When the input sits in the bottom of a fraction, the output change becomes one fraction minus another with different bottoms. Think of converting two currencies into the same currency before subtracting: you cannot compare the counts while they use different-sized units. Give the fractions one common bottom by multiplying their original bottoms. Multiply each top by the missing bottom too, so its value stays unchanged. Then subtract only the tops, keeping the second top in parentheses. In the reciprocal examples, the new top has an h factor. Dividing by the input step means multiplying by the fraction 1 over h, and that whole factor can cancel. Keep every forbidden zero bottom.
For x = 1 and h = 1, the dashed ruler has slope −1 = −21 · 2. It connects curve points and is not part of f.
Reminder
Multiplication by 1.xx = 1 only for x ≠ 0. This is why multiplying both top and bottom by x preserves a fraction.
A common denominator.13 − 15 = 515 − 315 = 215. Rename fractions before subtracting.
Dividing by a nonzero letter. A ÷ h = A · 1h for h ≠ 0. Dividing reverses multiplication.
Why it works. A common bottom names the same-sized pieces, so their counts can be subtracted. Multiplying top and bottom by the same nonzero expression multiplies by 1 and preserves the value. For 2x, the common-bottom top is 2x − 2(x + h) = 2x − 2x − 2h = −2h. This shows the cancellation directly, rather than assuming every kind of function must look like a polynomial. Dividing by h multiplies by 1h, because that undoes multiplication by h. The restrictions remain necessary even after a factor disappears.
RuleFor f(x) = 2x, combine the output difference over x(x + h), then divide by h to obtain −2x(x + h). Require h ≠ 0, x ≠ 0, and x + h ≠ 0. Other fraction bottoms need their own product as the common bottom.
The same idea, five ways
Say it
Say 'make the output fractions use one bottom, subtract their tops, then divide by the step'.
Write it
A fractional difference quotient uses the usual endpoint-rate formula with an extra common-denominator step.
In math
For f(x) = 2x: rate = −2x(x + h)
h ≠ 0; x ≠ 0; x + h ≠ 0
For h > 0: [x, x + h]; for h < 0: [x + h, x]
Like
Convert two amounts to matching-sized units before subtracting, then share the difference over the input change.
See itFor x = 1 and h = 1, the dashed ruler has slope −1 = −21 · 2. It connects curve points and is not part of f.
The same idea, other ways
With numbers first
13 − 15 = 515 − 315 = 215. The common bottom changes the piece size but not either fraction's value. The worked numerical part shows how each missing factor is chosen.
Thirds and fifths need one shared piece size before their counts can be subtracted.
Keep a whole bottom together
The fraction 2x + h means 2 divided by the whole x + h. The bar groups that bottom like parentheses. You do not split it into 2x plus 2h.
2x + h means 2 ÷ (x + h)
whole bottom stays grouped
The fraction bar groups the full denominator.
Name all three zero risks
For 2x, the starting output needs x ≠ 0, the ending output needs x + h ≠ 0, and the rate needs h ≠ 0. These are three separate restrictions, even though the final formula has no isolated h in its bottom.
start: x ≠ 0
end: x + h ≠ 0
step: h ≠ 0
Check the original divisions before simplifying.
.1Common-denominator arithmetic before letters
The top of a fraction is its numerator, counting pieces. The bottom is its denominator, naming their size. Rename both fractions with the same bottom before subtracting.
Never subtract fraction bottoms.
Multiplying a top and bottom by the same nonzero value preserves the fraction.
Thirds and fifths need one shared piece size before their counts can be subtracted.
Worked exampleCommon bottoms with numbers before letters
This asks how to subtract two fractions with different bottoms. Find 13 − 15.
Thirds and fifths need one shared piece size before their counts can be subtracted.
Use common bottom 3 · 5 = 15.Neither bottom divides the other; their product gives a bottom both fractions can use.
13 = 515, and 15 = 315.Multiply each top and bottom by its missing factor: 5 for the first fraction and 3 for the second.
