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  "body": "\n## Proof\n\nWe compare any two values of $f$ with the mean value theorem.\n\n1. Let $x, y \\in I$ with $x \u003c y$.\n2. Since $I$ is an [[interval|0TK8Y3J5RW]], every $z$ with $x \\le z \\le y$ lies in $I$. So $[x, y] \\subseteq I$.\n3. The function $f$ is differentiable at each point of $I$. So it is [[continuous|07ZMX4ANJS]] at each point of $I$, and in particular on $[x, y]$.\n4. The open interval $(x, y)$ lies in $I$, so $f$ is differentiable on $(x, y)$, with derivative $0$ there.\n5. By the [[mean value theorem|0BARQY52N5]], there is $c \\in (x, y)$ with $f(y) - f(x) = f'(c)(y - x)$.\n6. Since $f'(c) = 0$, we get $f(y) = f(x)$.\n7. So $f$ takes the same value at any two points of $I$. Fix $x_0 \\in I$. Then $f(x) = f(x_0)$ for all $x \\in I$, so $f$ is constant on $I$.\n\n## Idea\n\nThe mean value theorem writes the change $f(y) - f(x)$ as a derivative times $y - x$. If the derivative is zero everywhere, every such change is zero. The interval hypothesis puts the whole segment $[x, y]$ inside the domain.\n",
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