Question

If $f(x) = x e^{x^{1-x}}$, then $f(x)$ is:

• A. Increasing in $\mathbb{R}$
• B. Decreasing in $\mathbb{R}$
• C. Decreasing in $\left[\frac{1}{2}, 1 \right]$
• D. Increasing in $\left[-\frac{1}{2}, 1 \right]$

Answer 1

The Correct Option is D

1. Understand the problem:

We need to determine the behavior of the function \( f(x) = x e^{x(1-x)} \) by analyzing where it is increasing or decreasing.

2. Find the first derivative:

Using the product rule and chain rule:

\[ f'(x) = \frac{d}{dx}\left(x e^{x(1-x)}\right) = e^{x(1-x)} + x \cdot e^{x(1-x)} \cdot (1 - 2x) \]

Simplify the expression:

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

3. Analyze the sign of \( f'(x) \):

Since \( e^{x(1-x)} > 0 \) for all real \( x \), the sign of \( f'(x) \) depends on \( 1 + x - 2x^2 \).

4. Solve \( 1 + x - 2x^2 > 0 \):

Find the roots of the quadratic equation:

\[ -2x^2 + x + 1 = 0 \implies x = \frac{-1 \pm \sqrt{1 + 8}}{-4} = \frac{-1 \pm 3}{-4} \]

Roots: \( x = -1/2 \) and \( x = 1 \).

The quadratic \( -2x^2 + x + 1 \) is a downward-opening parabola. It is positive between its roots:

\[ f'(x) > 0 \text{ when } -\frac{1}{2} < x < 1 \]

5. Determine increasing/decreasing intervals:

The function \( f(x) \) is:

• Increasing on \( \left( -\frac{1}{2}, 1 \right) \) (since \( f'(x) > 0 \))
• Decreasing elsewhere (since \( f'(x) < 0 \) for \( x < -\frac{1}{2} \) or \( x > 1 \))

Correct Answer: (D) Increasing in \(\left[ -\frac{1}{2}, 1 \right]\)

Answer 2

Differentiate $f(x)$ using the product and chain rules: \[ f(x) = x e^{x^{1-x}} \] Let $u = x^{1-x}$, then: \[ \ln u = (1-x) \ln x \implies \frac{u'}{u} = -\ln x + \frac{1-x}{x} = -\ln x + \frac{1}{x} - 1 \] Thus, \[ u' = u \left(-\ln x + \frac{1}{x} - 1\right) \] Now, differentiate $f(x)$: \[ f'(x) = e^{u} + x e^{u} \left(-\ln x + \frac{1}{x} - 1\right) = e^{u} \left(1 + x \left(-\ln x + \frac{1}{x} - 1\right)\right) = e^{u} \left(1 - x \ln x + 1 - x\right) = e^{u} \left(2 - x \ln x - x\right) \] Analyzing the derivative: - For $x > 1$, the term $-x \ln x - x$ becomes more negative, making $f'(x)$ negative (decreasing). - For $x = 1$, $f'(1) = e^{0} (2 - 0 - 1) = e^0 (1) = 1 > 0$. - For $x < 1$, particularly in the interval $\left[-\frac{1}{2}, 1 \right]$, the derivative is positive (increasing).

Thus, $f(x)$ is increasing in $\left[-\frac{1}{2}, 1 \right]$ and decreasing elsewhere. 

Hence, the correct answer is Increasing in $\left[-\frac{1}{2}, 1 \right]$.