The function `f(x) = x^(4) - 4x` is strictly

Options

A. decreasing in `[1, infty)`
B. increasing in `(1, infty)`
C. increasing in `(-infty, 1]`
D. increasing in [-1,1]

Answer

To determine whether the function \( f(x) = x^4 - 4x \) is strictly increasing or decreasing, we will follow these steps:

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

\[ f'(x) = \frac{d}{dx}(x^4 - 4x) = 4x^3 - 4 \]

### Step 2: Factor the derivative
Next, we can factor the derivative to make it easier to analyze.

\[ f'(x) = 4(x^3 - 1) = 4(x - 1)(x^2 + x + 1) \]

### Step 3: Analyze the factors
Now we need to analyze the sign of \( f'(x) \). The factor \( x^2 + x + 1 \) is always positive since its discriminant is negative:

\[ D = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 < 0 \]

Thus, \( x^2 + x + 1 > 0 \) for all \( x \).

### Step 4: Determine the critical points
The critical point occurs when \( f'(x) = 0 \):

\[ 4(x - 1)(x^2 + x + 1) = 0 \implies x - 1 = 0 \implies x = 1 \]

### Step 5: Test intervals around the critical point
We will test the sign of \( f'(x) \) in the intervals \( (-\infty, 1) \) and \( (1, \infty) \).

1. **For \( x < 1 \)** (e.g., \( x = 0 \)):
\[ f'(0) = 4(0 - 1)(0^2 + 0 + 1) = 4(-1)(1) = -4 < 0 \]
Thus, \( f(x) \) is **decreasing** on \( (-\infty, 1) \).

2. **For \( x > 1 \)** (e.g., \( x = 2 \)):
\[ f'(2) = 4(2 - 1)(2^2 + 2 + 1) = 4(1)(4 + 2 + 1) = 4(1)(7) = 28 > 0 \]
Thus, \( f(x) \) is **increasing** on \( (1, \infty) \).

### Step 6: Conclusion
From our analysis, we conclude:
- The function \( f(x) \) is strictly decreasing on the interval \( (-\infty, 1) \).
- The function \( f(x) \) is strictly increasing on the interval \( [1, \infty) \).

### Summary of Results
- **Decreasing**: \( (-\infty, 1) \)
- **Increasing**: \( [1, \infty) \)