Question

The function $x^x$, $x > 0$ is strictly increasing at:

• A. $\forall x \in \mathbb{R}$
• B. $x < \frac{1}{e}$
• C. $x > \frac{1}{e}$
• D. $x < 0$

Answer 1

The Correct Option is C

1. Understand the problem:

We need to determine where the function \( f(x) = x^x \) (for \( x > 0 \)) is strictly increasing.

2. Use logarithmic differentiation:

Let \( y = x^x \). Take the natural logarithm:

\[ \ln y = x \ln x \]

Differentiate implicitly with respect to \( x \):

\[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \]

Thus:

\[ f'(x) = y (\ln x + 1) = x^x (\ln x + 1) \]

3. Determine where \( f'(x) > 0 \):

Since \( x^x > 0 \) for \( x > 0 \), the sign of \( f'(x) \) depends on \( \ln x + 1 \):

\[ \ln x + 1 > 0 \implies \ln x > -1 \implies x > e^{-1} = \frac{1}{e} \]

Correct Answer: (C) \( x > \frac{1}{e} \)

Answer 2

Rewrite $y = x^x$ as $y = e^{x \ln x}$. Differentiate: \[ \frac{dy}{dx} = y (\ln x + 1). \] The sign of $\ln x + 1$ determines monotonicity. Solve $\ln x + 1 > 0$: \[ \ln x > -1 \implies x > \frac{1}{e}. \] Thus, $x^x$ is strictly increasing for $x > \frac{1}{e}$.