What is the <strong>GREATEST</strong> value of f(x)?
Direction: Consider the following for the two (02) items that follow : Let f(x)=\cos 2x+x on [-\pi/2,\pi/2].
- A. \frac{\sqrt{3}}{2}-\frac{\pi}{12}
- B. \frac{\sqrt{3}}{2}+\frac{\pi}{12} ✓
- C. \frac{\sqrt{3}}{2}+\frac{\pi}{9}
- D. \frac{\sqrt{3}}{2}+\frac{\pi}{6}
Correct Answer: B. \frac{\sqrt{3}}{2}+\frac{\pi}{12}
Explanation
Differentiating gives f'(x) = -2\sin 2x + 1 = 0, so \sin 2x = 1/2. In [-\pi/2, \pi/2], 2x \in [-\pi, \pi], which means 2x = \pi/6 or 5\pi/6. This gives critical points x = \pi/12 and x = 5\pi/12. Evaluating f(x) at x = \pi/12 gives \cos(\pi/6) + \pi/12 = \frac{\sqrt{3}}{2} + \frac{\pi}{12}, which represents the maximum value on this interval.
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