Question

The interval in which the function \( f(x) = x^x, \, x>0 \), is strictly increasing is:

• A. \( \left( 0, \frac{1}{e} \right] \)
• B. \( \left[ \frac{1}{e^2}, 1 \right) \)
• C. \( (0, \infty) \)
• D. \( \left[ \frac{1}{e}, \infty \right) \)

Answer 1

The Correct Option is D

Function:

\[ f(x) = x^x, \quad x > 0. \]

Applying the natural logarithm:

\[ \ln f(x) = x \ln x. \]

Differentiating both sides with respect to \(x\):

\[ \frac{1}{f(x)} \frac{df}{dx} = \ln x + 1 \implies \frac{df}{dx} = f(x)(\ln x + 1) = x^x(1 + \ln x). \]

For \( f(x) \) to be strictly increasing, the derivative must be positive:

\[ \frac{df}{dx} > 0 \implies x^x(1 + \ln x) > 0. \]

Since \( x^x > 0 \) for \( x > 0 \), this simplifies to:

\[ 1 + \ln x > 0. \]

Solving for \(x\):

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

Therefore, the function \( f(x) \) is strictly increasing on the interval:

\[ \left[\frac{1}{e}, \infty\right). \]