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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 $x_0$?)

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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 $\|x-x_0\|$.

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