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 find the largest interval of \( x \) in which the function \( f(x) = x^3 - 3x + 1 \) is decreasing, we will follow these steps:

### Step 1: Find the derivative of the function
We first differentiate the function \( f(x) \) to find \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(x^3 - 3x + 1) = 3x^2 - 3 \]

### Step 2: Set the derivative equal to zero
To find the critical points where the function may change from increasing to decreasing, we set the derivative equal to zero:
\[ 3x^2 - 3 = 0 \]
\[ 3(x^2 - 1) = 0 \]
\[ x^2 - 1 = 0 \]
\[ (x - 1)(x + 1) = 0 \]
Thus, the critical points are:
\[ x = -1 \quad \text{and} \quad x = 1 \]

### Step 3: Determine the intervals for increasing and decreasing
We will test the intervals determined by the critical points: \( (-\infty, -1) \), \( (-1, 1) \), and \( (1, \infty) \).

1. **Interval \( (-\infty, -1) \)**:
Choose \( x = -2 \):
\[ f'(-2) = 3(-2)^2 - 3 = 12 - 3 = 9 \quad (\text{positive, increasing}) \]

2. **Interval \( (-1, 1) \)**:
Choose \( x = 0 \):
\[ f'(0) = 3(0)^2 - 3 = -3 \quad (\text{negative, decreasing}) \]

3. **Interval \( (1, \infty) \)**:
Choose \( x = 2 \):
\[ f'(2) = 3(2)^2 - 3 = 12 - 3 = 9 \quad (\text{positive, increasing}) \]

### Step 4: Identify the largest interval of decrease
From our analysis, we see that the function is decreasing in the interval \( (-1, 1) \).

### Step 5: Identify \( a \) and \( b \)
Here, \( a = -1 \) and \( b = 1 \).

### Step 6: Calculate \( a + 2b \)
Now we calculate \( a + 2b \):
\[ a + 2b = -1 + 2(1) = -1 + 2 = 1 \]

### Final Answer
Thus, the value of \( a + 2b \) is:
\[ \boxed{1} \]