 + 36x + 2 \) is increasing, we need to find its derivative and analyze when this derivative is greater than zero.

### Step-by-Step Solution:

1. **Find the Derivative**:
We start by differentiating the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(9x^2) + \frac{d}{dx}(36x) + \frac{d}{dx}(2) \]
\[ f'(x) = 3x^2 - 18x + 36 \]

2. **Set the Derivative Greater Than Zero**:
To find where the function is increasing, we set the derivative greater than zero:
\[ 3x^2 - 18x + 36 > 0 \]

3. **Simplify the Inequality**:
We can simplify this inequality by dividing all terms by 3:
\[ x^2 - 6x + 12 > 0 \]

4. **Calculate the Discriminant**:
We need to determine the nature of the roots of the quadratic equation \( x^2 - 6x + 12 \) by calculating its discriminant \( D \):
\[ D = b^2 - 4ac = (-6)^2 - 4 \cdot 1 \cdot 12 = 36 - 48 = -12 \]

5. **Analyze the Discriminant**:
Since the discriminant \( D < 0 \), the quadratic has no real roots. This means the quadratic expression \( x^2 - 6x + 12 \) does not change sign and is either always positive or always negative.

6. **Determine the Sign of the Quadratic**:
The coefficient of \( x^2 \) is positive (1), which means the quadratic opens upwards. Therefore, \( x^2 - 6x + 12 > 0 \) for all \( x \).

7. **Conclusion**:
Since \( f'(x) > 0 \) for all \( x \), the function \( f(x) \) is increasing for all real numbers:
\[ \text{The function } f(x) \text{ is increasing for } x \in \mathbb{R}. \]