# Showing a function is unbounded

Let $$f(x)=x\cos x+\sin x$$. Show that for every $$M>0$$ and every $$k\geq 1$$, there is $$x_0>M$$ such that

$$f(x)\geq k$$, $$\forall x\in [x_0,x_0+\frac{1}{k}]$$.

If we replace $$x$$ by $$2\pi n$$ then

$$|f(2\pi n)=|2\pi n|$$.

Can this help?

Can I get a help to start it?

• Your idea seems very relevant! Indeed, consider that $\sin x\ge 0$ for $x\in[2\pi n,2\pi n+1]$ so $f(x)\ge x\cos x$. What else might you be able to say, for that interval. (Also, note that $1/k\le 1$ for all $k\ge 1$.) Commented Dec 17, 2021 at 3:54

Let $$x_0=2\pi n$$ where $$n\in\Bbb Z^+$$ is large enough that

$$2\pi n(\cos 1)-1\ge k$$ and $$2\pi n>M.$$

If $$x\in [x_0,x_0+1/k]$$ then $$\cos x=\cos (x-2\pi n)\ge \cos 1>0$$ so $$f(x)=x\cos x+\sin x\ge 2\pi n\cos x-|\sin x|\ge$$ $$\ge 2\pi n(\cos 1)-|\sin x|\ge$$ $$\ge 2\pi n(\cos 1)-1\ge k.$$

Let $$M>0$$ (large). Choose $$M < x= 2k \pi + l$$ , $$k \geq 1$$ and $$l \in (0,\frac{1}{k})$$.

We know that

$$\cos{(2k \pi + l)}= \cos{(2k \pi )} \cos{(l)} - \sin{( l)} \sin{(2k \pi)} = \cos(l).$$

and

$$\sin{(2k \pi + l)}= \sin{(2k \pi )} \cos{(l)} - \sin{( l)} \cos{(2k \pi)} = \sin(l).$$

Therefore,

$$f(2 k \pi + l) = (2 k \pi + l) \cos(2 k \pi + l) + \sin(2 k \pi + l)$$

$$f(x)= f(2 k \pi + l) = (2 k \pi + l) \cos(l) + \sin(l) >k\quad x=2k\pi + l \in [x,x+\frac{1}{k}].$$