The function `f(x)=2x^(3)-15x^(2)+36x+1` is increasing in the interval

Options

A. `x le 2" or "x ge 3`
B. `x lt 2" or "x gt3`
C. `x ge 2" or "x le 3`
D. `x gt 2" or "x lt 3`

Answer

To determine the intervals where the function \( f(x) = 2x^3 - 15x^2 + 36x + 1 \) is increasing, we need to follow these steps:

### Step 1: Find the derivative of the function
We start by finding the first derivative \( f'(x) \) of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 1) \]

Using the power rule, we differentiate each term:

\[ f'(x) = 6x^2 - 30x + 36 \]

### Step 2: Set the derivative equal to zero
To find the critical points, we set the derivative equal to zero:

\[ 6x^2 - 30x + 36 = 0 \]

### Step 3: Simplify the equation
We can simplify this equation by dividing all terms by 6:

\[ x^2 - 5x + 6 = 0 \]

### Step 4: Factor the quadratic equation
Next, we factor the quadratic equation:

\[ (x - 2)(x - 3) = 0 \]

### Step 5: Find the critical points
Setting each factor equal to zero gives us the critical points:

\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]

### Step 6: Test intervals around the critical points
To determine where the function is increasing, we need to test the intervals defined by the critical points \( x = 2 \) and \( x = 3 \). The intervals to test are:

1. \( (-\infty, 2) \)
2. \( (2, 3) \)
3. \( (3, \infty) \)

We can choose test points from each interval:
- For \( x < 2 \), let’s choose \( x = 0 \).
- For \( 2 < x < 3 \), let’s choose \( x = 2.5 \).
- For \( x > 3 \), let’s choose \( x = 4 \).

### Step 7: Evaluate the derivative at the test points
1. **For \( x = 0 \)**:
\[ f'(0) = 6(0)^2 - 30(0) + 36 = 36 \quad (\text{positive}) \]
This means \( f(x) \) is increasing on \( (-\infty, 2) \).

2. **For \( x = 2.5 \)**:
\[ f'(2.5) = 6(2.5)^2 - 30(2.5) + 36 = 6(6.25) - 75 + 36 = 37.5 - 75 + 36 = -1.5 \quad (\text{negative}) \]
This means \( f(x) \) is decreasing on \( (2, 3) \).

3. **For \( x = 4 \)**:
\[ f'(4) = 6(4)^2 - 30(4) + 36 = 6(16) - 120 + 36 = 96 - 120 + 36 = 12 \quad (\text{positive}) \]
This means \( f(x) \) is increasing on \( (3, \infty) \).

### Step 8: Conclusion
The function \( f(x) \) is increasing in the intervals \( (-\infty, 2) \) and \( (3, \infty) \).

### Final Answer
The function \( f(x) = 2x^3 - 15x^2 + 36x + 1 \) is increasing in the intervals \( (-\infty, 2) \) and \( (3, \infty) \).

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