Are the following conditions enough for showing monotonicity in $[M,\infty]$

Question

Given:

1. $f$ has a derivative for every $x \in (0,\infty)$

2. $f'(x) > x$ for every $x>0$

Can I prove that there exists an $M \in \mathbb R$, such that $f(x+1) - f(x)$ is a monotonic increasing function in $[M,\infty]$?

From Lagrange I know that in every $I = [x,x+1]$ where $x>0$, $f(x+1) - f(x) > x$.

$f$ has derivative so $f'(x+1) - f'(x)$ is defined for $x>0$.

But I am not sure how to continue..

I know from beautiful answers to a previous question of mine that if $M = 0$ the statement is not true. Shouldn't it be true for some $[M,\infty]$ though?

Thanks!

Answer 1

Consider a function $f$ that for large $x$ (say $x>\pi$) is given by the formula $f(x) = x^2+\cos(\pi x)$. Then $f'(x) =2x-\pi\sin(\pi x)=x+(x-\pi\sin(\pi x))>x$ for $x>\pi$. Also $f'(x+1)-f'(x) = 2+2\pi\sin(\pi x)$ is negative whenever $x$ is close enough to $2k+{3\over 2}$, $k=1,2,\ldots$, so that $f(x+1)-f(x)$ is not monotone increasing on any half-line $[M,\infty)$.

Answer 2

WRONG ANSWER, as pointed out in the comments of this answer……... I think you have already answered correctly, you have to show that $g(x)=f(x+1)-f(x)$ is monotone increasing; using, as you said, lagrange you can deduce that $g(x)=f(x+1)-f(x) \gt x \\ \text{now you can use the fact that f has derivative to deriva both hands of the above inequality}\\ g'(x)=f'(x+1)-f'(x) \gt (x)'=1\\ \text{so you can conclude that g is monotone increasing, and therefore } f'(x+1)-f(x) \text{ is monotone increasing}$