A. Rate of Change · Limits and Rates of Change

Instantaneous Rates With and Without Simplifying

A1.6 A1.5About 70 minutes

Learning goals

I can calculate an instantaneous rate of change of a polynomial at a point by substituting values of h that approach 0.
I can calculate the same rate by simplifying the difference quotient first and then letting h approach 0.
I can compare the two methods and explain what each one shows.
I can find the instantaneous rate of change of x², x³ and x⁴ at a general point x = a.

Words to know

substitute first (numerical method)
Evaluate (f(a + h) − f(a))/h for h = 0.1, 0.01, 0.001, … (and negative values) and look for the value the results approach.

simplify first (algebraic method)
Expand and simplify the difference quotient to an expression in h with no h in the denominator, then let h → 0.

error term
In a simplified difference quotient such as 6 + h, the part that depends on h (here, h). It measures how far the secant slope is from the tangent slope.

binomial expansion
Expanding (a + h)ⁿ, for example (a + h)³ = a³ + 3a²h + 3ah² + h³.

round-off error
Error caused by a calculator keeping only a limited number of digits.

Two roads to the same place

You now have two ways to find an instantaneous rate of change from the difference quotient:

Method A (substitute first): plug in values of h that approach 0, like 0.1, 0.01, 0.001 and −0.001, and watch the results.
Method B (simplify first): expand and simplify (f(a + h) − f(a))/h algebraically, then let h → 0.

The curriculum asks you to compare these, for polynomial functions such as x² and x³. Doing both side by side shows exactly what the limit is doing, and why algebra beats tables when you want an exact answer.

Worked example

Example 1: f(x) = x² at x = 3, both ways (curriculum example)

Find the instantaneous rate of change of f(x) = x² at x = 3 (a) by substituting values of h first and (b) by simplifying first. Compare.

(a) The difference quotient is ((3 + h)² − 9)/h. With h = 0.1: (3.1² − 9)/0.1 = (9.61 − 9)/0.1 = 6.1.
Continue: h = 0.01 → 6.01; h = 0.001 → 6.001; h = −0.1 → 5.9; h = −0.01 → 5.99; h = −0.001 → 5.999. The values approach 6 from both sides. But h = 0 itself gives 0/0.
(b) Expand: (9 + 6h + h² − 9)/h = (6h + h²)/h = 6 + h, for h ≠ 0.
Now let h → 0: 6 + h → 6. No table needed.
Compare: substituting h = 0.01 into 6 + h gives 6.01, exactly what Method A gave. The simplified form explains the table: each value is 6 plus the error h.

Both methods give 6. Method A suggests it; Method B proves it, and shows the secant slope is always exactly h more than the tangent slope.

Method A builds a table and spots a trend. Method B turns the difference quotient into 6 + h, which gives every table value at once and the exact limit.

Graph the difference quotient q(h) as a function of h: it is the line 6 + h with a hole at h = 0. The limit is the height of the hole, 6.

Worked example

Example 2: f(x) = x³ at x = 2, both ways

Find the instantaneous rate of change of f(x) = x³ at x = 2 both ways.

Substitute first: h = 0.1 gives 12.61; h = 0.01 gives 12.0601; h = 0.001 gives 12.006 001; h = −0.001 gives 11.994 001; h = −0.01 gives 11.9401; h = −0.1 gives 11.41.
The values approach 12.
Simplify first: (2 + h)³ = 8 + 12h + 6h² + h³, so the quotient is (12h + 6h² + h³)/h = 12 + 6h + h².
Let h → 0: 12.
Check one table value: h = 0.1 gives 12 + 0.6 + 0.01 = 12.61 ✓.

The rate is 12. Here the error term is 6h + h², so the secants approach more slowly than for x² (for small h, the error is about 6h, not h).

What each method is good for

Substitute first works for any function you can evaluate, even one with no nice algebra (like 2ˣ or data from a sensor), and it shows the trend. But it never reaches h = 0, and with very tiny h a calculator can lose digits.

Simplify first gives the exact value and a formula for the error. But it needs algebra that works out, which is always true for polynomials.

Worked example

Example 3: A general point

Simplify first to find the instantaneous rate of change of f(x) = x² and of g(x) = x³ at any point x = a.

f: ((a + h)² − a²)/h = (2ah + h²)/h = 2a + h → 2a.
g: ((a + h)³ − a³)/h = (3a²h + 3ah² + h³)/h = 3a² + 3ah + h² → 3a².
So at x = 3, the slope of y = x² is 2(3) = 6 (matching Example 1), and at x = 2, the slope of y = x³ is 3(2²) = 12 (matching Example 2).

Slope of y = x² at x = a: 2a. Slope of y = x³ at x = a: 3a². Method A would need a new table for every a; Method B handles all of them at once. The sibling unit Derivatives and Differentiation Rules turns these into slope functions.

