Question:

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

Updated On: Nov 27, 2024
  • \( \left( 0, \frac{1}{e} \right] \)
  • \( \left[ \frac{1}{e^2}, 1 \right) \)
  • \( (0, \infty) \)
  • \( \left[ \frac{1}{e}, \infty \right) \)
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The Correct Option is D

Solution and Explanation

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:

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

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