What is the <strong>LEAST</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. -(1+\frac{\pi}{2}) ✓
- B. -(\frac{1}{2}+\frac{\pi}{2})
- C. -(1+\frac{\pi}{4})
- D. -2(\frac{1}{2}-\frac{\pi}{4})
Correct Answer: A. -(1+\frac{\pi}{2})
Explanation
To find the least value, we check the endpoints x = -\pi/2 and x = \pi/2 as well as the critical points. f(-\pi/2) = \cos(-\pi) - \pi/2 = -1 - \pi/2 = -(1 + \pi/2). This is smaller than the value at x = \pi/2 which is -1 + \pi/2, and the critical points which are positive. Thus, the minimum value is -(1 + \pi/2).
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