Worked example

Example 4: A polynomial with several terms

Find the instantaneous rate of change of f(x) = x³ − 4x + 1 at x = 1 by simplifying first. Check with h = 0.01.

f(1) = 1 − 4 + 1 = −2.
f(1 + h) = (1 + 3h + 3h² + h³) − 4(1 + h) + 1 = −2 − h + 3h² + h³.
f(1 + h) − f(1) = −h + 3h² + h³, so the quotient is −1 + 3h + h².
Let h → 0: −1.
Check (substitute first): f(1.01) = 1.030 301 − 4.04 + 1 = −2.009 699, so (−2.009 699 − (−2))/0.01 = −0.9699, and −1 + 0.03 + 0.0001 = −0.9699 ✓.

−1: at x = 1 the function is decreasing, with tangent slope −1.

Just for fun

Method A is like checking if the cookies are done by opening the oven every 30 seconds: you learn a lot, but you never quite get there. Method B is reading the recipe, which tells you exactly when they will be ready.

Hands-on activity

Spreadsheet versus algebra

You need: a free spreadsheet (Google Sheets or LibreOffice) or a scientific calculator, and paper.

In column A, enter h = 0.1, 0.01, 0.001, 0.0001, and the same values with negative signs.
In column B, enter the unsimplified difference quotient for f(x) = x³ at a = 2: =((2+A1)^3-8)/A1, and fill down.
On paper, simplify the quotient to 12 + 6h + h². In column C, enter =12+6*A1+A1^2 and fill down.
Compare columns B and C. Are they equal?
Now add rows with h = 1E-10, 1E-13 and 1E-15. Compare the columns again.

What to look for: columns B and C agree for the ordinary values of h, which confirms your algebra. For extremely tiny h, column B may start to drift or even show 0, while column C stays at 12 to many decimal places. Column B subtracts two nearly equal numbers and runs out of digits (round-off error). Simplifying first avoids the problem entirely.

Your turn

Substitute first: evaluate ((2 + h)² − 4)/h for h = 0.1 and h = 0.01.

Simplify ((2 + h)² − 4)/h and find the instantaneous rate of change of x² at x = 2.

Substitute first: evaluate ((1 + h)³ − 1)/h for h = 0.1.

Simplify ((1 + h)³ − 1)/h and check your answer to the last question.

Which method can give an exact answer, and why?

Find the instantaneous rate of change of f(x) = x² + 5x at x = −2 by simplifying first.

For f(x) = x² at x = 3, the quotient is 6 + h. How small must |h| be for the secant slope to be within 0.001 of 6?

More practice: warm-up

Expand (3 + h)².

Expand (2 + h)³.

Simplify (6h + h²)/h for h ≠ 0.

Evaluate 6 + h at h = 0.01.

Why can't you substitute h = 0 into ((3 + h)² − 9)/h directly?

More practice: on your own

For f(x) = x⁴ at x = 1, compute the quotient with h = 0.01 (substitute first) and then simplify. Do they agree? What is the rate?

Find the instantaneous rate of change of f(x) = 2x³ at x = −1 by simplifying first.

Find the instantaneous rate of change of f(x) = x³ − 4x + 1 at x = 2.

Use the general result for y = x² (slope 2a) to find the slope at x = −4. Would Method A have given the same?

A ball's height is s(t) = 20t − 5t² metres. Find its velocity at t = 1 s both ways.

Challenge

For f(x) = x³ at a = 2, verify that Method A with h = −0.1 matches 12 + 6h + h². Then explain why the secants from the left give values below 12 and those from the right give values above 12.

A calculator keeps 10 significant digits. Explain what it would give for ((3 + h)² − 9)/h with h = 10⁻¹², and why simplifying first avoids the problem.

Find b so that the tangent to y = x³ + bx at x = 1 is horizontal. Simplify first.

Check your understanding

Quiz

Substituting h = 0.1, 0.01 and 0.001 into ((3 + h)² − 9)/h gives:

((3 + h)² − 9)/h simplifies to:

What is the main advantage of simplifying the difference quotient before letting h → 0?

The difference quotient for f(x) = x³ at x = 2 simplifies to:

What is the instantaneous rate of change of f(x) = x³ at x = a?

Using h = 10⁻¹⁵ in the unsimplified difference quotient on a calculator can give a wrong answer because:

For parents and teachers

Big idea (A1.6): for polynomials, the instantaneous rate at (a, f(a)) can be found by substituting smaller and smaller h into the difference quotient, or by simplifying first. Both agree; simplifying gives the exact value and shows the error term (for x² at x = 3, the secant slope is exactly 6 + h).

How to help: Do the spreadsheet activity together. Ask your teen to point to the error term in a simplified quotient and explain what it means for the secant.

Common difficulties:

Expanding (a + h)³ incorrectly. Use (a + h)³ = a³ + 3a²h + 3ah² + h³, or Pascal's triangle (1, 3, 3, 1) from MCR3U.
Forgetting to subtract f(a) before dividing by h.
Dividing only one term by h.
Trusting calculator results for extremely small h.