For the function \(f(x) = x^{1/x}, x \gt 0\), which of the following are correct?
(A) \(x = 0\) is the only point where extremum may occur.
(B) The given function is maximum at \(x = e\).
(C) The function has no extreme value for \(x \gt 0\).
(D) The maximum value of the function \(f(x)\) is \(e^{1/e}\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B) and (D) only
Let \(y = x^{1/x}\). Taking logarithms,
\( \displaystyle \log y = \frac{\log x}{x} \)
By logarithmic differentiation:
\( \displaystyle \frac{1}{y}\frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \log x}{x^2} = \frac{1 - \log x}{x^2} \)
\( \displaystyle f'(x) = x^{1/x} \cdot \frac{1 - \log x}{x^2} \)
Since \(x^{1/x} \gt 0\) and \(x^2 \gt 0\), the sign of \(f'(x)\) is the sign of \(1 - \log x\):
\( \displaystyle f'(x) = 0 \text{ only at } x = e \)
\( \displaystyle f'(x) \gt 0 \text{ for } x \lt e \)
\( \displaystyle f'(x) \lt 0 \text{ for } x \gt e \)
So the only critical point is \(x = e\); f has a maximum there, with \(f(e) = e^{1/e}\).
Checking the statements
(A) \(x = 0\) is the only point where extremum may occur. — Incorrect. x = 0 is not in the domain x > 0; the only critical point is x = e.
(B) The given function is maximum at \(x = e\). — Correct. f′ changes sign from + to − at x = e.
(C) The function has no extreme value for \(x \gt 0\). — Incorrect. It has a maximum at x = e.
(D) The maximum value of the function \(f(x)\) is \(e^{1/e}\). — Correct. \(f(e) = e^{1/e}\).