515 − 315 = 215.With equal bottoms, subtract only the tops and keep the common bottom.
Answer
215.
Check13 = 515, and 215 + 315 = 515, so adding the subtracted amount recovers the starting fraction.
.2A sum in the bottom
A bottom such as x + 2 remains one grouped expression. Substitute into the entire bottom and use both resulting bottoms in the common product.
For 3x + 2, the bottoms are x + 2 and x + h + 2.
Require x ≠ −2 and x + h ≠ −2, as well as h ≠ 0.
For the curve 3x + 2, the dashed ruler between x = 2 and x = 5 has slope −328 and is not part of f.
Worked exampleDifference quotient of f(x) = 52x + 3
Let f(x) = 52x + 3. Find and simplify the difference quotient f(x + h) − f(x)h. State every restriction on x and h.
Plug in: f(x + h) = 52(x + h) + 3 = 52x + 2h + 3.Every x, including the one below the fraction bar, is replaced by the whole new input x + h. The 2 multiplies both x and h.
Set up the subtraction: f(x + h) − f(x) = 52x + 2h + 3 − 52x + 3.The difference quotient starts with the new output minus the original output.
Use the common bottom (2x + 2h + 3)(2x + 3): 5(2x + 3) − 5(2x + 2h + 3)(2x + 2h + 3)(2x + 3).The common bottom is the product of the two original bottoms. Each top is multiplied by the factor its own bottom is missing. The second top stays grouped so the minus sign applies to all of it.
Distribute: 10x + 15 − 10x − 10h − 15(2x + 2h + 3)(2x + 3) = −10h(2x + 2h + 3)(2x + 3).The outside minus changes the sign of every term in the second product. Then 10x − 10x = 0 and 15 − 15 = 0, which leaves only −10h.
Divide by h: −10h(2x + 2h + 3)(2x + 3) · 1h = −10(2x + 2h + 3)(2x + 3).Dividing by h is the same as multiplying by 1h. Here h is a whole factor of the numerator and h ≠ 0, so it cancels.
State the restrictions: h ≠ 0, 2x + 3 ≠ 0 (so x ≠ −32), and 2x + 2h + 3 ≠ 0 (so x + h ≠ −32).The original quotient divides by h. Both f(x) and f(x + h) must be defined, so neither bottom can be zero. These restrictions still hold after cancelling.
Answer
f(x + h) − f(x)h = −10(2x + 2h + 3)(2x + 3), with h ≠ 0, x ≠ −32, and x + h ≠ −32.
Check Try x = 0 and h = 1. Directly: f(1) = 55 = 1 and f(0) = 53, so 1 − 531 = −23. With the formula: −10(0 + 2 + 3)(0 + 3) = −1015 = −23. The two results match.
Work to write
f(x + h) = 52x + 2h + 3
5(2x + 3) − 5(2x + 2h + 3)(2x + 2h + 3)(2x + 3)
−10h(2x + 2h + 3)(2x + 3)
−10(2x + 2h + 3)(2x + 3)
h ≠ 0, x ≠ −32, x + h ≠ −32
f(x + h) − f(x)h = −10(2x + 2h + 3)(2x + 3), with h ≠ 0, x ≠ −32, and x + h ≠ −32.
.3A fixed starting input
The same method works when the start is the number 2. The bottom at the start is then 3, and the bottom at the end is 3 + h.
For b(x) = 1x + 1 starting at 2, rate = −13(3 + h).
Require h ≠ 0 and h ≠ −3.
Starting at 2 with h = 3, the dashed ruler has slope −118; it is not part of b.
Worked exampleDifference quotient of 6x − 2 starting at x = 5
Let f(x) = 6x − 2. Find and simplify the difference quotient f(5 + h) − f(5)h. State every restriction on h.
