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  "body": "\n## Proof\n\nWe apply the mean value theorem on each subinterval.\n\n1. Let $x, y \\in I$ with $x \\le y$. We must show $f(x) \\le f(y)$.\n2. If $x = y$, then $f(x) = f(y)$. So assume $x \u003c y$.\n3. Since $I$ is an [[interval|0TK8Y3J5RW]], every $z$ with $x \\le z \\le y$ lies in $I$. So $[x, y] \\subseteq I$.\n4. Each $z \\in (x, y)$ is an interior point of $I$, because the open interval $(x, y)$ contains $z$ and lies in $I$.\n5. So $f$ is differentiable at each point of $(x, y)$, with $f'(z) \\ge 0$ there.\n6. The function $f$ is continuous on $I$, so it is continuous on $[x, y]$.\n7. By the [[mean value theorem|0BARQY52N5]], there is $c \\in (x, y)$ with $f(y) - f(x) = f'(c)(y - x)$.\n8. Here $f'(c) \\ge 0$ by step 5, and $y - x \u003e 0$. So $f(y) - f(x) \\ge 0$.\n9. Thus $f(x) \\le f(y)$ whenever $x \\le y$ in $I$, so $f$ is increasing on $I$.\n\n## Idea\n\nThe mean value theorem writes $f(y) - f(x)$ as a value of $f'$ times $y - x$. A nonnegative derivative and a positive length give a nonnegative change. So $f$ never goes down.\n",
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