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Let $F$ be a differentiable function $$ F:\mathbb{R} \to \mathbb{R}$$

and $$ \lim_{x \to -\infty} F(x) = \lim_{x \to \infty} F(x) = a \in \mathbb{R}$$

Prove that there exists a ${x_0} \in \mathbb{R} $ such that $f'({x_0}) = 0$

$\textbf{My Try} $

let $ R \in \mathbb{R}$ and $R >0$

$ \lim_{x \to -\infty} F(x)= a $ $\Rightarrow$ it exists an ${x_1}$ so that for all $x <-R$ ,$ x<{x_1}$ and $f({x_1})= a + \epsilon$

$ \lim_{x \to \infty} F(x)= a $ $\Rightarrow$ it exists an ${x_2}$ so that for all $x >R$ ,$ x>{x_2}$ and $f({x_2})= a + \epsilon$

because ${x_1} <{x_2}$ $\Rightarrow$ it exist an ${x_0} \in ({x_1},{x_2})$ so that $F'(x_0) = \frac{F({x_2})-F({x_1})}{{x_2}-{x_1}}=\frac{a + \epsilon-(a + \epsilon)}{{x_2}-{x_1}}=0 \Rightarrow F'x_{0})=0$

Is that correct?

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    Your quantifiers and "variables" are wrong. You should start by fixing $\varepsilon>0$ (for simplicity, take $\varepsilon =1$ if you like): then this gives you the existence of $R \in\mathbb{R}$ such that for all $x< R$ we have $a-\varepsilon \leq f(x) \leq a+\varepsilon$. You (a) cannot choose $R$ first, and (b) what is this $x_1$ doing there? – 2017-02-04

1 Answers 1

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Your argument is not rigorous enough since such $x_1$ may not exists for an arbitrary $\epsilon$.

You may consider the following hint:

WLOG, we assume $F$ is not a constant function, otherwise it is trivial.

Then you can show $F$ obtains either the global maximum (if there exists $y\in \mathbb{R}$ such that $F(y)>a$) or global minimum (if there exists $y\in \mathbb{R}$ such that $F(y)

Since $F$ is defined on $\mathbb{R}$ (there is no end-point of the domain) and $F$ is differentiable, the global extremum must be a local extremum and at that point, $F'(x_0)=0$.

Please let me know if there is anything unclear.