Plug in: f(5 + h) = 6(5 + h) − 2 = 6h + 3, and f(5) = 65 − 2 = 63.Every x, including the one below the fraction bar, is replaced by the whole new input. f(5) is left as 63 so that both outputs are fractions with visible bottoms.
Subtract: f(5 + h) − f(5) = 6h + 3 − 63 = 6 · 3 − 6(h + 3)3(h + 3).The product of the two bottoms, 3(h + 3), is a common bottom. Each top is multiplied by the factor its bottom lacks. The second top, 6(h + 3), stays grouped so the minus sign applies to all of it.
Simplify the numerator: 18 − 6(h + 3) = 18 − 6h − 18 = −6h. The difference is −6h3(h + 3).Distributing the minus over the grouped term changes the sign of each piece. The constants 18 and −18 then cancel.
Divide by h: −6h3(h + 3) · 1h = −63(h + 3) = −2h + 3.Dividing by h is the same as multiplying by 1h. Since h ≠ 0, the whole factor h cancels. Finally −63 reduces to −2.
State the restrictions: h ≠ 0 and h ≠ −3.The quotient has h in its bottom, so h ≠ 0. The input 5 + h must not equal 2, which would make the bottom of f zero, so h ≠ −3. Both restrictions remain after simplifying.
Answer
f(5 + h) − f(5)h = −2h + 3, for h ≠ 0 and h ≠ −3.
Check Take h = 1. Then f(6) = 64 = 1.5 and f(5) = 2, so the quotient is 1.5 − 21 = −0.5. The formula gives −21 + 3 = −0.5. The two values match.
Work to write
f(5 + h) = 6h + 3 and f(5) = 63
6h + 3 − 63 = 18 − 6(h + 3)3(h + 3)
Numerator: 18 − 6h − 18 = −6h
−6h3(h + 3) · 1h = −2h + 3
Restrictions: h ≠ 0, h ≠ −3
f(5 + h) − f(5)h = −2h + 3, for h ≠ 0 and h ≠ −3.
.4The input in both top and bottom
Every occurrence of the input must change. A common bottom can create two products on top, so multiply out both before combining their difference.
For xx + 1, substitute x + h above and below the bar.
The common top simplifies to h, giving rate 1(x + 1)(x + h + 1).
Require h ≠ 0, x ≠ −1, and x + h ≠ −1.
The curve is 1 − 1x + 1, equal to xx + 1; the dashed ruler has slope 118 for x = 2 and h = 3.
Worked exampleRung 5: the input also appears in the top
This asks for an average-rate formula when the input occurs above and below the fraction bar. Find the difference quotient of f(x) = xx + 1.
The curve is 1 − 1x + 1, equal to xx + 1; the dashed ruler has slope 118 for x = 2 and h = 3.
f(x) = xx + 1
common bottom:
(x + 1)(x + h + 1)
combined top simplifies to h
When the input occurs in both the top and bottom, substitute it in both places.The curve is 1 − 1x + 1, equal to xx + 1; the dashed ruler has slope 118 for x = 2 and h = 3.
Require h ≠ 0, x ≠ −1, and x + h ≠ −1.The step and both original function bottoms must be nonzero.
f(x + h) = x + hx + h + 1, and f(x) = xx + 1.The new input replaces both occurrences of x in the first output.
Use common bottom (x + 1)(x + h + 1). The first top becomes (x + h)(x + 1), and the second becomes x(x + h + 1).Each top gets multiplied by the whole bottom missing from its own fraction.
The combined top is (x + h)(x + 1) − x(x + h + 1).After the bottoms match, subtract the entire second top.
(x + h)(x + 1) = x · x + x · 1 + h · x + h · 1 = x2 + x + xh + h.Each term of the first parentheses multiplies each term of the second.
x(x + h + 1) = x2 + xh + x.The outside x multiplies all three terms inside.
The combined top is (x2 + x + xh + h) − (x2 + xh + x) = x2 + x + xh + h − x2 − xh − x = h.Distribute the outside minus, then cancel the matched x2, xh, and x terms.
