# Fermat's Theorem for Extrema

When looking at Local Maxima and Minima, and Absolute Maxima and Minima, we assumed that if the point $(a, f(a))$ is an extreme (that is, the point is either a local maximum or local minimum), then $f'(a) = 0$. We will now prove this theorem commonly known as one of Fermat's theorems.

Theorem 1 (Fermat's Theorem for Extrema): If $f$ is a differentiable function and the point $(a, f(a))$ is an extrema on $f$, then $f'(a) = 0$ provided that $f'(a)$ exists. |

**Proof of Theorem:**Suppose that $f$ has a local maximum at $(a, f(a))$. By the definition of a local maximum, we know that $f(a) ≥ f(x)$ when $x$ is close to $a$, that is $f(a) ≥ f(a + h)$ as $h \to 0$ and thus $f(a + h) - f(a) ≤ 0$. Dividing both sides by $h > 0$ and taking the right-hand limit of both sides we obtain that:

- Since $f'(a)$ exists, we know that $f'(a) ≤ 0$. Now suppose that $h < 0$. We therefore get that $\frac{f(c + h) - f(c)}{h} ≥ 0$, and taking the lefthand limit of both sides we get that:

- Once again, since $f'(a)$ exists, we know that $f'(a) ≥ 0$. Since $f'(a) ≥ 0$ AND $f'(a) ≤ 0$, then $f'(a) = 0$. $\blacksquare$

It is important to note that the converse of this theorem is not necessarily true, that is, $f(a) = 0$ does not imply that $(a, f(a))$ is an extrema. For example, consider the function $f(x) = x^3$ whose derivative is $f'(x) = 3x^2$. Clearly, $f'(x) = 0$ at $x = 0$ or rather, the coordinates $(0, 0)$, however, just by looking at the graph of the function $f$:

The point $(0,0)$ is neither a local maxima or local minima. We will now look at another important definition.

Definition: If $f$ is a function, then $a \in D(f)$ (where the notation $D(f)$ represents the domain of the function $f$) is said to be a Critical Point or Critical Number of $f$ when $f'(a) = 0$ or $f'(a)$ does not exist. |