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 $(a,b)$, $(a,\mathrm{\infty })$, $(-\mathrm{\infty },b)$, or $(-\mathrm{\infty },\mathrm{\infty })$). Suppose the derivative of $f$ exists and is positive everywhere on $I$, i.e., $f^{\prime }(x)>0$ for all $x\in I$. Then, $f$ is an increasing function on $I$, i.e.:

$x_1,x_2\in I,x_1<x_2⟹f(x_1)<f(x_2)$

Related facts

Similar facts

• Zero derivative implies locally constant
• Negative derivative implies decreasing

Converse

• Increasing and differentiable implies nonnegative derivative that is zero only at isolated points

Facts used

1. Lagrange mean value theorem

Proof

General version

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

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 $f^{\prime }(x_3)$ equals this difference quotient.	Fact (1)	$x_1<x_2$, $f$ is defined and continuous on an interval 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 $f^{\prime }(x_3)$ equals the difference quotient.
2	The difference quotient $\frac{f(x_2)-f(x_1)}{x_2-x_1}$ is positive.		$f^{\prime }(x)$ is positive for all $x\in I$.	Step (1)	[SHOW MORE] By Step (1), there exists $x_3$ such that $f^{\prime }(x_3)$ equals the difference quotient. From the given data, $f^{\prime }(x_3)$ 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.