The output change is h(x + 1)(x + h + 1). Divide by h: h(x + 1)(x + h + 1) · 1h = 1(x + 1)(x + h + 1).The common nonzero multiplier h cancels from the multiplied fraction.
Answer
1(x + 1)(x + h + 1), with h ≠ 0, x ≠ −1, and x + h ≠ −1.
Check At x = 2 and h = 3, f(5) = 56 and f(2) = 23. Direct rate: (56 − 46) ÷ 3 = 16 · 13 = 118, matching 13 · 6.
Strategy: step by step
1. Plug in: replace every input occurrence, including those below a fraction bar, by the whole new input.
2. Subtract: use the product of the original bottoms as a common bottom, multiply each top by its missing factor, and keep the second top grouped.
3. Simplify: distribute the outside minus and combine the numerator terms.
4. Cancel a valid h factor: multiply by 1h, cancel only a whole nonzero factor, and keep all original restrictions.
Strategy
The fractional difference quotient in four moves
1
Is the bottom exactly x for the original function?
YesIts two output bottoms are x and x + h; use x(x + h).
NoWrite the actual two whole bottoms and use their product.
↓
2
Does the input appear in the numerator too?
YesReplace it there as well; multiply out the new top products carefully.
NoKeep the constant numerator as given.
↓
3
Has the minus reached the entire second numerator?
YesCombine matching numerator terms.
NoKeep that numerator in parentheses and distribute the outside −1.
↓
4
Is the starting input a fixed number?
YesUse its value for the starting bottom and start-plus-h for the ending bottom.
NoKeep the general starting letter.
↓
5
Does any original bottom or h equal 0?
YesExclude that input or step before cancellation.
NoCancel a whole common h factor if one is present.
Plug in and record every nonzero-bottom condition.
Subtract using a common bottom for the two output fractions.
Simplify the combined top while leaving the bottom as a product.
Divide by the step through multiplication by 1h, and cancel only a whole factor.
Worked exampleDifference quotient with x in the denominator
This asks for a reusable average-rate formula when the input is in a fraction bottom. Find f(x + h) − f(x)h for f(x) = 2x.
For x = 1 and h = 1, the dashed ruler has slope −1 = −21 · 2. It connects curve points and is not part of f.
Require h ≠ 0, x ≠ 0, and x + h ≠ 0. Write f(x + h) = 2x + h and f(x) = 2x.The outer division by h and both original function outputs need nonzero denominators.
The output change is 2x + h − 2x. Use common bottom x(x + h).The two bottoms are different, so both fractions must be renamed using the same-sized pieces before subtracting.
2x + h = 2xx(x + h).Multiply the first top and bottom by x; since x ≠ 0, this multiplies the value by xx = 1.
2x = 2(x + h)x(x + h).Multiply the second top and bottom by x + h; its nonzero condition makes this multiplication by 1.
Subtract the tops: 2x − 2(x + h)x(x + h).The common bottom stays fixed. Keep the whole second top grouped so the minus reaches both pieces.
2x − (2x + 2h)x(x + h) = 2x − 2x − 2hx(x + h) = −2hx(x + h).Distribute 2 inside the second top, then distribute the outside minus to +2x and +2h. The +2x and −2x cancel.
Divide by the input change h: −2hx(x + h) ÷ h = −2hx(x + h) · 1h.Dividing by nonzero h is multiplying by its reciprocal, so this multiplication compares the output change with the input step.
−2hx(x + h) · 1h = −2hh · x(x + h) = −2x(x + h).Multiply tops and bottoms. The whole numerator −2h and denominator both contain the nonzero factor h, which cancels.
Answer
−2x(x + h), with h ≠ 0, x ≠ 0, and x + h ≠ 0.
Check Use x = 2 and h = 3. f(5) = 25 and f(2) = 1, so the direct rate is (25 − 1) ÷ 3 = −35 · 13 = −15. The result gives −22 · 5 = −15.
Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.
Rung 1Rung 1: the unit reciprocal
This asks for the average-rate expression for the reciprocal function. Find the difference quotient of f(x) = 1x.
The dashed ruler through the two endpoint curve points has slope −110 and is not part of f.
Require h ≠ 0, x ≠ 0, and x + h ≠ 0. The output change is 1x + h − 1x.All three original divisions need nonzero bottoms.
Use common bottom x(x + h): 1x + h = xx(x + h) and 1x = x + hx(x + h).Multiply the first top and bottom by x and the second top and bottom by x + h; both multipliers are nonzero.
Subtract: x − (x + h)x(x + h) = x − x − hx(x + h) = −hx(x + h).The second top stays in parentheses until its minus reaches both x and h.
Divide by h: −hx(x + h) · 1h = −hh · x(x + h).Multiplying by the reciprocal divides the output change by the input step.
−hh · x(x + h) = −1x(x + h).The nonzero factor h cancels; the minus sign remains.
Answer
−1x(x + h), also written −1x(x + h).
Require h ≠ 0, x ≠ 0, and x + h ≠ 0.
Check At x = 2 and h = 3, (15 − 12) ÷ 3 = (210 − 510) · 13 = −310 · 13 = −110. The result gives −12 · 5 = −110.
Rung 2Rung 2: a constant numerator of 2
This asks for a reusable average-rate formula when the input is in a fraction bottom. Find f(x + h) − f(x)h for f(x) = 2x.
For x = 1 and h = 1, the dashed ruler has slope −1 = −21 · 2. It connects curve points and is not part of f.
Require h ≠ 0, x ≠ 0, and x + h ≠ 0. Write f(x + h) = 2x + h and f(x) = 2x.The outer division by h and both original function outputs need nonzero denominators.
The output change is 2x + h − 2x. Use common bottom x(x + h).The two bottoms are different, so both fractions must be renamed using the same-sized pieces before subtracting.
2x + h = 2xx(x + h).Multiply the first top and bottom by x; since x ≠ 0, this multiplies the value by xx = 1.
2x = 2(x + h)x(x + h).Multiply the second top and bottom by x + h; its nonzero condition makes this multiplication by 1.
Subtract the tops: 2x − 2(x + h)x(x + h).The common bottom stays fixed. Keep the whole second top grouped so the minus reaches both pieces.
2x − (2x + 2h)x(x + h) = 2x − 2x − 2hx(x + h) = −2hx(x + h).Distribute 2 inside the second top, then distribute the outside minus to +2x and +2h. The +2x and −2x cancel.
Divide by the input change h: −2hx(x + h) ÷ h = −2hx(x + h) · 1h.Dividing by nonzero h is multiplying by its reciprocal, so this multiplication compares the output change with the input step.
−2hx(x + h) · 1h = −2hh · x(x + h) = −2x(x + h).Multiply tops and bottoms. The whole numerator −2h and denominator both contain the nonzero factor h, which cancels.
Answer
−2x(x + h), with h ≠ 0, x ≠ 0, and x + h ≠ 0.
Check Use x = 2 and h = 3. f(5) = 25 and f(2) = 1, so the direct rate is (25 − 1) ÷ 3 = −35 · 13 = −15. The result gives −22 · 5 = −15.
Rung 3Difference quotient when the bottom is x + 2
Let f(x) = 3x + 2. Find and simplify the difference quotient f(x + h) − f(x)h. State every restriction on x and h.
Plug in: f(x + h) = 3(x + h) + 2 = 3x + h + 2, so the difference quotient is 3x + h + 2 − 3x + 2h.Every x, including the one inside the bottom x + 2, is replaced by the whole new input x + h.
Subtract using the common bottom (x + h + 2)(x + 2): 3x + h + 2 − 3x + 2 = 3(x + 2) − 3(x + h + 2)(x + h + 2)(x + 2).The common bottom is the product of the two original bottoms. Each top is multiplied by the factor its bottom is missing. The second top, 3(x + h + 2), stays grouped so the minus applies to all of it.
