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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 $x \u003c y$. We must show $f(x) \u003c f(y)$.\n2. Since $I$ is an [[interval|0TK8Y3J5RW]], every $z$ with $x \\le z \\le y$ lies in $I$. So $[x, y] \\subseteq I$.\n3. Each $z \\in (x, y)$ is an interior point of $I$, because the open interval $(x, y)$ contains $z$ and lies in $I$.\n4. So $f$ is differentiable at each point of $(x, y)$, with $f'(z) \u003e 0$ there.\n5. The function $f$ is continuous on $I$, so it is continuous on $[x, y]$.\n6. By the [[mean value theorem|0BARQY52N5]], there is $c \\in (x, y)$ with $f(y) - f(x) = f'(c)(y - x)$.\n7. Here $f'(c) \u003e 0$ by step 4, and $y - x \u003e 0$. So $f(y) - f(x) \u003e 0$.\n8. Thus $f(x) \u003c f(y)$ whenever $x \u003c y$ in $I$, so $f$ is [[strictly increasing|0TR9SR9FTB]] on $I$.\n\n## Idea\n\nThe mean value theorem turns the sign of the derivative into the sign of a difference of values. A positive slope at one point between $x$ and $y$ forces $f(y) \u003e f(x)$. Continuity is needed only so that the theorem also applies when $x$ or $y$ is an endpoint of $I$.\n",
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