In the interval (-1, 1), the function ` f(x) = x^(2) - x + 4` is :

Options

A. increasing
B. decreasing
C. neither increasing nor decreasing
D. None of the above

Answer

To determine the behavior of the function \( f(x) = x^2 - x + 4 \) in the interval (-1, 1), we will follow these steps:

### Step 1: Find the derivative of the function
The first step is to find the derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(x^2 - x + 4) \]

Calculating the derivative:

\[ f'(x) = 2x - 1 \]

### Step 2: Determine where the derivative is zero
Next, we need to find the critical points by setting the derivative equal to zero.

\[ 2x - 1 = 0 \]

Solving for \( x \):

\[ 2x = 1 \implies x = \frac{1}{2} \]

### Step 3: Analyze the sign of the derivative
Now, we will analyze the sign of \( f'(x) \) in the interval (-1, 1). We will check the intervals created by the critical point \( x = \frac{1}{2} \).

1. For \( x < \frac{1}{2} \) (for example, \( x = 0 \)):
\[ f'(0) = 2(0) - 1 = -1 \quad (\text{negative}) \]

2. For \( x > \frac{1}{2} \) (for example, \( x = 1 \)):
\[ f'(1) = 2(1) - 1 = 1 \quad (\text{positive}) \]

### Step 4: Conclusion about the function's behavior
From the analysis:
- In the interval \( (-1, \frac{1}{2}) \), \( f'(x) < 0 \), indicating that the function is **decreasing**.
- In the interval \( (\frac{1}{2}, 1) \), \( f'(x) > 0 \), indicating that the function is **increasing**.

Since the function changes from decreasing to increasing at \( x = \frac{1}{2} \), we conclude that:

**The function \( f(x) = x^2 - x + 4 \) is neither strictly increasing nor strictly decreasing in the interval (-1, 1).**

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