Extreme Values of Functions
- Calculus 3.4
Extreme Values
Let $f$ be a function with domain $D$. Then $f$ has an absolute maximum value on $D$ at a point $c$ if $f(x) \le f(c)$ for all $x \in D$. Similarly, $f$ has an absolute minimum value on $D$ at a point $c$ if $f(x) \ge f(c)$ for all $x \in D$. Maximum and minimum values are collectively referred to as extreme values. Absolute maxima and minima are also called global maxima and minima.
Extreme Value Theorem
The extreme value theorem(EVT) states that:
Let $f$ be a continuous function on a closed interval $I$, then $f$ always has global maximum and minimum values on $I$.
This means that on a closed interval $I$, we can define the notations $\min$ and $\max$ as:
\[ \min_{t \in I} f(t) \le f(x) \le \max_{t \in I} f(t) \]
which holds for all $x \in I$. Also, we can introduce the notation $\argmin$ and $\argmax$ as:
\[ \argmin_{x \in I} f(x) = \Set{ x \in I | f(x) = \min_{t \in I} f(t) } \nt \argmax_{x \in I} f(x) = \Set{ x \in I | f(x) = \max_{t \in I} f(t) } \]
If argmin and argmax are singletons, we rather define them as the unique elements of those sets. We can define these functions more generally in real analysis using $\sup$ and $\inf$, but omitting them here. The proof of the EVT also requires knowledge from real analysis, also omitted here.
Local Extreme Values
A function $f$ has a local maximum at a point $c$ within its domain $D$ if there exists a neighborhood $N \in \mathcal{N}_c \subset D$ such that $f(x) \le f(c)$ for all $x \in N$. Similarly, a function $f$ has a local minimum at a point $c$ within its domain $D$ if there exists a neighborhood $N \in \mathcal{N}_c \subset D$ such that $f(x) \ge f(c)$ for all $x \in N$.
The First Derivative Test for Local Extreme Values
If $f$ has a local maximum or minimum at an internal point $c$ of its domain and $f^\prime(c)$ exists, then $f^\prime(c) = 0$.
To prove that $f^\prime(c) = 0$, we show that $f^\prime(c) \le 0$ and then that $f^\prime(c) \ge 0$. To begin, suppose that $f$ has a local maximum at $c$ so that $f(x) - f(c) \le 0$ for all $x \in N \in \mathcal{N}_c$ for some neighborhood $N$ of $c$. Then we have:
\[ f^\prime(c) = \lim_{x \to c^+} \frac{f(x) - f(c)}{x - c} \le 0 \nt f^\prime(c) = \lim_{x \to c^-} \frac{f(x) - f(c)}{x - c} \ge 0 \]
Thus, we conclude that $f^\prime(c) = 0$. The proof for a local minimum is similar.
Critical Points
A point $c$ in the domain of a function $f$ is called a critical point of $f$ if either $f^\prime(c) = 0$ or $f^\prime(c)$ does not exist.
Finding Extreme Values on a Closed Interval
- Find all critical points of $f$ in the closed interval.
- Evaluate $f$ at each critical point and at the endpoints of the interval.
- Take the largest and smallest of these values as the absolute maximum and minimum values of $f$ on the interval.
