- #36
- 28,084
- 19,596
I thought it might be worth adding an outline solution to this.
We need to show that ##f(x) > 0 \ ## when ##x \ne 0##, where ##f(x) = \frac 1 2 x^2 - x\cos x + \sin x##.
Note that ##f## is symmetric about the y-axis (##f(x) = f(-x)##), so it is enough to show that ##f(x) > 0## for ##x > 0##. Note also that ##f(0) = 0##.
##f'(x) = x + x\sin x = x(1 + \sin x)##
Therefore ##f'(x) \ge 0## for ##x > 0##, hence (by the mean value theorem) ##f(x) \ge 0## for ##x > 0##.
Note that ##f'(x) = 0## when ##x = \frac 3 2 \pi + 2n\pi##. In particular, ##f'(x) > 0## for ##x \in (0, \pi]##. Therefore, by the mean value theorem, ##f(x) > 0## for ##x \in (0, \pi]##. And, of course, ##f(\pi) > 0##.
Finally, for ##x > \pi## we have ##f'(x) \ge 0##, hence ##f(x) \ge f(\pi) > 0##; and the result follows.
We need to show that ##f(x) > 0 \ ## when ##x \ne 0##, where ##f(x) = \frac 1 2 x^2 - x\cos x + \sin x##.
Note that ##f## is symmetric about the y-axis (##f(x) = f(-x)##), so it is enough to show that ##f(x) > 0## for ##x > 0##. Note also that ##f(0) = 0##.
##f'(x) = x + x\sin x = x(1 + \sin x)##
Therefore ##f'(x) \ge 0## for ##x > 0##, hence (by the mean value theorem) ##f(x) \ge 0## for ##x > 0##.
Note that ##f'(x) = 0## when ##x = \frac 3 2 \pi + 2n\pi##. In particular, ##f'(x) > 0## for ##x \in (0, \pi]##. Therefore, by the mean value theorem, ##f(x) > 0## for ##x \in (0, \pi]##. And, of course, ##f(\pi) > 0##.
Finally, for ##x > \pi## we have ##f'(x) \ge 0##, hence ##f(x) \ge f(\pi) > 0##; and the result follows.