3
$\begingroup$

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?

$\endgroup$

2 Answers 2

6
$\begingroup$

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 $h=0$.

$\endgroup$
3
  • $\begingroup$ Could you please show why $\cos x=-1$ (I see why $h=0$.) $\endgroup$
    – user84413
    Commented Mar 8, 2015 at 20:35
  • $\begingroup$ @user84413: if $h>0$ and $|\cos x|<1$ then the RHS is positive for sure, since $|\sin h|\leq h$. So the only possibilities are given by $\cos x=\pm 1$, and it is quite easy to check that $\cos x=+1$ does not lead to any solution different from $h=0$, too. $\endgroup$ Commented Mar 8, 2015 at 20:37
  • $\begingroup$ @user84413: you're welcome. $\endgroup$ Commented Mar 8, 2015 at 20:40
5
$\begingroup$

differentiate: you get $ 1 + \cos (x)$ this is positive except on a discrete set of points. Integrate it and you get a strictly increasing function.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .