$\Big| \dfrac{df}{dx}(x)\Big|\leq 5$ for all $x$

Question

$f:\mathbb{R}\rightarrow \mathbb{R}$ 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

1. $(5,6)$

2. $[-5,5]$

3. $(-\infty,-5)\cup (5,\infty)$

4. $[-4,4]$

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.

Answer 1

Take $f(x)=5x$ for $5$ and $f(x)=-5x$ for $-5$.

Answer 2

Mean value Theorem:

$\dfrac{f(1) -f(0)}{1-0} = f'(t)$, with $t \in (0,1).$

Hence:

$|f(1)| = |f'(t)| \le 5.$

$\Rightarrow: f(1) \in [-5,5].$

Answer 3

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!)

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"...$f(1)=\int_{0}^{1}f'(x)dx$. 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 $f(x)=-5x$."