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  "body": "\n## Proof\n\nWe subtract a linear function from $f$ so that Rolle's theorem applies.\n\n1. Let $r = \\dfrac{f(b) - f(a)}{b - a}$. This is a real number because $b - a \u003e 0$.\n2. Define $g \\colon [a, b] \\to \\R$ by $g(x) = f(x) - r (x - a)$.\n3. The function $x \\mapsto r(x - a)$ is [[continuous|0P27W9RW0G]] on $[a, b]$, and it has [[derivative|0F1WR9C83K]] $r$ at every point.\n4. So $g$ is continuous on $[a, b]$, as a difference of continuous functions.\n5. By the sum and [[constant multiple rule|0Z9JB71BS0]], $g$ is differentiable on $(a, b)$ with $g'(x) = f'(x) - r$.\n6. We have $g(a) = f(a)$ and $g(b) = f(b) - (f(b) - f(a)) = f(a)$. So $g(a) = g(b)$.\n7. By [[Rolle's theorem|0194DY35BG]], there is $c \\in (a, b)$ with $g'(c) = 0$.\n8. By step 5, $f'(c) - r = 0$. Hence $f'(c) = \\dfrac{f(b) - f(a)}{b - a}$.\n\n## Idea\n\nThe line through $(a, f(a))$ and $(b, f(b))$ has slope $r$. Subtracting it from $f$ gives a function with equal values at the endpoints. Rolle's theorem gives a point where its derivative vanishes, and there $f'$ equals $r$.\n",
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