---
tag: 931VMFD34H
kind: proof
name: Proof that a function with nonnegative derivative is increasing
msc: [26A24]
proves: 026BJBQNGJ
uses: [0BARQY52N5, 0TK8Y3J5RW]
sources:
  - url: https://proofwiki.org/wiki/Real_Function_with_Positive_Derivative_is_Increasing
---

## Proof

We apply the mean value theorem on each subinterval.

1. Let $x, y \in I$ with $x \le y$. We must show $f(x) \le f(y)$.
2. If $x = y$, then $f(x) = f(y)$. So assume $x < y$.
3. Since $I$ is an [[interval|0TK8Y3J5RW]], every $z$ with $x \le z \le y$ lies in $I$. So $[x, y] \subseteq I$.
4. Each $z \in (x, y)$ is an interior point of $I$, because the open interval $(x, y)$ contains $z$ and lies in $I$.
5. So $f$ is differentiable at each point of $(x, y)$, with $f'(z) \ge 0$ there.
6. The function $f$ is continuous on $I$, so it is continuous on $[x, y]$.
7. By the [[mean value theorem|0BARQY52N5]], there is $c \in (x, y)$ with $f(y) - f(x) = f'(c)(y - x)$.
8. Here $f'(c) \ge 0$ by step 5, and $y - x > 0$. So $f(y) - f(x) \ge 0$.
9. Thus $f(x) \le f(y)$ whenever $x \le y$ in $I$, so $f$ is increasing on $I$.

## Idea

The 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.