How fast something changed between two moments is a subtraction followed by a division: the change in the output over the change in the input. Done exactly it is a fraction, and it is the slope of the straight line through two points of the graph. What the fraction depends on is the step h as well as the place, and that dependence is the whole subject of this course.
Core ideas
Three ideas. The first is the formula, the second is the picture, and the third is the one the rest of the course is built on.
Between t = a and t = a + h the input changes by h and the output changes by f(a + h) − f(a). The average rate is the second divided by the first. Both parts matter: the subtraction says how much the output moved, and the division says how much input it took to move it, which is what makes it a rate.
Plot the points (a, f(a)) and (a + h, f(a + h)). The straight line through them is a secant, and its slope is rise over run, which is exactly the quotient. So an average rate is a slope you could draw, and every fact about slopes from Algebra applies to it.
For a line the rate is the same whatever h is, which is what makes a line a line. For a curve it is not: a different h is a different secant with a different slope. A single point does not yet have a rate, only a family of them, one for each step. The next lessons ask what that family looks like.
The material
The definition, a first computation, the picture it draws, and the case where the step does not matter.
For a function f, a starting input a and a step h ≠ 0, the average rate of change of f over [a, a + h] is (f(a + h) − f(a))/h.
It is also called a difference quotient: a difference in the numerator, divided by the step h.
The lab takes the function as text and the place and step as exact numbers and prints the quotient as a fraction. Nothing is rounded: f(a + h) is found by substituting a fraction into the function, and the subtraction and division are done on fractions.
Take f(t) = t² and start at a = 1. With h = 1 the input goes from 1 to 2 and the output from 1 to 4, so the rate is (4 − 1)/1 = 3. Halve the step to h = 1/2: the output at 3/2 is 9/4, the change is 9/4 − 1 = 5/4, and dividing by 1/2 gives 5/2. A third halving gives 9/4.
The rates are not the same: 3, then 5/2, then 9/4. The function did not change; the step did. Each number is the slope of a different secant through the same point (1, 1).
Look at the three rates as 2 + 1, 2 + 1/2, 2 + 1/4. Each is 2 plus the step. That is not a coincidence of these three numbers; it can be derived. The output at 1 + h is (1 + h)² = 1 + 2h + h², so the change is 2h + h², and dividing by h gives 2 + h. For this function at this point the slope of the secant through (1, 1) and (1 + h, (1 + h)²) is 2 + h, for every nonzero h.
Let f(t) = t³ − t and start at a = 2. Then f(2) = 6, and with h = 1 the output at 3 is 24, so the rate is (24 − 6)/1 = 18.
With h = 1/2 the output at 5/2 is 105/8; the change is 105/8 − 6 = 57/8, and dividing by 1/2 gives 57/4. The rate fell from 18 to 57/4, which is 14.25, as the step halved. The lab continues the column and prints the last row as a single fraction.
Try f(t) = 3t + 1 at a = 5. The output at 5 is 16, and at 5 + h it is 16 + 3h, so the change is 3h and the quotient is 3. It is 3 for h = 1, for h = 1/2, for every step the lab can take. A function whose average rate does not depend on the step is exactly a line, and the rate is its slope. A curve is the opposite case: its rate depends on the step, and so its secants have different slopes.
That is what makes the next question worth asking. If a curve has no single rate over a step, is there a rate at a single place? The answer takes the next several lessons, and every one of them starts from the quotient on this page.
Interactive
Average rates as exact fractions
Method
Four moves, always in this order, so that the division is never forgotten.
Find f(a) and f(a + h) as exact numbers. Substitute the fraction for t and square, cube or multiply it out; leave nothing as a decimal.
f(a + h) − f(a) is the change. Its sign is the direction: positive means the output rose, negative that it fell.
The step is h, the change in the input. A step of 1/2 is divided by as 1/2, which doubles the change. This is the move that turns a change into a rate, and the one that is easiest to leave out.
The fraction you have is the slope of the secant through (a, f(a)) and (a + h, f(a + h)). If the answer does not look like a slope you could draw, recheck the evaluation.
The function is f(t) = t² and the start is a = 1, so f(1) = 1. Three steps, each halving the last, each done with fractions.
The three rates, 3 and 5/2 and 9/4, are 2 plus the step in each case, and the general line says why. The slope of the secant through (1, 1) and (1 + h, (1 + h)²) is 2 + h.
For a rehearsal, take f(t) = t² at a = 2. Work out the rate for h = 1 by hand, then the general quotient ((2 + h)² − 4)/h, and compare it with what the lab prints for the same function.
You should be able to evaluate f(a) and f(a + h) as fractions, subtract and divide without dropping the division, draw or describe the secant, and say what changes when h does.
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