Consider the function $$u(x)=\begin{cases}x \qquad \qquad x\in[0,\frac{1}{2}]\\-x+1 \qquad x \in[\frac{1}{2},1]\end{cases}.$$
Certainly $u$ is continuous in $[0,1]$, derivable at each point except at $x=\frac{1}{2}$.
I wonder if there is a formal way to prove that there is no $\phi\in C^1((0,1))$ such that $\phi(\frac{1}{2})=u(\frac{1}{2})$ and its graph is locally under the graph of $u$, ie there exist $\epsilon>0$ such that $\phi(y)<u(y)$ for all $y \in B_\epsilon(\frac{1}{2})=\{y\in (0,1): |y-\frac{1}{2}|<\epsilon\}$, $y\neq \frac{1}{2}$.
Assume by contrary that $\epsilon>0$ exists such that $$\left|x-{1\over 2}\right|<\epsilon\to \phi(x)\le u(x)$$since $\phi(0.5)=u(0.5)$ we can write$$\phi(x)-\phi(0.5)\le u(x)-u(0.5)$$for $0.5<x<0.5+\epsilon$ we obtain $${\phi(x)-\phi(0.5)\over x-0.5}\le {u(x)-u(0.5)\over x-0.5}$$ and by tending $x\to 0.5^+$ since $\phi\in C^1[(0,1)]$ we have $$\phi'(0.5)\le u'^+(0.5)=-1$$where $u'^+(0.5)$ is the right derivative of $u$ in $0.5$. Similarly $0.5-\epsilon<x<0.5$ yields to $${\phi(x)-\phi(0.5)\over x-0.5}\ge {u(x)-u(0.5)\over x-0.5}$$and $$\phi'(0.5)\ge u'^-(0.5)=1$$which leads to the impossible inequality $1\le-1$. Therefore such $\epsilon$ does not exist and the proof is complete.