For the function \(f(x) = x^x, x \gt 0\), which of the following are TRUE?
(A) \(f'(x) = x^x(1+ \log x)\)
(B) \(x=e\) is the critical point
(C) \(f\) is increasing in \(\left(\displaystyle\frac{1}{e},\infty\right)\)
(D) \(f\) is increasing in \((0, \infty)\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (C) only
Let \(y = x^x\). Taking logarithms of both sides:
\(\log y = x\log x\)
Differentiating with respect to x:
\( \displaystyle \frac{1}{y}\frac{dy}{dx} = \log x + 1 \)
\(f'(x) = x^x(1 + \log x)\)
Since \(x^x \gt 0\), the sign of f'(x) is the sign of \(1 + \log x\):
\( \displaystyle 1 + \log x = 0 \text{ at } x = \frac{1}{e} \)
\( \displaystyle f'(x) \lt 0 \text{ for } 0 \lt x \lt \frac{1}{e} \)
\( \displaystyle f'(x) \gt 0 \text{ for } x \gt \frac{1}{e} \)
So \(f'(x) = x^x(1 + \log x)\), the only critical point is \(x = \frac{1}{e}\), and f is decreasing on \(\left(0, \frac{1}{e}\right)\) and increasing on \(\left(\frac{1}{e}, \infty\right)\).
Checking the statements
(A) \(f'(x) = x^x(1+ \log x)\) — Correct. Logarithmic differentiation gives exactly this.
(B) \(x=e\) is the critical point — Incorrect. \(f'(e) = 2e^e \ne 0\); the critical point is \(x = \frac{1}{e}\).
(C) \(f\) is increasing in \(\left(\frac{1}{e},\infty\right)\) — Correct. \(1 + \log x \gt 0\) for \(x \gt \frac{1}{e}\).
(D) \(f\) is increasing in \((0, \infty)\) — Incorrect. f is decreasing on \(\left(0, \frac{1}{e}\right)\).