$\Big| \dfrac{df}{dx}(x)\Big|\leq 5$ for all $x$ 
$f:\mathbb{R}\rightarrow \mathbb{R}$ is such that $f(0)=0$ and $\Big| \dfrac{df}{dx}(x)\Big|\leq 5$ for all $x$. We can conclude that $f(1)$ is in

*

*$(5,6)$


*$[-5,5]$


*$(-\infty,-5)\cup (5,\infty)$


*$[-4,4]$

The answer would be 2. (You can also find a solution here and here and here)
My question is: Is there an example of $f$ for which $f(1)$ can be $5$ or $-5$?
For example if we take $f(x)=5\sin x$, $f(1)$ is not $5$(obviously), but we can not take $f(x)=5\sin x +c$, where $c\neq 0$, since $f(0)=0$.
Can someone help me here? I am very bad at finding examples. I guess someone has to start with choosing $f(0)=0$ and $f(1)=5$, but what would be the next steps? Thanks.
Added:
So stupid, thanks to Kenny Lau, I understood that. What about if I add an extra condition that $f$ has non-constant derivative.

In the comment section of Kenny Lau's answer, an user Shalop wrote exactly the answer what I wanted, and I think he is right. I told him to post as an answer so that I can accept it, but he did not do that. So, I am writing exactly the same thing what he wrote. (If he reconsiders to an answer in detail, and posts an answer, I will delete my answer!)

"...$f(1)=\int_{0}^{1}f'(x)dx$. So if $f(1)=5$, that implies that $f'(x)=5$ a.e (by this question since $5-f'\geq 0$ and $\int (5-f')=0$). Hence $f(x)=5x$. Symmetrically, if $f(1)=-5$, then $f(x)=-5x$."
