Increasing and differentiable implies nonnegative derivative that is not identically zero on any interval

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 everywhere on $I$. Suppose further that $f$ is an increasing function on $I$, i.e.:

$x_1,x_2\in I,x_1<x_2⟹f(x_1)<f(x_2)$

Then, $f^{\prime }(x)\ge 0$ for all $x\in I$. Further, there is no sub-interval of $I$ such that $f^{\prime }(x)=0$ for all $x$ in the sub-interval.

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 everywhere on the interior of $I$. Further, suppose $f$ is an increasing function on $I$, i.e.:

$x_1,x_2\in I,x_1<x_2⟹f(x_1)<f(x_2)$

Then, $f^{\prime }(x)\ge 0$ for all $x$ in the interior of $I$. Further, there is no sub-interval of $I$ such that $f^{\prime }(x)=0$ for all $x$ in the sub-interval.

Related facts

Converse

• Nonnegative derivative that is not identically zero on any interval implies increasing

Other similar facts

• Positive derivative implies increasing
• Zero derivative implies locally constant
• Negative derivative implies decreasing