Simplify the top: 3(x + 2) − 3(x + h + 2) = 3x + 6 − 3x − 3h − 6 = −3h. The difference is −3h(x + h + 2)(x + 2).Distributing the outside minus changes the sign of every term in the second group. The 3x terms cancel and the 6 terms cancel, leaving only −3h.
Divide by h: −3h(x + h + 2)(x + 2) · 1h = −3(x + 2)(x + h + 2).Dividing by h is multiplying by 1h. h is a whole factor of the top, and h ≠ 0, so it cancels. The original restrictions still hold after the cancellation.
Record the restrictions: h ≠ 0, x ≠ −2, and x + h ≠ −2.h ≠ 0 because we divided by h. The bottom x + 2 cannot be 0, so x ≠ −2. The bottom x + h + 2 cannot be 0, so x + h ≠ −2.
Answer
f(x + h) − f(x)h = −3(x + 2)(x + h + 2), with h ≠ 0, x ≠ −2, and x + h ≠ −2.
Check Take x = 1 and h = 1. Directly: f(2) = 34 and f(1) = 33 = 1, so f(2) − f(1)1 = 34 − 1 = −14. From the formula: −3(1 + 2)(1 + 1 + 2) = −33 · 4 = −14. The two values match.
Work to write
f(x + h) = 3x + h + 2
3(x + 2) − 3(x + h + 2)(x + h + 2)(x + 2)
3x + 6 − 3x − 3h − 6 = −3h
−3h(x + h + 2)(x + 2) · 1h
−3(x + 2)(x + h + 2)
h ≠ 0, x ≠ −2, x + h ≠ −2
f(x + h) − f(x)h = −3(x + 2)(x + h + 2), with h ≠ 0, x ≠ −2, and x + h ≠ −2.
Rung 4Difference quotient of f(x) = 4x − 1 starting at x = 3
Let f(x) = 4x − 1. Find and simplify the difference quotient f(3 + h) − f(3)h. State every restriction on h.
Plug in: f(3 + h) = 4(3 + h) − 1 = 4h + 2 and f(3) = 43 − 1 = 42.Every x, including the one below the fraction bar, is replaced by the whole new input. The fraction 42 stays unreduced so that both terms are still fractions with their own bottoms.
Subtract over the common bottom 2(h + 2): 4h + 2 − 42 = 4·2 − 4(h + 2)2(h + 2) = 8 − (4h + 8)2(h + 2).The product of the original bottoms is a common bottom. Each top is multiplied by the factor its bottom is missing. The second top stays grouped so that the minus sign applies to all of it.
Simplify: 8 − 4h − 82(h + 2) = −4h2(h + 2) = −2hh + 2.Distributing the minus gives 8 − 4h − 8. The constants cancel and leave −4h. Then the common factor 2 is reduced.
Divide by h: −2hh + 2 · 1h = −2h + 2, with h ≠ 0 and h ≠ −2.Dividing by h is the same as multiplying by 1h. The whole factor h cancels only because h ≠ 0. The restriction h ≠ −2 comes from the original bottom (3 + h) − 1 ≠ 0, and it stays even though h itself has cancelled.
Answer
f(3 + h) − f(3)h = −2h + 2, with h ≠ 0 and h ≠ −2.
Check Take h = 1. Then f(4) = 43 and f(3) = 2, so f(4) − f(3)1 = 43 − 2 = −23. The formula gives −21 + 2 = −23, which matches. The general result −4(x − 1)(x + h − 1) at x = 3 also gives −42(h + 2) = −2h + 2.
Work to write
f(3 + h) = 4h + 2, f(3) = 42
4h + 2 − 42 = 8 − 4(h + 2)2(h + 2)
= −4h2(h + 2) = −2hh + 2
−2hh + 2 · 1h = −2h + 2
h ≠ 0, h ≠ −2
f(3 + h) − f(3)h = −2h + 2, with h ≠ 0 and h ≠ −2.
