How do you expand (2+h)³?

Use the binomial expansion (a+b)³ = a³ + 3a²b + 3ab² + b³ with a = 2 and b = h. That gives (2+h)³ = 8 + 12h + 6h² + h³. You can also multiply (2+h)(2+h)(2+h) by hand, but the binomial expansion is faster and less error-prone.

Why does the 8 cancel in the numerator?

Because f(2) = 2³ = 8 is subtracted from f(2+h) = 8 + 12h + 6h² + h³. The constant 8 in the expansion is exactly the value of f at x = 2, so subtracting f(2) zeros it out. Every difference quotient does this: the f(a) term is built to cancel the constant piece of f(a+h).

What is the simplified difference quotient for f(x) = x³ at x = 2?

h² + 6h + 12. After expanding (2+h)³ = 8 + 12h + 6h² + h³, subtracting f(2) = 8, and dividing every remaining term by h, the result is h² + 6h + 12.

Why do we need the difference quotient before learning derivatives?

Because the derivative f′(a) is defined as the limit of the difference quotient [f(a+h) − f(a)] / h as h approaches 0. Until a student can fluently expand, subtract, cancel, and divide a difference quotient, the limit definition of the derivative is just a symbol. Stewart Ch.1 Q2 is the algebra drill that makes the limit work.

AP Calculus & Pre-Calc · Stewart Ch.1

Difference Quotient of f(x) = x³ at x = 2 (Stewart Ch.1 Q2)

Q: What is the difference quotient?

The difference quotient is the expression [f(a+h) − f(a)] / h. It measures the average rate of change of f over the interval from x = a to x = a + h, and is the algebraic object whose limit as h → 0 defines the derivative f′(a).

If f(x) = x³, evaluate the difference quotient [f(2 + h) − f(2)] / h and simplify your answer. This is Stewart's first real algebra-meets-calculus problem. The answer is h² + 6h + 12, and the four steps that get you there are the foundation of every derivative computation that follows.

2026-05-29 · By the School of Math team · 226 W 79th St, UWS

Stewart Calculus Chapter 1 Question 2 — difference quotient of f(x) = x cubed at x = 2, four step walkthrough showing the expansion of (2+h)^3 and the simplified answer h squared plus 6h plus 12, worked at SOMATH on the Upper West Side.

Walkthrough from the SOMATH classroom · Watch on YouTube

Direct answer. For f(x) = x³, the difference quotient at x = 2 simplifies to h² + 6h + 12. Reached in four steps: expand (2 + h)³ = 8 + 12h + 6h² + h³; subtract f(2) = 8; cancel the 8s in the numerator; divide every remaining term by h.

Why Stewart asks this question second

The first problem in the textbook tests whether you can read a function — input, output, zeros, domain, range. The second problem tests whether you can manipulate a function — substitute, expand, subtract, divide. These are the two languages of calculus: the geometric one (graphs) and the algebraic one (symbols). Every later topic depends on switching fluently between them.

The difference quotient is also the single most important algebraic object in Chapter 2. The derivative f′(a) = lim_h→0 [f(a + h) − f(a)] / h is exactly the limit of what you compute here. By making you simplify the difference quotient first, Stewart sets up the limit so it actually has somewhere to go — divide by h first, then take the limit, and the answer falls out.

Step 1 — Compute f(2 + h)

Substitute 2 + h into f:

f(2 + h) = (2 + h)³

Now expand using the binomial expansion (a + b)³ = a³ + 3a²b + 3ab² + b³ with a = 2 and b = h:

(2 + h)³ = 2³ + 3(2)²h + 3(2)h² + h³ = 8 + 12h + 6h² + h³

You can also multiply (2 + h)(2 + h)(2 + h) by hand. We teach the binomial expansion first because it scales — the same pattern works for (x + h)³, (x + h)⁴, and beyond, which is exactly what you need for the next dozen Stewart problems.

Step 2 — Compute f(2)

The other half of the numerator. Substitute x = 2 directly:

f(2) = 2³ = 8

Notice the 8 here matches the 2³ = 8 constant term in the expansion above. That is not a coincidence — it is the entire reason the difference quotient simplifies.

