Theorem: If $f$ has a local maximum or minimum at $c$ and $f'(c)$ exists, then $f'(c) = 0$.

Proof: Suppose $f$ has a local maximum at $c$ (the minimum case is similar). Then there exists $\delta > 0$ such that $f(c) \geq f(x)$ for all $x \in (c - \delta, c + \delta)$.

Consider the difference quotient from the right: $\frac{f(c+h) - f(c)}{h} \quad \text{for } 0 < h < \delta$

Since $f(c) \geq f(c+h)$, we have $f(c+h) - f(c) \leq 0$. With $h > 0$, this gives: $\frac{f(c+h) - f(c)}{h} \leq 0$

Taking the limit as $h \to 0^+$: $f'_+(c) = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} \leq 0$

Similarly, for the left-hand limit with $-\delta < h < 0$: $\frac{f(c+h) - f(c)}{h} \geq 0 \quad \text{(negative numerator, negative denominator)}$

Taking the limit as $h \to 0^-$: $f'_-(c) = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h} \geq 0$

Since $f'(c)$ exists, we have $f'_+(c) = f'_-(c) = f'(c)$. Therefore: $f'(c) \leq 0 \quad \text{and} \quad f'(c) \geq 0$

This forces $f'(c) = 0$. $\square$

Theorem: Suppose $f''$ is continuous near $c$ and $f'(c) = 0$.

• If $f''(c) > 0$, then $f$ has a local minimum at $c$
• If $f''(c) < 0$, then $f$ has a local maximum at $c$

Proof: We prove the minimum case; the maximum case is analogous.

Assume $f'(c) = 0$ and $f''(c) > 0$. Since $f''$ is continuous and $f''(c) > 0$, there exists $\delta > 0$ such that $f''(x) > 0$ for all $x \in (c - \delta, c + \delta)$.

Because $f''(x) > 0$ on this interval, $f'$ is strictly increasing on $(c - \delta, c + \delta)$.

For $x \in (c - \delta, c)$: Since $f'$ is increasing and $f'(c) = 0$, we have $f'(x) < f'(c) = 0$. Thus $f$ is decreasing on $(c - \delta, c)$.

For $x \in (c, c + \delta)$: Since $f'$ is increasing and $f'(c) = 0$, we have $f'(x) > f'(c) = 0$. Thus $f$ is increasing on $(c, c + \delta)$.

Therefore, $f$ decreases to the left of $c$ and increases to the right of $c$, which means $f$ has a local minimum at $c$. $\square$