Notes - Analysis II HT23, Extrema
Flashcards
Suppose $f : E \to \mathbb R$ and $x _ 0 \in E$. What does it mean for $x _ 0$ to be a local maximum or local minimum?
Can you state Fermat’s theorem on extrema (the one that connects local extrema to derivatives)?
If $f:(a, b) \to \mathbb R$ and $x _ 0 \in (a, b)$ is a local extremum and $f$ is differentiable at $x _ 0$, then $f’(x _ 0) = 0$.
Suppose $f : [a, b] \to \mathbb R$ is continuous. By the boundedness theorem, $f$ has a maximum and minimum. What are the possibilities for the location of the maximum and minimum?
- At $a$ or $b$
- At some point $x _ 0$ where $f’(x _ 0) = 0$
- At some point $x _ 0$ where $f’(x _ 0)$ is undefined
When proving Fermat’s theorem on extrema, i.e. that
If $f:(a, b) \to \mathbb R$ and $x _ 0 \in (a, b)$ is a local extremum and $f$ is differentiable at $x _ 0$, then $f’(x _ 0) = 0$.
If you assume $x _ 0$ is a local maximum, how do you know that
\[\frac{f(x) - f(x_0)}{x - x_0} \le 0\]
on some interval $(x _ 0, x _ 0 + \delta)$?
If $f:(a, b) \to \mathbb R$ and $x _ 0 \in (a, b)$ is a local extremum and $f$ is differentiable at $x _ 0$, then $f’(x _ 0) = 0$.
There exists some $\delta$ such that for all $0 < x - x _ 0 < \delta$, $f(x) \le f(x _ 0)$, so rearranging and dividing through by the positive $x - x _ 0$ gives
\[\frac{f(x) - f(x_0)}{x - x_0} \le 0\]Proofs
Prove Fermat’s theorem on extrema:
If $f:(a, b) \to \mathbb R$ and $x _ 0 \in (a, b)$ is a local extremum and $f$ is differentiable at $x _ 0$, then $f’(x _ 0) = 0$.
If $f:(a, b) \to \mathbb R$ and $x _ 0 \in (a, b)$ is a local extremum and $f$ is differentiable at $x _ 0$, then $f’(x _ 0) = 0$.
Todo.