Topic 2.1 Notes – Defining Average and Instantaneous Rates of Change at a Point

Average Rate of Change

Think “slope between two points.” If a function $f$ is defined on $[a,b]$, the average rate of change over that interval is

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

That’s just rise over run.

What it represents

• Change in output divided by change in input
• Slope of the secant line through $(a,f(a))$ and $(b,f(b))$
• Units: output units per input unit

Equivalent forms you’ll see:

• $\dfrac{f(a+h)-f(a)}{h}$
• $\dfrac{f(x)-f(a)}{x-a}$

They all mean the same thing: slope between two distinct x-values.

How it looks on a graph

On a graph, this is the slope of the secant line connecting the two points on the curve.

Secant line representing average rate of change on $[a,b]$

If the secant line slopes upward, the average rate is positive. Downward means negative.

Quick example

Let $f(x)=3x^2-1$. Find the average rate of change on $[1,4]$.

1. $f(4)=3(16)-1=47$
2. $f(1)=3(1)-1=2$
3. $\dfrac{47-2}{4-1}=\dfrac{45}{3}=15$

That 15 is the slope of the secant line.

Common slip-ups

• Reversing the subtraction (it changes the sign).
• Forgetting the denominator is change in x, not y.
• Using derivative rules when the question only asks for average rate.

Instantaneous Rate of Change and the Derivative at a Point

Now shrink the interval. Let the second point move closer and closer to $a$.

The instantaneous rate of change at $x=a$ is defined by a limit:

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

Equivalent form:

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

If this limit exists, the function is differentiable at $a$.

What this means

• It is the slope of the tangent line at $x=a$.
• It’s the exact rate of change at that single input.
• In motion language, if $f$ is position, then $f'(a)$ is velocity at time $a$.

Geometric picture

In the diagrams below, the secant line through $(a, f(a))$ and a nearby point moves closer and closer to $x=a$.

As $h\to 0$, the secant line becomes the tangent line.

This idea is huge. The derivative is just the limit of average rates of change.

Computing a Derivative from the Limit Definition

Sometimes your teacher or the AP exam will say “use the definition.” That means no shortcut rules.

Let’s find $f'(a)$ for $f(x)=x^2+5x$.

Start with
$$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$

1. $f(a+h)=(a+h)^2+5(a+h)=a^2+2ah+h^2+5a+5h$
2. Subtract $f(a)=a^2+5a$
   → numerator becomes $2ah+h^2+5h$
3. Factor: $h(2a+h+5)$
4. Cancel the $h$
5. Take the limit as $h\to 0$

Result:
$$f'(a)=2a+5$$

Two things matter here:

• You must expand correctly.
• You must factor out and cancel the $h$ before plugging in 0.

If you plug in $h=0$ too early, you get $0/0$ and freeze.

On free-response questions, showing the limit setup and algebra earns points. Don’t skip steps.

Average vs Instantaneous Rate of Change

Here’s how to keep them straight.

When a question says:

• “Average rate of change on $[2,6]$” → use the slope formula.
• “Instantaneous rate at $x=2$” or “find $f'(2)$” → use the derivative.

They are connected, but not interchangeable.