The absolute minimum value of \(f(x) = x - 2\log x\) on the interval [1, e] is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(2 - 2\log 2\)
On a closed interval the absolute minimum is the least of the values at the critical points inside the interval and at the end points.
\( \displaystyle f'(x) = 1 - \frac{2}{x} = 0 \), so x = 2
Since 1 < 2 < e, the point x = 2 lies in the interval. The values are
\( f(1) = 1 - 0 = 1 \)
\( f(2) = 2 - 2\log 2 \approx 0.614 \)
\( f(e) = e - 2 \approx 0.718 \)
The least of these is \(2 - 2\log 2\).
Why the other options are incorrect
Option (1) \(e - 2\) — Incorrect. This is the value at the end point x = e, about 0.718, which is more than f(2) ≈ 0.614.
Option (2) 1 — Incorrect. This is f(1), the largest of the three values, so it is the absolute maximum.
Option (3) \(2 + 2\log 2\) — Incorrect. It has the wrong sign before 2 log 2; f(2) = 2 − 2 log 2.