Prove that this function is nonnegative

[Mark44]

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.

 

[sergey_le]

why x=3/2+πk is not a solution?

 

[PeroK]

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.

 

[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

 

[sergey_le]

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]

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.