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Let $f: \mathbb{R} \to \mathbb{R}$ be a derivable function.

We know that:

$\lim_{x\to-\infty}\ f'(x) = \lim_{x\to+\infty}\ f'(x) = \frac1 2$

Prove that such function is surjective.

Ok my thoughts so far are:

Let's suppose that such a function has an upper bound. Then let $s=\sup(f)$ be such upper bound. Since it is continuous, there has to be a neighbourhood of $+\infty$ in which the derivative remains positive, and we can take it big enough to include $s$ (?). Then we can conclude that in such neighbourhood the function is increasing, which goes against the fact that s is an upper bound. Therefore the function has to not have an upper bound. A similar reasoning can be done for the lower bound and we can conclude that the function in surjective.

Pretty sure it's completely wrong, but still a try.

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    Prove that $y=\dfrac12x$ is asymptotic with function. – 2017-01-21

1 Answers 1

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We may choose $M>0$ and $N<0$ such that if $x\in (-\infty, N)\cup (M,\infty),$ then $f'(x)>0. $

Now, consider separately the intervals $(-\infty, N),\ [N,M],\ (M,\infty). $

On the first and third of these, $f$ is increasing and continuous, hence surjective.

Wlog $f(M)\ge f(N)$. Then, the IVT implies that $f([N,M])$ is onto $[f(N),f(M)],\ $

so $f$ is surjective, as required.

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    I don't think it's correct to say WLOG $f(M)\geq f(N)$. However, note that if $f(N)\geq f(M)$, then $f(x)$ is already surjective. – 2017-01-21
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    Yes, it's ok to use the wlog but if you do not like the wlog algorithm, then just consider each case separately. – 2017-01-21