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

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 $\left[-\frac{1}{2}, 1 \right]$.