If `f(x)=x e^(x(x-1))`, then `f(x)` is

Options

A. increasing on`[-1/2,1]`
B. decreasing on `R`
C. increasing on `R`
D. decreasing on `[-1/2,1]`

Explanation

To determine the monotonicity of the function \( f(x) = x e^{x(x-1)} \), we will follow these steps:

### Step 1: Differentiate the function
To analyze whether the function is increasing or decreasing, we first need to find its derivative \( f'(x) \).

Using the product rule, which states that if \( f(x) = u(x)v(x) \), then \( f'(x) = u'v + uv' \), we can differentiate \( f(x) \).

Let:
- \( u = x \)
- \( v = e^{x(x-1)} \)

Now, we need to find \( u' \) and \( v' \):
- \( u' = 1 \)
- To find \( v' \), we use the chain rule:
- \( v = e^{g(x)} \) where \( g(x) = x(x-1) \)
- \( g'(x) = 2x - 1 \)
- Thus, \( v' = e^{g(x)} \cdot g'(x) = e^{x(x-1)}(2x - 1) \)

Now applying the product rule:
\[ f'(x) = u'v + uv' = 1 \cdot e^{x(x-1)} + x \cdot e^{x(x-1)}(2x - 1) \]
\[ f'(x) = e^{x(x-1)} + x e^{x(x-1)}(2x - 1) \]

### Step 2: Factor the derivative
We can factor out \( e^{x(x-1)} \):
\[ f'(x) = e^{x(x-1)} \left( 1 + x(2x - 1) \right) \]
\[ = e^{x(x-1)} \left( 1 + 2x^2 - x \right) \]
\[ = e^{x(x-1)} \left( 2x^2 - x + 1 \right) \]

### Step 3: Analyze the sign of the derivative
Since \( e^{x(x-1)} \) is always positive for all real \( x \), we only need to analyze the quadratic \( 2x^2 - x + 1 \).

To determine the roots of the quadratic, we calculate the discriminant:
\[ D = b^2 - 4ac = (-1)^2 - 4 \cdot 2 \cdot 1 = 1 - 8 = -7 \]
Since the discriminant is negative, the quadratic \( 2x^2 - x + 1 \) has no real roots and is always positive (as the coefficient of \( x^2 \) is positive).

### Step 4: Conclusion about monotonicity
Since \( f'(x) \) is positive for all \( x \) (as both factors are positive), we conclude that the function \( f(x) \) is increasing for all real \( x \).

### Final Answer
The function \( f(x) = x e^{x(x-1)} \) is increasing on the entire real line \( (-\infty, \infty) \).

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