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

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.