Average rate, instantaneous rate AB & BC

Over the interval $[a, b]$ the average rate of change of $f$ is

$$\frac{f(b) - f(a)}{b - a},$$

the slope of the secant line through $(a, f(a))$ and $(b, f(b))$. Written with an increment $h$, so that $b = a + h$, the same number is

$$\frac{f(a + h) - f(a)}{h}.$$

These are the same expression in two costumes. The second is the one that survives into the definition of the derivative, because it has a single moving part: the increment $h$.

An average rate is a fact about an interval. It says nothing on its own about any single moment inside that interval, and in particular it need not match the rate at either endpoint.

Fix the left endpoint at $a$ and let the interval shrink. The average rates form a list of numbers, and the value they close in on is the instantaneous rate of change of $f$ at $a$:

$$\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}.$$

An equivalent statement drops the increment and lets the second input slide toward $a$ directly:

$$\lim_{x \to a} \frac{f(x) - f(a)}{x - a}.$$

Both are called the difference quotient, and both produce the same number. Which one is easier depends on the problem: the $h$ form is the workhorse for algebra, and the $x$ form is convenient when a specific point is already named.

Neither limit ever evaluates at the instant. Setting $h = 0$ or $x = a$ gives $\frac{0}{0}$, so the whole point is to simplify first and take the limit afterward. Topic 1.6 is the toolkit for that simplification.

A rate is a change in output divided by the matching change in input. Most of the errors in this topic are a numerator and a denominator that do not describe the same interval. Three checks catch nearly all of them:

1. The denominator is a difference, not an input. Over $[a, a+h]$ the input changed by $h$, so the denominator is $h$, never $a$ and never $a + h$.
2. The subtraction runs the same direction on top and bottom. $\frac{f(x) - f(a)}{x - a}$ is a rate; $\frac{f(x) - f(a)}{a - x}$ is its negative.
3. A limit symbol is present, and it matches the variable. With $h$ in the quotient the limit is $h \to 0$; with $x$ it is $x \to a$. A quotient with no limit at all is an average rate, not an instantaneous one.

Worked, on $f(t) = t^2 + 3t$ at $a = 1$. Expand the numerator first:

$$f(1 + h) - f(1) = \left[(1 + h)^2 + 3(1 + h)\right] - 4 = 5h + h^2.$$

Divide, which is legal because $h \neq 0$ throughout the limit process, and only then let $h$ go to zero:

$$\frac{5h + h^2}{h} = 5 + h \longrightarrow 5.$$

Take $f(t) = 100t - 5t^2$ metres after $t$ seconds. On $[1, 3]$ the average rate is

$$\frac{f(3) - f(1)}{3 - 1} = \frac{255 - 95}{2} = 80 \text{ m/s}.$$

The instantaneous rate at $t = 1$ is a different number. The quotient simplifies to $90 - 5h$, so the limit is $90$ m/s, and the same work at $t = 3$ gives $70$ m/s. The average sits between them, which is what an average does; it does not equal either one.

The units come free from the quotient: output units over input units. Metres over seconds gives metres per second. Litres over minutes gives litres per minute. If a question asks what a rate means, the sentence to write names the quantity, the units, and the point or interval, in that order.

The last habit is to keep the two ideas apart in language. "The average velocity on $[1, 3]$ was 80 m/s" and "the velocity at $t = 3$ was 70 m/s" are both true here, and swapping the words makes both false.