The First Derivative Test for Differentiable Functions

Table of Contents

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

(c - \delta, c)

and

f'(x) \leq 0

(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

(c - \delta, c)

and

f'(x) \geq 0

(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:

(1)

\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}

(2)

\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 $(c, c + \delta)$ .

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