The First Derivative Test for Differentiable Functions

# The First Derivative Test for Differentiable Functions

Theorem 1 (The First Derivative Test): Let $f : [a, b] \to \mathbb{R}$, $c \in (a, b)$, and let $f$ be differentiable on $(a, b)$.a) If there exists a $\delta > 0$ for which $(c - \delta, c + \delta) \subseteq [a, b]$, $f'(x) \geq 0$ on $(c - \delta, c)$ and $f'(x) \leq 0$ on $(c, c + \delta)$ then $f$ attains a local maximum value at $c$.b) If there exists a $\delta > 0$ for which $(c - \delta, c + \delta) \subseteq [a, b]$, $f'(x) \leq 0$ on $(c - \delta, c)$ and $f'(x) \geq 0$ on $(c, c + \delta)$ then $f$ attains a local minimum value at $c$. |

**Proof of a)**For each $x \in (c - \delta, c)$ and for each $y \in (c, c + \delta)$ - since $f$ is differentiable on $(a, b)$ we have that $f$ is continuous on $[a, b]$ and so $f$ is continuous on $(x, c)$ and $(c, y)$. Furthermore, since $f$ is differentiable on $(a, b)$ we have that $f$ is differentiable on $(x, c)$ and $(c, y)$.- By The Mean Value Theorem for Differentiable Functions we have that there exists a point $c_1 \in (x, c)$ such that:

\begin{align} \quad f'(c_1) = \frac{f(c) - f(x)}{c - x} \quad \Leftrightarrow \quad f'(c_1)(c - x) = f(c) - f(x) \quad (*) \end{align}

- And similarly we have that there exists a point $c_2 \in (c, y)$ such that:

\begin{align} \quad f'(c_2) = \frac{f(y) - f(c)}{y - c} \quad \Leftrightarrow \quad f'(c_2)(y - c) = f(y) - f(c) \quad (**) \end{align}

- Since $c_1 \in (x, c) \subset (c - \delta, c)$ we have that $f'(c_1) \geq 0$ by hypothesis. So $(*)$ implies that $f(c) - f(x) \geq 0$ for all $x \in (c - \delta, c)$, i.e., $f(c) \geq f(x)$ on $(c - \delta, c)$.

- Since $c_2 \in (c, y) \subseteq (c, c + \delta)$ we have that $f'(c_2) \leq 0$ by hypothesis. So $(**)$ implies that $f(y) - f(c) \leq 0$, i.e., $f(y) \leq f(c)$ on $(c, c + \delta)$.

- So $f$ attains a local maximum at $f(c)$. $\blacksquare$

**Proof of b)**Analogous to (a).