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Precalculus

The difference quotient

The average rate of change takes one interval at a time. Can a single expression give it for every interval at once?

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The difference quotient

When you reach for it

The average rate of change from a formula answers one question at a time: name an interval, evaluate the rule at both ends, divide the change in output by the width. Ask about a second interval and the work starts over.

The difference quotient does that work once, for every interval. Call the left end x and the width h, so the interval runs from x to x + h. The average rate on it is then a single expression in x and h, and any particular interval is just a pair of numbers put into it.

Reach for it when you want to see how the average rate depends on where the interval sits and how wide it is, and above all when you want to watch it as the width shrinks. That last use, narrower and narrower intervals around one input, is how a rate at a single instant is found.

The procedure

On the interval from x to x + h the outputs are f(x) and f(x + h), and the width is (x + h) - x = h. The average rate of change, change in output over change in input, is therefore

\frac{f(x + h) - f(x)}{h}, \qquad h \ne 0.

This is the difference quotient of f. Geometrically it is the slope of the secant line through (x, f(x)) and (x + h, f(x + h)), exactly as in the slope from two points.

x y
x
x + h
f(x)
f(x + h)
h
Figure 1The difference quotient is the slope of the secant through (x, f(x)) and (x + h, f(x + h)). Drawn here with x = 1 and h = 2.

Finding it for a given rule is three moves, each with its own reason.

Evaluate at x + h. The whole expression x + h goes in wherever x stands in the rule, as in evaluating a function at an expression. For f(x) = x^2 that gives f(x + h) = x^2 + 2xh + h^2.

Subtract f(x). For a polynomial, every term that does not contain h cancels: here x^2 goes, leaving 2xh + h^2. This is no coincidence. When h = 0 the two inputs are equal, so f(x + h) - f(x) is 0; a polynomial in h that is 0 at h = 0 has no constant term, so every surviving term carries a factor of h.

Divide by h. Because h \ne 0, the common factor h cancels from top and bottom: \displaystyle \frac{2xh + h^2}{h} = 2x + h. That cancellation is the point of the exercise. The original quotient has h alone in its denominator and makes no sense at h = 0; the simplified form 2x + h can be evaluated for any small h without dividing by anything, which is what shrinking intervals need.

For rational and root rules the factor of h is hidden, and the subtraction step needs a further move to expose it: a common denominator for a fraction, the conjugate for a square root. Examples 4 and 5 show both.

The width h may be negative. Then the interval runs from x + h up to x, and the quotient still gives its average rate, because swapping the two ends changes the sign of both the top and the bottom.

Worked examples

Example 1

A line

Find the difference quotient of \displaystyle g(x) = 7 - \frac{5}{2}x.

Solution

Put x + h in for x:

g(x + h) = 7 - \frac{5}{2}(x + h) = 7 - \frac{5}{2}x - \frac{5}{2}h.

Subtracting g(x) removes 7 and \displaystyle -\frac{5}{2}x, leaving \displaystyle g(x + h) - g(x) = -\frac{5}{2}h. Divide by h, allowed since h \ne 0:

\frac{g(x + h) - g(x)}{h} = -\frac{5}{2}.

No x and no h remain: every interval, wherever it starts and however wide, has average rate \displaystyle -\frac{5}{2}. That is what it means for the graph to be a line of slope \displaystyle -\frac{5}{2}; every secant is the line itself.

Example 2

A quadratic rule

Find and simplify the difference quotient of f(x) = -x^2 + 6x - 1.

Solution

Evaluate at x + h, expanding (x + h)^2 = x^2 + 2xh + h^2 and distributing the minus sign over all three terms:

\begin{aligned} f(x + h) &= -(x + h)^2 + 6(x + h) - 1 \\ &= -x^2 - 2xh - h^2 + 6x + 6h - 1. \end{aligned}

Subtract f(x) = -x^2 + 6x - 1. The terms -x^2, 6x and -1 cancel, as the procedure promised:

f(x + h) - f(x) = -2xh - h^2 + 6h.

Every term carries h. Factor it out and cancel it, since h \ne 0:

\begin{aligned} \frac{f(x + h) - f(x)}{h} &= \frac{h(-2x - h + 6)}{h} \\ &= -2x - h + 6. \end{aligned}

Check with x = 1 and h = 2, the interval [1, 3] in the figure above: f(1) = 4 and f(3) = 8, so the average rate is \displaystyle \frac{8 - 4}{2} = 2. The formula gives -2 - 2 + 6 = 2.

