Welcome back to Professor Baker's Math Class! In this session, we are shifting gears from simply calculating derivatives to using them to analyze the behavior of functions. Specifically, we covered Sections 4-1 (Maximum and Minimum Values) and 4-2 (The Mean Value Theorem). These concepts are the bedrock of optimization—finding the highest peak or the lowest valley in a dataset.

Understanding Extrema: Peaks and Valleys

In Section 4-1, we defined what it means for a function to have an extreme value. Visually, think of a graph with hills and dips.

  • Absolute Maximum: The highest point on the entire domain. mathematically, $f(c) \ge f(x)$ for all $x$.
  • Local Maximum: A peak relative to its immediate neighbors (the top of a specific hill, even if there is a taller mountain elsewhere).

The same logic applies to minimums. But how do we find these points without graphing every single function?

The Key: Critical Numbers

This brings us to one of the most important definitions in Calculus I: the Critical Number. A critical number of a function $f$ is a number $c$ in the domain such that either:

  1. $f'(c) = 0$ (The tangent line is horizontal), or
  2. $f'(c)$ does not exist (A sharp corner or cusp).

According to Fermat's Theorem, if a function has a local max or min at $c$, then $c$ must be a critical number. Let's look at an example from our class notes:

Example: Find the critical numbers of $f(x) = x^3 - 3x^2 + 1$.
First, take the derivative:
$$f'(x) = 3x^2 - 6x$$
Set it to zero to find horizontal tangents:
$$3x(x - 2) = 0$$
This gives us critical numbers at $x = 0$ and $x = 2$.

The Closed Interval Method

When looking for the Absolute Max and Min on a closed interval $[a, b]$, we use a foolproof three-step process:

  1. Find the values of $f$ at the critical numbers inside the interval.
  2. Find the values of $f$ at the endpoints ($a$ and $b$).
  3. Compare the values. The largest is the absolute maximum; the smallest is the absolute minimum.

We applied this in class to a physics application regarding the Space Shuttle Discovery, where we differentiated the velocity function $v(t)$ to find the maximum acceleration $a(t)$.

The Mean Value Theorem (Section 4-2)

Section 4-2 introduces a theoretical heavyweight: The Mean Value Theorem (MVT). It states that if a function is continuous on $[a, b]$ and differentiable on $(a, b)$, there exists a number $c$ such that:

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

In plain English: There is at least one point on the curve where the instantaneous rate of change (slope of the tangent) equals the average rate of change (slope of the secant line) connecting the endpoints.

Class Example:
Given $f(x) = x^3 - x$ on the interval $[0, 2]$.
First, find the average slope:
$$\frac{f(2) - f(0)}{2 - 0} = \frac{6 - 0}{2} = 3$$
Now, set the derivative equal to this average slope:
$$f'(c) = 3c^2 - 1 = 3$$
Solving for $c$, we get $3c^2 = 4$, which simplifies to $c = \frac{2}{\sqrt{3}}$ (since $c$ must be in the interval).

These theorems allow us to predict the behavior of functions with certainty. Keep practicing your derivatives, especially with fractional exponents, as they frequently appear when finding critical numbers!