$f:\mathbb R\to \mathbb R$ be a differentiable function such that $f(n)=n$ for all $n\in \mathbb Z$ then $\begin{align} 1.\ lim_{x\to \infty}f(x)=\infty \\2.\ f \ is \ unbounded \\3.\ f(\mathbb R)=\mathbb R \\4. \ \exists x_n\in (n,n+1) \ such \ that \ f'(x_n)=1 \end{align}$.
$\textbf{Which are the correct?}$
$\textbf{My attempts:}$
$1,2,4$ are correct .
Because:
(1)If, $x\in \mathbb Z$ then we are done!. Now take, $x\in (n,n+1)$, as $x\to \infty \implies n\to \infty$, since $f$ continuous so, $f(n)\to \infty \implies f(x)\to \infty$.
(2) By (1) we can say (2) is correct.
(4) Take the interval $[n,n+1]$, since $f$ is differentiable so by Lagranges MVT $\exists x\in (n,n+1)$ such that $\frac{f(n+1)-f(n)}{(n+1)-n}=f'(x_n)$. So, we are done!.
But (3) I could not prove or disprove. Can anyone please help me in this regard?
Thanks in advance.
Best Answer
As already mentioned in the comments $1.$ need not be true ,
For $3.$ Hint: Use the intermediate value theorem since if $f$ differentiable then its continuous .