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Given:

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

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

Can I prove that $f(x+1) - f(x)$ is a monotonic increasing function?

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, How can I show that $f'(x+1) - f'(x) > 0$?

Thanks!

Edit: Thank you al for your answers! If I want to prove a weaker statement, that there exists some $M \in \mathbb R$ such that $f'(x+1) - f'(x)$ is increasing in $(M,\infty)$. Is it, then, can be proved?

Thanks again!

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  • $\begingroup$ @Gono You need $f'(x) > x$ as well... $\endgroup$ – gt6989b Jun 21 '18 at 12:41
  • $\begingroup$ read $f'(x) > 0$ instead, so I deleted my comment ;) $\endgroup$ – Gono Jun 21 '18 at 12:42
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No, you can't.

Counterexample: $f(x)=\left\{ \begin{array}{ll} 2x,&0<x<\frac{3}{2},\\ \frac{2}{3}x^2+\frac{3}{2},&x\geq \frac{3}{2}. \end{array} \right.$

Note how $f'(1.1)-f'(0.1)=0$.

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  • $\begingroup$ Your function is discontinuois at $3/2$. $\endgroup$ – Serg Jun 21 '18 at 13:10
  • $\begingroup$ @Serg Thank you for your remark. It should be continuous now. $\endgroup$ – The Phenotype Jun 21 '18 at 13:12
  • $\begingroup$ How about $f(x+1)-f(x)$ being monotonically increasing? $\endgroup$ – Adrian Keister Jun 21 '18 at 13:17
  • $\begingroup$ @AdrianKeister OP states $f'(x+1) - f'(x) > 0$, so I assume he wants strictly increasing. $\endgroup$ – The Phenotype Jun 21 '18 at 13:18
  • $\begingroup$ Oh, you're right. I misread it. $\endgroup$ – Adrian Keister Jun 21 '18 at 13:19
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You can't. Have a look at $f(x)=\frac12x^2-x^{-1}$. Then $f'(x)=x+\frac 1{x^2}>x$ for all $x>0$ and $f(x+1)-f(x)=\frac12(2x+1)+\frac 1{x^2+x}$. As this tends $\to+\infty$ as $x\to 0^+$, it cannot be increasing on all of $(0,\infty)$.

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It's wrong. A counterexample: $f: x\in [0, \infty[\mapsto 2x^3-4x^2+7x\in\mathbb{R}$.

We have $f'(x) = 6x^2-8x+7$, $f(x+1)-f(x) = 6x^2-2x+5$.

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