Prove that this function is nonnegative

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  • #26
PeroK
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nothing special.
Please direct me
This is painful. Things to try:

1) Sketch a graph of the function.

2) Analyse the derivative.

3) Analyse the points where the derivative is zero: local min, max, saddle point?

4) Look at the second derivative ##f''(x)##.

5) Use the fact that ##f## never decreases. If it did ##f'(x)## would be less than zero at some points.

6) What happens for small ##x##?

7) Have you noticed the function is symmetric about the y-axis?

You have to be prepared to do some of the thinking here.
7)
 
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  • #27
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nothing special.
Please direct me
I already did.
what are the values of f(x) at the points where f'(x) = 0?
You found f'(x) in post #1 -- f'(x) = x(1 + sin(x)).

Are you saying you don't know how to solve the equation x(1 + sin(x)) = 0?
 
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  • #28
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I already did.

You found f'(x) in post #1 -- f'(x) = x(1 + sin(x)).

Are you saying you don't know how to solve the equation x(1 + sin(x)) = 0?
You mean I have to show that there is no f' axis with x axis?
 
  • #29
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You mean I have to show that there is no f' axis with x axis?
I have no idea what that means.
I'm asking whether you can solve the equation x(1 + sin(x)) = 0. This is something you should have done in an algebra/trig class.
 
  • #30
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I have no idea what that means.
I'm asking whether you can solve the equation x(1 + sin(x)) = 0. This is something you should have done in an algebra/trig class.
Yes I know how to solve it
x(1 + sin(x)) = 0 when
x=0 or x=3/2+πk
 
  • #31
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Yes I know how to solve it
x(1 + sin(x)) = 0 when
x=0 or x=3/2+πk
x = 0, yes, but x=3/2+πk is not a solution.
 
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why x=3/2+πk is not a solution?
 
  • #33
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why x=3/2+πk is not a solution?
I guess you meant
I guess you meant ##x = \frac{3\pi}{2} + 2\pi k##?
So, @sergey_le, now you know that f'(x) = 0 for x = 0 or for ##x = \frac{3\pi}{2} + 2\pi k##.
What are the values of f(x) at these x values?

You also know that f'(x) > 0 for ##x \ne 0## and for ##x \ne \frac{3\pi}{2} + 2\pi k##.

Remember that you're trying to show that f(x) > 0 for x > 0 as part of the problem.
 
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  • #34
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I guess you meant ##x = \frac{3\pi}{2} + 2\pi k##?
Yes I'm sorry.
I don't know why I make such mistakes
 
  • #35
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I guess you meant

So, @sergey_le, now you know that f'(x) = 0 for x = 0 or for ##x = \frac{3\pi}{2} + 2\pi k##.
What are the values of f(x) at these x values?

You also know that f'(x) > 0 for ##x \ne 0## and for ##x \ne \frac{3\pi}{2} + 2\pi k##.

Remember that you're trying to show that f(x) > 0 for x > 0 as part of the problem.
Thanks so much I finally realized
 
  • #36
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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.
 
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