# Prove that this function is nonnegative

PeroK
Homework Helper
Gold Member
nothing special.
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)

Math_QED
Mark44
Mentor
nothing special.
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?

sergey_le and Math_QED

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?

Mark44
Mentor
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.

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

Mark44
Mentor
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.

why x=3/2+πk is not a solution?

PeroK
Homework Helper
Gold Member
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.

Last edited by a moderator:
sergey_le
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

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

PeroK
Homework Helper
Gold Member
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.

JD_PM and sergey_le