I have a function $f(x)=x+\sin x$ and I want to prove that it is strictly increasing. A natural thing to do would be examine $f(x+\epsilon)$ for $\epsilon > 0$, and it is equal to $(x+\epsilon)+\sin(x+\epsilon)=x+\epsilon+\sin x\cos \epsilon + \sin \epsilon \cos x$.
Now all I need to prove is that $x+\epsilon+\sin x\cos \epsilon + \sin \epsilon \cos x – \sin x – x$ is always greater than $0$ but it's a dead end for me as I don't know how to proceed. Any hints?
Best Answer
Let $f(x)=x+\sin x$. Then $f'(x)=1+\cos x\geq 0$ and:
$$ f(x+h)-f(x) = h\, f'(\xi),\quad \xi\in(x,x+h) $$ by Lagrange's theorem, hence $f(x+h)-f(x)\geq 0$.
In order to prove that the inequality is strict, we can notice that: $$ f(x+h)-f(x-h) = 2h + 2\cos x \sin h $$ can be zero only if $\cos x=-1$ and $\sin h=h$, i.e. for $h=0$.