If the largest interval of x in which the function `f(x)=x^(3)-3x+1` is decreasing is (a, b), then the value of `a+2b` is equal to

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To solve the problem, we need to find the interval where the function \( f(x) = x^3 - 3x + 1 \) is decreasing. This can be determined by analyzing the derivative of the function.

### Step-by-Step Solution:

1. **Find the derivative of the function**:
\[ f'(x) = \frac{d}{dx}(x^3 - 3x + 1) \]
Using the power rule:
\[ f'(x) = 3x^2 - 3 \]

2. **Set the derivative less than zero**:
To find where the function is decreasing, we need to solve:
\[ f'(x) < 0 \]
This gives us:
\[ 3x^2 - 3 < 0 \]

3. **Factor the inequality**:
We can factor out the 3:
\[ 3(x^2 - 1) < 0 \]
This simplifies to:
\[ x^2 - 1 < 0 \]
Factoring further, we have:
\[ (x - 1)(x + 1) < 0 \]

4. **Determine the critical points**:
The critical points from the factors are:
\[ x = -1 \quad \text{and} \quad x = 1 \]

5. **Test intervals around the critical points**:
We will test the intervals determined by the critical points:
- Interval 1: \( (-\infty, -1) \)
- Interval 2: \( (-1, 1) \)
- Interval 3: \( (1, \infty) \)

- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ f'(-2) = 3(-2)^2 - 3 = 12 - 3 = 9 \quad (\text{positive}) \]
- For \( -1 < x < 1 \) (e.g., \( x = 0 \)):
\[ f'(0) = 3(0)^2 - 3 = -3 \quad (\text{negative}) \]
- For \( x > 1 \) (e.g., \( x = 2 \)):
\[ f'(2) = 3(2)^2 - 3 = 12 - 3 = 9 \quad (\text{positive}) \]

6. **Identify the decreasing interval**:
From the tests, we find that \( f'(x) < 0 \) in the interval:
\[ (-1, 1) \]

7. **Assign values to \( a \) and \( b \)**:
Here, \( a = -1 \) and \( b = 1 \).

8. **Calculate \( a + 2b \)**:
\[ a + 2b = -1 + 2(1) = -1 + 2 = 1 \]

### Final Answer:
The value of \( a + 2b \) is \( \boxed{1} \).