Question

If \( f(x) = x e^{x(1-x)}, \, x \in \mathbb{R} \), then \( f(x) \) is:

• A. increasing on \( \left[-\frac{1}{2}, 1\right] \)
• B. decreasing on \( \mathbb{R} \)
• C. increasing on \( \mathbb{R} \)
• D. decreasing on \( \left[-\frac{1}{2}, 1\right] \)

Hint:

When checking intervals of increase or decrease, always find the derivative, set it to zero to get critical points, and solve the inequality for where the derivative is positive or negative.

Answer 1

The Correct Option is A

We are asked to determine the interval where the function \[ f(x) = x e^{x(1-x)} \] is increasing. Step 1: Find the derivative of \( f(x) \) Using the product rule: \[ f'(x) = \frac{d}{dx} \left(x e^{x(1-x)}\right) = e^{x(1-x)} + x . e^{x(1-x)} . (1-2x) \] Simplify: \[ = e^{x(1-x)} \left(1 + x(1-2x)\right) \] Simplify inside the bracket: \[ = e^{x(1-x)} \left(1 + x - 2x^2\right) = e^{x(1-x)} \left(-2x^2 + x + 1\right) \] Step 2: Find the interval where \( f'(x) \geq 0 \) Since \( e^{x(1-x)} \) is always positive, the sign of \( f'(x) \) depends on: \[ -2x^2 + x + 1 \geq 0 \] Step 3: Solve the inequality Consider: \[ -2x^2 + x + 1 = 0 \] Use quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] with \( a = -2, b = 1, c = 1 \) \[ \Delta = 1^2 - 4(-2)(1) = 1 + 8 = 9 \] \[ x = \frac{-1 \pm 3}{-4} \] So, \[ x = \frac{-1 + 3}{-4} = \frac{2}{-4} = -\frac{1}{2},
x = \frac{-1 - 3}{-4} = \frac{-4}{-4} = 1 \] Step 4: Determine interval of increase Since it's a downward opening parabola (as \( a = -2 \)), the inequality \[ -2x^2 + x + 1 \geq 0 \] holds between the roots: \[ x \in \left[-\frac{1}{2}, 1\right] \] % Final Answer Hence, \( f(x) \) is increasing on \( \left[-\frac{1}{2}, 1\right] \).