Rung 5Rung 5: the input also appears in the top
This asks for an average-rate formula when the input occurs above and below the fraction bar. Find the difference quotient of f(x) = xx + 1.
The curve is 1 − 1x + 1, equal to xx + 1; the dashed ruler has slope 118 for x = 2 and h = 3.
f(x) = xx + 1
common bottom:
(x + 1)(x + h + 1)
combined top simplifies to h
When the input occurs in both the top and bottom, substitute it in both places.The curve is 1 − 1x + 1, equal to xx + 1; the dashed ruler has slope 118 for x = 2 and h = 3.
Require h ≠ 0, x ≠ −1, and x + h ≠ −1.The step and both original function bottoms must be nonzero.
f(x + h) = x + hx + h + 1, and f(x) = xx + 1.The new input replaces both occurrences of x in the first output.
Use common bottom (x + 1)(x + h + 1). The first top becomes (x + h)(x + 1), and the second becomes x(x + h + 1).Each top gets multiplied by the whole bottom missing from its own fraction.
The combined top is (x + h)(x + 1) − x(x + h + 1).After the bottoms match, subtract the entire second top.
(x + h)(x + 1) = x · x + x · 1 + h · x + h · 1 = x2 + x + xh + h.Each term of the first parentheses multiplies each term of the second.
x(x + h + 1) = x2 + xh + x.The outside x multiplies all three terms inside.
The combined top is (x2 + x + xh + h) − (x2 + xh + x) = x2 + x + xh + h − x2 − xh − x = h.Distribute the outside minus, then cancel the matched x2, xh, and x terms.
The output change is h(x + 1)(x + h + 1). Divide by h: h(x + 1)(x + h + 1) · 1h = 1(x + 1)(x + h + 1).The common nonzero multiplier h cancels from the multiplied fraction.
Answer
1(x + 1)(x + h + 1), with h ≠ 0, x ≠ −1, and x + h ≠ −1.
Check At x = 2 and h = 3, f(5) = 56 and f(2) = 23. Direct rate: (56 − 46) ÷ 3 = 16 · 13 = 118, matching 13 · 6.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this:2x + h = 2x + 2h.
At x = 1 and h = 1 the left side is 22 = 1, but the right side is 2 + 2 = 4.
✓ Instead: Keep the entire denominator grouped: 2x + h = 2 ÷ (x + h).
✗ Not this: Subtract fraction tops and bottoms directly: 23 − 25 = 0−2 = 0.
The original fractions have different piece sizes. Their actual difference is 1015 − 615 = 415.
✓ Instead: Make the bottoms match, then subtract only the tops.
✗ Not this: The bottom x(x + h) is required for every function with a fraction.
For 3x + 2, the actual output bottoms are x + 2 and x + h + 2.
✓ Instead: Use the product (x + 2)(x + h + 2) for that function.
✗ Not this: After canceling h, allow h = 0, x = 0, or x + h = 0.
Those choices make an original division undefined. A shortened expression cannot restore missing original outputs or a zero input span.
✓ Instead: Preserve h ≠ 0, x ≠ 0, and x + h ≠ 0 for the reciprocal function.
Tips and tricks
The fraction bar groups the entire bottom: 2x + h means 2 ÷ (x + h).
Leave the common bottom as a product so its nonzero factors and any cancellations stay visible.
Show the combined top before opening the parentheses. In 2x − 2(x + h), the minus must reach both 2x and 2h.
Check with x = 2 and h = 3, provided every original bottom is nonzero.
A formula built from endpoint values does not describe values at an input where the original function is undefined; keep domain gaps in mind when interpreting the interval.
Trap. Subtracting bottoms, splitting a denominator sum, or losing the minus on the second numerator. Use a common bottom, subtract the whole second top, then divide by h.
