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  1. Suppose $f:\mathbb R\to \mathbb R$ is differentiable and there's a constant $c>0$ such that $f'(x)>c$ for all $x\in(a,\infty)$. Prove that $\displaystyle\lim_{x\to\infty}f(x)=+\infty$

  2. Suppose $f:\mathbb R\to \mathbb R$ differentiable twice, bounded and has a minimum on $x_0$. Prove that there's a point $y\in\mathbb R$ such that $f''(y)=0$.

  1. Differentiability implies continuity. From Fermat's theorem we know that there's no extramum point and since the derivative is always positive the function is monotonically increasing.

  2. From Fermat's, $f'(x_0)=0$, $f'$ can't be negative for $x>x_0$ so it's definitely positive but if it will always be positive it won't be bounded like in 1. so there's another point $x_1$ where $f'(x_1)=0$, and from Roll's we get there must be a point $y$ such that $f''(y)=0$.

I have a question about 2. what is the definition for the derivative or slope on the bound ? Is it because there's no two sided limits on a bound that there's no derivative there ? If so then how can I in claim that $f'(x_1)=0$ if $x_1$ is the bound itself ?

Note: no integrals.

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  • $\begingroup$ Should it be $c>0$ in the first question? $\endgroup$ – angryavian Jun 18 '14 at 20:00
  • $\begingroup$ Yes, fixed. ${}$ $\endgroup$ – GinKin Jun 18 '14 at 20:00
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    $\begingroup$ 1. A monotonically increasing function need not have infinite limit at $\infty$ (take $f(x)=1-e^{-x}$). 2. $f'(x)$ could be positive for all $x>x_0$; your $x_1$ need not exist. $\endgroup$ – David Mitra Jun 18 '14 at 20:22
  • $\begingroup$ @DavidMitra, Is that a counter example for 1 or I was just totally wrong ? and for 2. if it's always positive, then how there could be another point where $f'(x)=0$ ? $\endgroup$ – GinKin Jun 18 '14 at 20:36
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    $\begingroup$ I'm pointing out your arguments are, I'm afraid, flawed. 1. is true. You could use the Fundamental Theorem of Calculus to prove it ($ f(x)=\int_{a}^x f'(t)\,dt > c (x-a)$). In 2., there may not be another point $x$ where $f'(x)=0$. A function can have a second derivative of value $0$ at some point with, still, the first derivative always positive ($f''(x)=0$ at an "extremal point" $x$ of $f'$). See the answer below for a solution for 2. Essentially, assuming $f''(x_0)>0$, you need to argue that $f'$ isn't increasing on all of $[x_0,\infty)$. $\endgroup$ – David Mitra Jun 18 '14 at 20:53
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I would approach 2. as follows:

  • If $f''(x_0)=0$ there is nothing to be proved.
  • And, of course, $f''(x_0)<0$ is absurd since $f$ has a minimum at $x_0$.
  • So, we may suppose that $f''(x_0)>0$. Now, If $f''(t)\ne 0$ for all $t\geq x_0$ then by the intermediate value property (Darboux's Theorem) we will have $f''(t)>0$ for all $t\geq x_0$. Thus, $f'$ is strictly increasing on $[x_0,+\infty)$.

    So if $x_1>x_0$ and we set $c=f'(x_1)>f'(x_0)=0$ we obtain $f'(x)>c$ for all $x>x_1$ and this implies that $\lim_{x\to\infty} f(x)=+\infty $ according to 1. which is a contradiction since $f$ is assumed to be bounded. Consequently $f''$ must vanish at some point from the interval $[x_0,+\infty)$. ${}{}{}$

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    $\begingroup$ I may be misreading, but $f''$ need not be continuous; the Intermediate Value Theorem then doesn't apply. You could use Darboux's Theorem, though. $\endgroup$ – David Mitra Jun 18 '14 at 20:24
  • $\begingroup$ @DavidMitra, You are right, I didn't want to cite Darboux's Theorem because I thought $f$ is twice continuously differentiable. But Darboux's theorem will do. I will edit the post accordingly. Thanks. $\endgroup$ – Omran Kouba Jun 18 '14 at 20:31
  • $\begingroup$ We posted at 5 sec difference. If $f(x)=x^2, f'(x)=2x, f''(x)=2$ on the interval $(-1,1)$ then where there could be another point where $f'(x_1)=0$ which implies from roll there's $x_2\in (x_0,x_1): f''(x_2)=0$ ? $\endgroup$ – GinKin Jun 19 '14 at 9:21
  • $\begingroup$ Can't we bind it on some interval ? Or in other words does the function itself has to be bounded for all $x\in \mathbb R$ ? $\endgroup$ – GinKin Jun 19 '14 at 9:34
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    $\begingroup$ @GinKin Of course, that is the point, your example shows that the conclusion is not correct if we replace $\mathbb{R}$ by $(-1,1)$. $\endgroup$ – Omran Kouba Jun 19 '14 at 9:51

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