If \( f(x)=x\log_e x \), \( x\gt 0 \), then the minimum value of \( f(x) \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → -1/e
By the product rule,
\( \displaystyle f'(x)=x\cdot\frac{1}{x}+\log_e x \)
\( \displaystyle =1+\log_e x \)
At a turning point f′(x) = 0:
\( \displaystyle \log_e x=-1 \)
\( \displaystyle x=e^{-1}=\frac{1}{e} \)
By the second derivative test:
\( \displaystyle f''(x)=\frac{1}{x}\gt 0\ \text{at}\ x=\frac{1}{e} \)
So f has its minimum at \( x=\frac{1}{e} \), and the minimum value is
\( \displaystyle f\left(\frac{1}{e}\right)=\frac{1}{e}\log_e\left(\frac{1}{e}\right) \)
\( \displaystyle =\frac{1}{e}\times(-1)=-\frac{1}{e} \)
Why the other options are incorrect
Option (1) -e — Incorrect. The least value of \( x\log_e x \) is \( -\frac{1}{e} \), which is greater than −e, so f never equals −e.
Option (2) e — Incorrect. f(e) = e loge e = e, so e is a value of f but not its least value.
Option (4) 1/e — Incorrect. \( \frac{1}{e} \) is the point x where the minimum occurs, not the minimum value f(x).