Derivative of Monotone Function

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Theorem

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.


If $\forall x \in \openint a b: \map {f'} x \ge 0$, then $f$ is increasing on $\closedint a b$.
If $\forall x \in \openint a b: \map {f'} x > 0$, then $f$ is strictly increasing on $\closedint a b$.
If $\forall x \in \openint a b: \map {f'} x \le 0$, then $f$ is decreasing on $\closedint a b$.
If $\forall x \in \openint a b: \map {f'} x < 0$, then $f$ is strictly decreasing on $\closedint a b$.


Proof

Let $c, d \in \closedint a b: c < d$.

Then $f$ satisfies the conditions of the Mean Value Theorem on $\closedint c d$.

Hence:

$\exists \xi \in \openint c d: \map {f'} \xi = \dfrac {\map f d - \map f c} {d - c}$


If $\forall x \in \openint a b: \map {f'} x \ge 0$, then $\map {f'} \xi \ge 0$ and hence:

$\map f d \ge \map f c$

Thus $f$ is increasing on $\closedint a b$


If $\forall x \in \openint a b: \map {f'} x > 0$, then $\map {f'} \xi > 0$ and hence:

$\map f d > \map f c$

Thus $f$ is strictly increasing on $\closedint a b$.


The other cases follow similarly.

$\blacksquare$


Sources