롤의 정리와 평균값 정리

Since $f(x)$ is differentiable for all $x \in (0,10)$, then $f(x)$ is also continuous for all $x \in (0,10)$. If $f(x)$ were continuous on the closed interval $[0,10]$, then the MVT would guarantee $f'(x)=\dfrac{f(10)-f(0)}{10-0}=1$ for some $c \in (0,10)$; however, this is not the case. So, it must be that $f(x)$ is continuous for all $x \in (0,10)$, but not for all $x \in [0,10]$.

Since $f'(c)=0$ for $c \in (0,10)$, that means $f$ is constant on that interval. So, $f(x)$ is a function like this:

where the height of the constant function can be anything.

So, one possible answer is $f(x) = \left\{\begin{array}{lr} 0&x \neq 10\\ 10&x=10 \end{array}\right.$.