Derivative of a superlinear function is surjective Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be differentiable and superlinear, i.e. $$\lim_{x\rightarrow \pm \infty} \frac{f(x)}{|x|}=+\infty$$ then $f'(\mathbb{R})= \mathbb{R}$
Don't even know how to approach the problem. Thank you.
 A: Use the Mean Value Theorem (MVT): If $f$ is differentiable on $(a,b)$ with $a<b$ and $f$ is continuous on  $[a,b]$ then $(f(b)-f(a)/(b-a)=f'(c)$ for some $c \in (a,b)$. in this Q, $f$ is differentiable everywhere so $f$ is continuous everywhere. For brevity, let $$D(a,b)= (f(b)-f(a))/(b-a) \text{ for }a<b .$$  We will show that $$\text{for each } r \in R \text{ there exist } a, b \text{ with } a<b \text{ and }  D(a,b)=r.$$.First, for $r=0$  take    $b>0$ with $f(b)>f(0)$ and $a^*<0$ with $f(a^*)>f(b)$. This is possible because  $f(x) \to \infty$ as $ x\to -\infty$.By continuity of $f$,take $a \in (a^*,0)$ with $f(a)=f(b)$.So $$D(a,b)=0.$$Second,for $ r>0 $, take any $b^*$ and consider that, for $x<b^*$ we have $D(x,b^*) \to -\infty$ as $ x \to -\infty$  with $b^*$ fixed, so for some $a<b^*$ we have $D(a,b)<r$. But $D(a,x) \to \infty$  as $x \to  \infty$  with $a$ fixed, so $g(b^{**} ,a)>r$ for any sufficiently large $b^{**}$,in particular we can take $b^{**}>b^*$. Consider the line through the point $(a,f(a))$ with slope $ r$ which we describe by the equation $ y=g(x)$ ,that is, $$g(x)=f(a)+r(x-a) \text{  for } x \in R.$$We have $$f(b^*)-g(b^*)<0<f(b^{**})-g(b^{**})$$ and  $f-g$ is continuous so for some $b \in (b^*,b^{**})$ we have $f(b)=g(b)$, which by the def'n of $g$, is equivalent to $$D(a,b)=r.$$ Third,for $r<0$, apply the second part to the function $f^*(x)=f(-x)$  : The derivative of $f^*$ ,which is $ -f'$,takes all positive values so $f'$ takes all negative values.
A: Combining all the suggestions in the comments, we can solve the problem as follows: Define $g : \Bbb{R} \to \Bbb{R}$ by
$$ g(x) = \frac{f(x) - f(0)}{x} \quad \text{if } x \neq 0 \quad \text{and} \quad g(0) = f'(0). $$
On the one hand, $g$ is continuous and
$$ \lim_{x\to\infty} g(x) = +\infty, \quad \lim_{x\to-\infty} g(x) = -\infty. $$
Thus by the intermediate value theorem, we have $g(\Bbb{R}) = \Bbb{R}$. On the other hand, for $x \neq 0$, Mean Value Theorem tells us that there exists $c = c(x)$ such that
$$ g(x) = f'(c) \quad \text{for some } c. $$
This is also true when $x = 0$ by definition of $g$. So we have $g(\Bbb{R}) \subseteq f'(\Bbb{R})$. This completes the proof.
