If `f(x)=xe^(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]`

Answer

To determine whether the function \( f(x) = x e^{x(x-1)} \) is increasing or decreasing, we will follow these steps:

### Step 1: Find the derivative \( f'(x) \)
We start by differentiating \( f(x) \) using the product rule. The product rule states that if \( f(x) = u(x) v(x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \).

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

Now we differentiate \( u(x) \) and \( v(x) \):
- \( u'(x) = 1 \)
- To differentiate \( v(x) \), we use the chain rule:
\[ v'(x) = e^{x(x-1)} \cdot \frac{d}{dx}(x(x-1)) = e^{x(x-1)} \cdot (2x - 1) \]

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

### Step 2: Simplify the derivative
Now we simplify the expression:
\[ f'(x) = e^{x(x-1)}(1 + 2x^2 - x) \]
\[ = e^{x(x-1)}(2x^2 - x + 1) \]

### Step 3: Analyze the sign of \( f'(x) \)
Since \( e^{x(x-1)} \) is always positive for all real \( x \), the sign of \( f'(x) \) depends on the quadratic \( 2x^2 - x + 1 \).

### Step 4: Determine if the quadratic is always positive
To check if the quadratic \( 2x^2 - x + 1 \) is always positive, we calculate its discriminant \( D \):
\[ D = b^2 - 4ac = (-1)^2 - 4 \cdot 2 \cdot 1 = 1 - 8 = -7 \]
Since the discriminant is negative, the quadratic has no real roots and opens upwards (as the coefficient of \( x^2 \) is positive). Thus, \( 2x^2 - x + 1 > 0 \) for all \( x \).

### Step 5: Conclusion about \( f'(x) \)
Since \( f'(x) = e^{x(x-1)}(2x^2 - x + 1) \) is a product of two positive quantities, we conclude that:
\[ f'(x) > 0 \quad \text{for all } x \in \mathbb{R} \]
This means that \( f(x) \) is increasing for all real \( x \).

### Final Answer
Thus, the function \( f(x) = x e^{x(x-1)} \) is **increasing on \( \mathbb{R} \)**.

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