# Which interval do we use at the Mean Value Theorem?

Let $$f:\mathbb{R}\rightarrow \mathbb{R}$$ be differentiable in $$\mathbb{R}$$ such that $$f'(x)>1$$ for each $$x>0$$.

I want to show that there is $$\xi >2$$ such that $$\frac{f}{x}>\frac{1}{2}$$ for each $$x>\xi$$.



For that do we have to use the mean value theorem?

But for which interval? I got confused with the $$2$$ at $$\xi>2$$.Do we have to consider for each $$x>2$$ the intervals $$[2,x]$$ to apply that theorem?

Is yes, then we have that there exists $$\xi\in (2,x)$$ such that $$\frac{f(x)-f(2)}{x-2}=f'(\xi)>1$$

How do we continue from here?

Or do we consider the intervals $$[2,x]$$ instead?

First, prove that $$f(x) > x+f(0)$$ for $$x>0$$, using the MVT. From there you easily get to $$f(x)> \frac 12x$$ for $$x>-2f(0)$$.
• Ok! I proved the first part. For the second part do we apply again the MVT? Or do we just use an inequality for $f(0)$ ? – Mary Star Sep 6 at 11:14
• $$(1) \ \ \ \ \ \sum_{n[\xi]+1}^{\infty}\frac{1}{f(n)\log^2(n+1)} \\ (2) \ \ \ \ \ \sum_{n=0}^{\infty}\frac{n^2}{(f(x))^n} \ \ \text{ with } x>\xi$$ At (1) I used again the inequality $f(n)>\frac{n}{2} \Rightarrow \frac{2}{n+1}>\frac{1}{f(n)}$ and so $\frac{1}{f(n)\log^2(n+1)}<\frac{1}{(n+1)\log^2(n+1)}$. Using integral criterion we see that the series of the right side converges and so also the left one. – Mary Star Sep 6 at 12:55
• At (2) we have for $x>\xi$ that $f(x)>\frac{x}{2}$ and so $$\frac{n^2}{(f(x))^n}<n^2\cdot \left (\frac{2}{x}\right )^n$$ Using the ratio test for the series of $n^2\cdot \left (\frac{2}{x}\right )^n$ using the fact that $\frac{2}{x}<1$ since $x>\xi>2$ we see that the series of the right side converges and so also the left one. Is everythig correct? – Mary Star Sep 6 at 12:55