# $f$ is 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. 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.

• Should it be $c>0$ in the first question? Jun 18, 2014 at 20:00
• Yes, fixed. ${}$ Jun 18, 2014 at 20:00
• 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. Jun 18, 2014 at 20:22
• @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 $f'(x)=0$ ? Jun 18, 2014 at 20:36
• 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 $[x_0,\infty)$. Jun 18, 2014 at 20:53

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 $[x_0,+\infty)$. ${}{}{}$

• 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. Jun 18, 2014 at 20:24
• @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. Jun 18, 2014 at 20:31
• 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 $x_2\in (x_0,x_1): f''(x_2)=0$ ? Jun 19, 2014 at 9:21
• Can't we bind it on some interval ? Or in other words does the function itself has to be bounded for all $x\in \mathbb R$ ? Jun 19, 2014 at 9:34
• @GinKin Of course, that is the point, your example shows that the conclusion is not correct if we replace $\mathbb{R}$ by $(-1,1)$. Jun 19, 2014 at 9:51