Welcome to Calc 1: Sections 4.1 & 4.2!

In this section, we'll be diving into the fascinating world of finding maximum and minimum values of functions, and exploring some powerful theorems that help us do so. Get ready to level up your calculus skills!

Absolute and Local Extrema

Let's start with the basics. Suppose we have a function $f$ with a domain $D$.

  • Absolute Maximum: $f(c)$ is the absolute maximum value of $f$ on $D$ if $f(c) \ge f(x)$ for all $x$ in $D$.
  • Absolute Minimum: $f(c)$ is the absolute minimum value of $f$ on $D$ if $f(c) \le f(x)$ for all $x$ in $D$.

Now, let's talk about local extrema:

  • Local Maximum: $f(c)$ is a local maximum value of $f$ if $f(c) \ge f(x)$ when $x$ is near $c$.
  • Local Minimum: $f(c)$ is a local minimum value of $f$ if $f(c) \le f(x)$ when $x$ is near $c$.

The Extreme Value Theorem

This theorem is a real game-changer! It states that if $f$ is continuous on a closed interval $[a, b]$, then $f$ attains an absolute maximum value $f(c)$ and an absolute minimum value $f(d)$ at some numbers $c$ and $d$ in $[a, b]$. In simpler terms, a continuous function on a closed interval is guaranteed to have a highest and lowest point.

Fermat's Theorem

Here's another important theorem: If $f$ has a local maximum or minimum at $c$, and if $f'(c)$ exists, then $f'(c) = 0$. This tells us that at local extrema, the derivative is zero (or undefined, as we'll see with critical numbers).

Critical Numbers

A critical number of a function $f$ is a number $c$ in the domain of $f$ such that either $f'(c) = 0$ or $f'(c)$ does not exist. Critical numbers are crucial because they are potential locations for local maxima and minima.

Example

Let's consider the function $f(x) = 3x^4 - 16x^3 + 18x^2$ on the interval $[-1, 4]$. To find the absolute maximum and minimum, we first find the derivative:

$f'(x) = 12x^3 - 48x^2 + 36x = 12x(x^2 - 4x + 3) = 12x(x-1)(x-3)$

Setting $f'(x) = 0$, we find critical points at $x = 0, 1, 3$. Now, we evaluate $f(x)$ at these critical points and the endpoints of the interval:

  • $f(-1) = 37$ (Absolute Maximum)
  • $f(0) = 0$ (Local Minimum)
  • $f(1) = 5$ (Local Maximum)
  • $f(3) = -27$ (Absolute Minimum)
  • $f(4) = 32$

Rolle's Theorem

Rolle's Theorem states that if a function $f$ satisfies the following three hypotheses:

  1. $f$ is continuous on the closed interval $[a, b]$.
  2. $f$ is differentiable on the open interval $(a, b)$.
  3. $f(a) = f(b)$

Then there is a number $c$ in $(a, b)$ such that $f'(c) = 0$.

The Mean Value Theorem

The Mean Value Theorem states that if $f$ is a function that satisfies the following hypotheses:

  1. $f$ is continuous on the closed interval $[a, b]$.
  2. $f$ is differentiable on the open interval $(a, b)$.

Then there is a number $c$ in $(a, b)$ such that:

$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

Or equivalently:

$$f(b) - f(a) = f'(c)(b - a)$$

Keep practicing, and you'll master these concepts in no time. Good luck, and happy calculating!