# Real Function with Strictly Negative Derivative is Strictly Decreasing

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

## Proof

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

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

Hence:

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

Let $f$ be such that:

$\forall x \in \openint a b: \map {f'} x < 0$

Then:

$\map {f'} \xi < 0$

and hence:

$\map f d < \map f c$

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

$\blacksquare$