1

Let $f : \mathbb{R} \to \mathbb{R}$ be differentiable all over $\; \mathbb{R}$ .

Also, $\;\;\lim_\limits{x \to \infty}f(x) = \lim_\limits{x \to -\infty}f(x) = \infty$

prove that there exists $c\in\mathbb{R}$ such that $f'(c) = 0$.

First of all, I sincerely apologize for my poor formatting and usage of Latex.

Now to the point.

I want to prove that statement by showing that $f$ is not one-to-one and therefore show that there are $a,b \in\mathbb{R}$ such that $a\neq b$ and $f(a) = f(b)$ and afterwards use Rolle's theorem and finish.

How could I show that?

Bernard
  • 175,478
GoodWilly
  • 905
  • Tip: "$$$\lim_{x \to \infty}$$$" produces $\lim_{x \to \infty}$. – xbh Jun 30 '19 at 14:09
  • 6
    Denote $f(0) = k$. Since the limit at both $\infty$ and $-\infty$ of the function is $\infty$ then there is $n_1 < 0$ such that $f(n_1) > k+1$ and there is $n_2 > 0$ such that $f(n_2)>k+1$. From the intermediate value theorem you will have two different points $n_1<x_1 < 0, 0<x_2<n_2$ such that $f(x_1) = f(x_2) = k+1$ – Gabi G Jun 30 '19 at 14:09
  • 1
    Note that once you have shown the existence of $x_1<0<x_2$ such that $f(x_1)=f(x_2)$ as in the previous comment, you already have the $a$ and $b$ you were looking for, and there is no need to refer to the definition of a one-to-one function. – David K Jun 30 '19 at 14:15
  • The derivability is not mandatory, coercivity and continuity suffice. https://math.stackexchange.com/questions/2240269/coercive-continuous-function-on-a-closed-subset-has-a-global-minimum-proof/2240390#2240390 – zwim Jun 30 '19 at 15:54

2 Answers2

-1

Since $f(x)$ is differentiable over $R$, it follows that $f(x)$ is continuous over $R$; and so, because of this and the two limits given, we can find two real numbers $a, b$ such that $a \neq b$ and $f(a) = f(b)$ (and so, f is not 1-1 as you wanted to show). Now, finish off the problem by applying the Mean Value Theorem.

DDS
  • 3,199
  • 1
  • 8
  • 31
  • Well, the whole point is finding those $a,b$ such that $f(a) = f(b)$ so I guess that stating "because of this and the two limits given, we can find two real numbers $a,b$ such that $a≠b$ and $f(a)=f(b)$ " Doesn't work, because were directly using what we need to prove. – GoodWilly Jul 05 '19 at 15:14
-1

Let $M>0$. By the two limits $\exists c_1,c_2>0:f(x)>M, x\in (-\infty, -c_2]\cup [c_1,\infty)$. Suppose $f(c_1)\neq f(-c_2)$ (otherwise the problem has finished) and furthermore $f(c_1)\gt f(-c_2)$ without loss of generality. By applying again the definition of the second limit, for the positive $f(c_1)$ exists a $c_2'>0$, such that $f(x)\gt f(c_1), x\leq -c_2'$. Provided immediately that $f(-c_2')>f(c_1)$, the Intermediate Value Theorem gives us for $f(c_1)\in [f(-c_2),f(-c_2')]$ a $ξ\in [-c_2',-c_2]$, such that $f(ξ)=f(c_1)$, thus the Rolle' s Theorem in $[ξ,c_1]$ completes the proof. It is absolutely obvious that we found $ξ \neq c_1$ with $f(ξ)=f(c_1)$, so $f$ is not one to one.

SK_
  • 575