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

Given:

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

Taking the natural logarithm:

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

Differentiating:

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

For \( f(x) \) to be strictly increasing:

\[ \frac{dy}{dx} > 0 \implies 1 + \ln x > 0. \]

Solve:

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

Thus, the function is strictly increasing in:

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

Answer 2

To determine the interval where the function \(f(x) = x^x\) is strictly increasing for \(x>0\), we need to analyze its derivative. The function \(f(x) = x^x\) can be rewritten using the exponential function:

\(x^x = e^{x\ln x}\).

Finding the derivative of \(f(x)\) using the product and chain rule, we have:

\(\frac{d}{dx}(x^x) = \frac{d}{dx}(e^{x\ln x}) = e^{x\ln x} \cdot \frac{d}{dx}(x\ln x).\)

Now, compute \(\frac{d}{dx}(x\ln x)\) by applying the product rule:

\(\frac{d}{dx}(x\ln x) = 1\cdot\ln x + x\cdot\frac{1}{x} = \ln x + 1.\)

Therefore, the derivative is:

\(f'(x) = e^{x\ln x} \cdot (\ln x + 1) = x^x \cdot (\ln x + 1).\)

For the function to be strictly increasing, \(f'(x) > 0\).

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

Thus, \(\ln x + 1 > 0\) implies:

\(\ln x > -1.\)

Solving for \(x\), we get:

\(x > e^{-1} = \frac{1}{e}.\)

Therefore, the function is strictly increasing in the interval \(\left(\frac{1}{e}, \infty \right)\).

When considering closed intervals, note that as \(x\) approaches \(\frac{1}{e}\), \(\ln x + 1 = 0\), making the function neither increasing nor decreasing at that specific point. Therefore, the function starts increasing strictly from \(x = \frac{1}{e}\) onwards.

Hence, the interval where the function \(f(x)\) is strictly increasing is \(\left[ \frac{1}{e}, \infty \right)\). Therefore, the correct answer is \(\left[ \frac{1}{e}, \infty \right)\).