---
title: Average Rates of Change
tags: teaching, calculus
---
# 2.1. Average Rates of Change
## Definition and General discussion
For some general function $f(x)$ and two numbers $x_1$ and $x_2$ in its domain (with $x_1<x_2$), the average rate of change of $f$ over the interval $[x_1,x_2]$ is $$\frac{f(x_2)-f(x_1)}{x_2 - x_1}.$$

### Examples
1. Calculate the average rate of change of the function $g(x) = x^3 - x$ on the interval $[0.5,1]$.
&#9655; Solution. We are working with the function $g$, and we have $x_1=0.5$ and $x_2=1$. And so we calculate, $$\frac{g(0.5)-g(1)}{0.5-1} = \frac{-0.375 - 0}{-0.5} = 0.75.$$
2. Define a function $h(x) = \frac{x}{x^2-1}$. Find the average rate of change of $h$ over the interval $[2.2,b]$ for $b = 2.7, 2.3,$ and $b=2.25$. As these values for $b$ are getting closer to $2.2$, what is happening to the average rate of change? Does it appear to be approaching a number?
&#9655; Solution. Let's begin by computing values of $h(x)$ that are needed.
::: success
| $x$ | 2.2 | 2.25 | 2.3 | 2.7 |
| -------- | -------- | -------- | -------- | -------- |
| $h(x)$ | 0.573 | 0.554 | 0.536 | 0.429 |
:::
Now, we have that: $$\frac{h(2.7)-h(2.2)}{2.7-2.2} \approx -0.287;$$ $$\frac{h(2.3)-h(2.2)}{2.3-2.2} \approx -0.368;$$ $$\frac{h(2.25)-h(2.2)}{2.25-2.2} \approx -0.381.$$
The average rate of change is becoming more negative. It may be closing in on a number that is close to -0.4. However, it is not certain; more needs to be checked.
:::spoiler
Some more values of $\frac{h(b)-h(2.2)}{b-2.2}$.

| b | 2.205 | 2.201 | 2.2001 |
| -------- | -------- | -------- | -------- |
| av.rate of change | -0.3945 | -0.3957 | -0.3960 |

:::

## Secant Lines
The average rate of change of a function can be understood in terms of *secant lines* on the graph of the function.

For an interval $[x_1,x_2]$ and the graph $y=f(x)$ of a function $f(x)$, consider the line passing through the two points on the graph which have $x$-coordinate either $x_1$ or $x_2$ (we're supposing that $x_1$ and $x_2$ are in the domain of $f$ ). These would be the points $(x_1,f(x_1))$ and $(x_2,f(x_2))$. The line passing through them is called the *secant line of the graph over* $[x_1,x_2]$. The slope of the secant line is $$\frac{f(x_2)-f(x_1)}{x_2-x_1}.$$

:::info
**Takeaway**
The slope of the secant line of $y=f(x)$ over $[x_1,x_2]$ is equal to the average rate of change of $f$ over $[x_1,x_2]$.
:::

### Exercises
1. Estimate the average rates of change, over the intervals given, of the function that has the graph in the picture below.
a) $[-1,1]$
b) $[0,1.5]$
::: spoiler
Answers: a) -1; b) something between 2 and 2.34
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## Average velocity
When the input variable represents time, and the output $f(x)$ represents a position of some object (or a "distance traveled"), then the average rate of change is called the *average velocity* over the given interval.
### Exercises
2. A ball is thrown so that, $x$ seconds after being thrown, its height off of the ground (in meters) is $$h(x) = -4.9x^2+25x+5.$$ Find the average velocity of the ball between $4$ and $5$ seconds after being thrown (that is, over the interval $[4,5]$).
**Note:** This function only measures vertical position (height), so the average velocity being asked for is just the rate at which that height is changing (just what you would compute from $h(x)$), and does not take the sideways motion of the ball into account.

## Section 2.1 Book Exercises
* 10, 11:sparkles:, 12, 13:sparkles:, 14.
* 20, 21:sparkles:, 22, 23:sparkles:.

> :sparkles:Has an answer in the online textbook.