Step 3 — Substitute into the difference quotient

Plug both pieces back into [f(2 + h) − f(2)] / h:

(8 + 12h + 6h² + h³) − 8 all over h

The 8 in the expansion cancels with the −8 from f(2):

= (12h + 6h² + h³) / h

This is the move that makes the problem work. Every remaining term in the numerator has a factor of h, so the h in the denominator can actually cancel. If the constants had not lined up, the difference quotient would blow up when you eventually let h → 0 — instead, the cancellation leaves a clean polynomial in h.

Step 4 — Divide each term by h

Split the fraction term by term:

12h / h + 6h² / h + h³ / h = 12 + 6h + h²

Reorder by descending powers of h:

h² + 6h + 12

That's the simplified difference quotient. Stewart accepts either ordering; the descending-power ordering matches what you'll see when you turn this into a derivative in Chapter 2.

What this answer is hiding

The expression h² + 6h + 12 is not just an algebraic answer. It is the average rate of change of x³ over the interval from x = 2 to x = 2 + h. Plug in different values of h and watch what happens:

h = 1: rate = 1 + 6 + 12 = 19 (slope of secant from (2, 8) to (3, 27)).
h = 0.1: rate = 0.01 + 0.6 + 12 = 12.61.
h = 0.01: rate = 0.0001 + 0.06 + 12 = 12.0601.
h → 0: h² + 6h + 12 → 12.

That limit is the derivative. f′(2) = 12, which is consistent with the power rule f′(x) = 3x² applied at x = 2: 3(2)² = 12. Same answer, two roads — and Stewart wants you to walk both before he lets you skip ahead to the rule.

The three mistakes worth memorizing

Writing (2 + h)³ = 8 + h³. This is the most common error in all of Chapter 2. Cubing is not distributive over addition. Always expand fully using the binomial expansion or hand-multiplication.
Forgetting to subtract f(2). Students sometimes write the numerator as just f(2 + h) and divide by h. That gives (8 + 12h + 6h² + h³) / h, which has an 8/h term and does not simplify. The subtraction of f(a) is what makes the difference quotient finite.
Cancelling h before splitting the fraction. You cannot cancel h from a single term of the numerator and leave the rest alone. You must either factor h out of the entire numerator first — h(12 + 6h + h²) / h — or split the fraction term-by-term and divide each piece by h separately.

How SOMATH teaches this problem

Every AP Calc and pre-calc student at SOMATH does the difference quotient on three functions in their first week — x², x³, and 1/x — at three points each. Same four steps every time: expand f(a + h), compute f(a), subtract, divide by h. The point is not to memorize the answers but to make the procedure automatic, so when the limit shows up next chapter the student is not also wrestling with algebra.

We teach this in small groups of three to five at 226 W 79th St (Upper West Side, between Broadway and Amsterdam), with whiteboards on every wall and no worksheet stacks. Each technique gets the binomial expansion, the cancellation diagnosis, and the link to the derivative. By the time the AP Calc class meets limits, the difference quotient is muscle memory.

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FAQ

What is the difference quotient and why does it matter?

The difference quotient is [f(a + h) − f(a)] / h. Geometrically it is the slope of the secant line through the two points (a, f(a)) and (a + h, f(a + h)) on the graph of f. As h → 0, that secant line tilts toward the tangent line, and its slope becomes the derivative.

Is there a faster way using the power rule?

Yes — f′(x) = 3x², so f′(2) = 12. But Stewart Ch.1 Q2 asks for the difference quotient, not the derivative. The whole point is to do the algebra by hand so that when the limit definition is introduced, you understand where the answer comes from. The power rule comes later in the textbook precisely because it skips this step.

Why do we divide by h instead of just subtracting?

Because dividing by h converts a difference (f(2 + h) − f(2)) into a rate (change per unit of h). Without the division, you'd be measuring how much f changed, not how fast. Calculus is the study of rates, so every derivative starts with a division by the change in input.

What if h is negative?

The algebra is identical and the simplified answer h² + 6h + 12 still holds. The secant just goes the other direction — from (2 + h, f(2 + h)) to (2, f(2)) with h < 0. As h → 0 from either side, the difference quotient converges to the same limit (12), which is exactly what makes the derivative well-defined here.

What is the next Stewart problem after this one?

Stewart Ch.1 Q3 asks you to compute the difference quotient for a general x (instead of a specific point), using a function like 1/x. Same four steps, but the algebra trades polynomial expansion for getting a common denominator. We have walkthroughs of the surrounding problems linked at the bottom.

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Difference Quotient of f(x) = x3 at x = 2 (Stewart Ch.1 Q2)