The absolute maximum value of \(f(x) = x + 2\cos x\) on the interval \([0, \pi]\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{\pi}{6} + \sqrt{3}\)
Setting the derivative equal to zero:
\(f'(x) = 1 - 2\sin x = 0\)
\( \displaystyle \sin x = \frac{1}{2} \)
So the critical points in \([0, \pi]\) are \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).
Comparing f at these critical points and at the end points:
\(f(0) = 0 + 2 = 2\)
\( \displaystyle f\left(\frac{\pi}{6}\right) = \frac{\pi}{6} + 2 \times \frac{\sqrt{3}}{2} \)
\( \displaystyle = \frac{\pi}{6} + \sqrt{3} \approx 2.26 \)
\( \displaystyle f\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} - \sqrt{3} \approx 0.89 \)
\(f(\pi) = \pi - 2 \approx 1.14\)
The largest of these is \(\displaystyle\frac{\pi}{6} + \sqrt{3}\).
Why the other options are incorrect
Option (1) 2 — Incorrect. This is f(0), the value at the left end point, which is less than \(f\left(\frac{\pi}{6}\right) \approx 2.26\).
Option (2) \(\pi - 2\) — Incorrect. This is f(π), the value at the right end point, about 1.14.
Option (4) \(\displaystyle\frac{5\pi}{6} - \sqrt{3}\) — Incorrect. This is the value at the other critical point, a local minimum, about 0.89.