[Math] mean value theorem: $ | \cos x-1 | \leq | x | $

Question

I've been studying mean value theorem, and I was told that I should use MTV for solving problems with $\leq$. So I've been trying to show that $ | \cos x – 1 | \leq | x | $ for all x values, using MTV, but I just don't get how MTV can be used…
I've been stuck on this for days now and I would be really grateful for any help.

Answer 1

For $x\ne0$ you should know that MVT says $$\frac{\cos x-\cos0}{x-0}=\cos'\xi$$ for some $\xi\in(0,x)$ or $(x,0)$ depending on the sign of $x$.

Therefore $|\cos x-1|=|\cos x-\cos 0|=|x-0||\cos'\xi|=|x||\!\!-\!\sin\xi|\le|x|\cdot1=|x|.$

Verification for $x=0$ is trivial.