# Graphical interpretation of mean value theorem

I'm struggling understanding how this theorem works:

Let $$f\in C^1(B\subseteq\mathbb{R}^n;\mathbb{R}^m)$$. Let $$x_0,x\in\mathring{B}$$ such that the segment $$S\in\mathring{B}$$ of extremes $$x,x_0$$ . Then

$$||f(x)-f(x_0)||\leq\sup_{\phi\in S}||Df(\phi)||\cdot||x-x_0||$$

What does this mean graphically? What I suppose is $$\frac{||f(x)-f(x_0)||}{||x-x_0||}$$ is the secant line of extremes $$x,x_0$$, the slope of this line is $$\leq$$ of a tangent line calculated in $$\phi\in S$$ but I don't understand why we use $$sup$$ (shouldn't $$\sup_{\phi\in S}$$ be the extreme of the segment S which is $$x$$ or $$x_0$$?)

We don't take the supremum of $$\phi\in S$$, we take the supremum over $$\phi\in S$$ of $$\|Df(\phi)\|$$, which denotes the size of the derivative at $$\phi$$. Intuitively, if the size of the derivative were smaller than $$\frac{\|f(x)-f(x_0)\|}{x-x_0}$$ everywhere, then there would be no way to go from $$f(x)$$ to $$f(x_0)$$ over a segment of length $$\|x-x_0\|$$.