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.:

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

Related facts

Similar facts

• Zero derivative implies locally constant
• Negative derivative implies decreasing
• Nonnegative derivative that is zero only at isolated points implies increasing
• Increasing and differentiable implies nonnegative derivative

Opposite facts

• Positive derivative at a point not implies increasing around the point

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] Since $f$ is defined and continuous on an interval containing both $x_1$ and $x_2$, it is in particular defined and on $[x_1,x_2]$, which lies inside the open interval. Further, $f$ is differentiable on the interior of $I$, and hence on the open interval $(x_1,x_2)$, which is contained in the interior of $I$. Thus, the conditions needed to apply Fact (1) are available. Using Fact (1) gives the existence of $x_3$ such that $f'(x_3)$ equals the difference quotient.
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] By Step (1), there exists $x_3$ such that $f'(x_3)$ equals the difference quotient. From the given data, $f'(x_3)$ is positive. Combining, we obtain that the difference quotient itself is positive.
3	$f(x_1) < f(x_2)$		$x_1 < x_2$	Step (2)	[SHOW MORE] In Step (2), we obtained that the difference quotient is positive. The denominator of the expression is positive because $x_1 < x_2$. Thus, the numerator must also be positive, giving $f(x_1) < f(x_2)$ upon rearrangement.