Monotonicity and Sign of Derivative

If the sign of the derivative of a real function is unchanging, the function is monotonic, either increasing or decreasing depending on the sign. This should be intuitive from the way the derivative is defined, since a positive derivative at all points on some interval corresponds with an increasing tangent at all such points.

Positive Derivative Corresponds with Strictly Increasing Function

If \(f\) is continuous on \([a, b]\) and \(f'(x) > 0 \enspace \forall x \in (a, b)\), then for all \(y, z \in (a, b)\):

\[ y < z \implies f(y) < f(z).\]

That is \(f\) is monotonic increasing.

Proof

Because differentiability implies continuity, given that \(y, z \in (a, b)\) where \(f\) is differentiable, \(f\) is continuous on \([a, b]\).

Applying the mean value theorem to \(f\) over \([y, z]\), there exists a \(c \in (a, b)\) such that for all \(y, z \in (a, b)\):

\[ \frac{f(y) - f(z)}{y - z} = f'(c) > 0.\]

Given that \(y < z\), \(y - z < 0\) and therefore we must reverse the inequality through multiplication, yielding

\[ f(y) - f(z) < 0 \implies f(y) < f(z).\]

Negative Derivative Corresponds with Strictly Decreasing Function

If \(f\) is continuous on \([a, b]\) and \(f'(x) < 0 \enspace \forall x \in (a, b)\), then for all \(y, z \in (a, b)\):

\[ y < z \implies f(y) > f(z).\]

That is \(f\) is monotonic decreasing.

Non-Negative Derivative Corresponds with Non-Decreasing Function

If \(f\) is continuous on \([a, b]\) and \(f'(x) \geq 0 \enspace \forall x \in (a, b)\), then for all \(y, z \in (a, b)\):

\[ y < z \implies f(y) \leq f(z).\]

That is \(f\) is monotonic non-increasing.

Non-Positive Derivative Corresponds with Non-Increasing Function

If \(f\) is continuous on \([a, b]\) and \(f'(x) \leq 0 \enspace \forall x \in (a, b)\), then for all \(y, z \in (a, b)\):

\[ y < z \implies f(y) \geq f(z).\]

That is \(f\) is monotonic non-increasing.