# Positive derivative implies increasing

## Statement

### On an open interval

Suppose $f$ is a function on an open interval $I$ that may be infinite in one or both directions (i..e, $I$ is of the form $\! (a,b)$, $(a,\infty)$, $(-\infty,b)$, or $(-\infty,\infty)$). Suppose the derivative of $f$ exists and is positive everywhere on $I$, i.e., $\! f'(x) > 0$ for all $x \in I$. Then, $f$ is an increasing function on $I$, i.e.:

$x_1,x_2 \in I, x_1 < x_2 \implies f(x_1) < f(x_2)$

### On a general interval

Suppose $f$ is a function on an interval $I$ that may be infinite in one or both directions and may be open or closed at either end. Suppose $f$ is a continuous function on all of $I$ and that the derivative of $f$ exists and is positive everywhere on the interior of $I$, i.e., $f'(x) > 0$ for all $x \in I$ other than the endpoints of $I$ (if they exist). Then, $f$ is an increasing function on $I$, i.e.:

$x_1,x_2 \in I, x_1 < x_2 \implies f(x_1) < f(x_2)$

## Facts used

1. Lagrange mean value theorem

## Proof

### General version

Given: A function $f$ on interval $I$ such that $f'(x) > 0$ for all $x$ in the interior of $I$ and $f$ is continuous on $I$. Numbers $x_1 < x_2$ with $x_1, x_2 \in I$.

To prove: $\! f(x_1) < f(x_2)$

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Consider the difference quotient $\! \frac{f(x_2) - f(x_1)}{x_2 - x_1}$. There exists $x_3$ such that $x_1 < x_3 < x_2$ and $\! f'(x_3)$ equals this difference quotient. Fact (1) $x_1 < x_2$, $f$ is defined and continuous on an interval $I$ containing $x_1,x_2$, differentiable on the interior of the interval. [SHOW MORE]
2 The difference quotient $\! \frac{f(x_2) - f(x_1)}{x_2 - x_1}$ is positive. $f'(x)$ is positive for all $x$ in the interior of $I$. Step (1) [SHOW MORE]
3 $f(x_1) < f(x_2)$ $x_1 < x_2$ Step (2) [SHOW MORE]