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  "body": "\n## Statement\n\nLet $U \\subseteq \\mathbb{R}^n$ be open, let $f \\colon U \\to \\mathbb{R}$ be [[differentiable|0135XZZQ8M]] on $U$, and let $x, y \\in U$ be points such that the segment $\\{x + t (y - x) : 0 \\le t \\le 1\\}$ lies in $U$.\nThen there is a number $c \\in (0, 1)$ such that\n$$f(y) - f(x) = \\nabla f\\big(x + c (y - x)\\big) \\cdot (y - x).$$\n\n## Remarks\n\nThe proof applies the one-variable mean value theorem to $g(t) = f(x + t(y - x))$ on $[0, 1]$, with $g'(t) = \\nabla f(x + t(y - x)) \\cdot (y - x)$ by the [[chain rule for a curve|0A8TJ0VHC7]].\nThe theorem is false for maps into $\\mathbb{R}^m$ with $m \\ge 2$. For $f(t) = (\\cos t, \\sin t)$ on $[0, 2\\pi]$, $f(2\\pi) - f(0) = 0$, but $f'(t) \\ne 0$ for each $t$. The [[mean value inequality|02CYB184RC]] replaces it.\nBy the Cauchy–Schwarz inequality, it gives $|f(y) - f(x)| \\le \\sup_{z} \\|\\nabla f(z)\\| \\, \\|y - x\\|$, with $z$ on the segment.\n\n## Explanations\n\n### L1 human\n\nOn a straight path from $x$ to $y$, the average slope of $f$ equals the actual slope at some point between them.\nThis reduces the problem to one variable, by looking at $f$ only along the segment.\n",
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