Question

If \(f(x) = x e^{x(1-x)}\), 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:

The sign of derivative determines increasing/decreasing behaviour.

Answer 1

The Correct Option is A

To determine if the function \( f(x) = x e^{x(1-x)} \) is increasing or decreasing, we need to find the first derivative and analyze its sign over the intervals in question. 

The first derivative \( f'(x) \) will help us determine the intervals where the function is increasing (if \( f'(x) > 0 \)) or decreasing (if \( f'(x) < 0 \)).

Let us begin by differentiating the function:

Given,

\(f(x) = x e^{x(1-x)}\)

Applying product rule to differentiate \( f(x) = x \cdot g(x) \), where \( g(x) = e^{x(1-x)} \):

\(f'(x) = x \cdot \frac{d}{dx}(e^{x(1-x)}) + e^{x(1-x)} \cdot \frac{d}{dx}(x)\)

First, differentiate \( g(x) = e^{x(1-x)} \) using the chain rule:

\(\frac{d}{dx}(e^{x(1-x)}) = e^{x(1-x)} \cdot \frac{d}{dx}(x(1-x))\)

\(= e^{x(1-x)} \cdot (1 - 2x)\)

Now substitute back to find \( f'(x) \):

\(f'(x) = x \cdot e^{x(1-x)} \cdot (1 - 2x) + e^{x(1-x)}\)

\(= e^{x(1-x)} \cdot (x(1 - 2x) + 1)\)

On simplifying,

\(f'(x) = e^{x(1-x)} \cdot (1 - x - 2x^2)\)

Now, we analyze the sign of \( f'(x) \) in the interval \([-1/2, 1]\). Since \( e^{x(1-x)} \) is always positive, the sign of \( f'(x) \) is determined by \( 1 - x - 2x^2 \).

Consider the quadratic \( h(x) = -2x^2 - x + 1 \).

We can factor or use the discriminant to analyze this quadratic. The roots are found using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Where \( a = -2 \), \( b = -1 \), and \( c = 1 \).

\(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(-2)(1)}}{2(-2)}\)

\(x = \frac{1 \pm \sqrt{1 + 8}}{-4}\)

\(x = \frac{1 \pm 3}{-4}\)

So, roots are \( x = -1 \) and \( x = \frac{1}{2} \).

Quadratic function \( h(x) \) is an inverted parabola (opens downwards), implying that it is positive between the roots \(-1\) and \(\frac{1}{2}\).

Thus, \( f'(x) > 0 \) in the interval \((-1, \frac{1}{2})\), making the function \( f(x) \) increasing.

This indicates that, within the domain \([-1/2, 1]\), \( f(x) \) is increasing.

Hence, the correct answer is that \( f(x) \) is increasing on \([-1/2, 1]\).