Example 3

Reading the quotient

For f(x) = -x^2 + 6x - 1, use the difference quotient -2x - h + 6 from Example 2 to find the average rate of change on [2, 5]. Then find which intervals have an average rate of 0.

Solution

The interval [2, 5] starts at x = 2 and has width h = 3, so the average rate is -2(2) - 3 + 6 = -1. Directly: f(2) = 7 and f(5) = 4, and \displaystyle \frac{4 - 7}{3} = -1.

For the second question, regroup the quotient around the midpoint of the interval, \displaystyle x + \frac{h}{2}:

-2x - h + 6 = -2\left(x + \frac{h}{2}\right) + 6.

So the average rate depends only on where the interval’s middle is. It is 0 exactly when the midpoint is 3: every interval centered on x = 3, such as [2, 4] or [0.5, 5.5], and no other. That fits the graph, since 3 is the input of the vertex and a parabola is symmetric about it, so the two ends of a centered interval have equal outputs. Intervals with midpoint left of 3 have a positive average rate, and those with midpoint right of 3 a negative one.

Example 4

A rational rule

Find and simplify the difference quotient of \displaystyle f(x) = \frac{3}{2x + 1}.

Solution

Evaluating at x + h gives \displaystyle f(x + h) = \frac{3}{2(x + h) + 1} = \frac{3}{2x + 2h + 1}. To subtract, use the common denominator (2x + 2h + 1)(2x + 1), as in subtracting rational expressions. Over it, the numerator of f(x + h) - f(x) is

3(2x + 1) - 3(2x + 2h + 1) = -6h,

because 6x and 3 appear in both products and cancel. So

f(x + h) - f(x) = \frac{-6h}{(2x + 2h + 1)(2x + 1)}.

Dividing by h is multiplying by \displaystyle \frac{1}{h}, and the h cancels:

\begin{aligned} &\frac{f(x + h) - f(x)}{h} \\ &\quad = \frac{-6}{(2x + 2h + 1)(2x + 1)}. \end{aligned}

Check with x = 1 and h = 1: f(1) = 1 and \displaystyle f(2) = \frac{3}{5}, so the average rate is \displaystyle \frac{3}{5} - 1 = -\frac{2}{5}. The formula gives \displaystyle \frac{-6}{5 \cdot 3} = -\frac{2}{5}.

The result holds only where both ends are in the domain: \displaystyle x \ne -\frac{1}{2} and \displaystyle x + h \ne -\frac{1}{2}.

Example 5

A square root

Find and simplify the difference quotient of \displaystyle f(x) = \sqrt{3x + 1}.

Solution

The quotient is

\frac{\sqrt{3x + 3h + 1} - \sqrt{3x + 1}}{h},

and the factor of h is hidden inside the roots. Multiply top and bottom by the conjugate \displaystyle \sqrt{3x + 3h + 1} + \sqrt{3x + 1}, which is not 0 when 3x + 1 > 0 and so changes nothing. This is the move used to rationalize a denominator, applied to the numerator. By (A - B)(A + B) = A^2 - B^2, the numerator becomes

(3x + 3h + 1) - (3x + 1) = 3h,

so, cancelling h,

\begin{aligned} &\frac{f(x + h) - f(x)}{h} \\ &\quad = \frac{3}{\sqrt{3x + 3h + 1} + \sqrt{3x + 1}}. \end{aligned}

Check with x = 0 and h = 1: f(0) = 1 and f(1) = 2, average rate 1. The formula gives \displaystyle \frac{3}{2 + 1} = 1. The rule needs 3x + 1 \ge 0 and 3x + 3h + 1 \ge 0, with the two not both 0.

Example 6

Bayarmaa’s dive

Bayarmaa, a cliff diver, springs upward off a ledge. Her height above the water t seconds later is s(t) = 20 + 3t - 5t^2 meters, for 0 \le t \le 2. Find the average rate of change of her height on [t, t + h], then use it for intervals starting at t = 1 with h = 1, 0.5, 0.1, 0.01 and -0.5.

Solution

Here the input is t, so t + h replaces it:

\begin{aligned} s(t + h) &= 20 + 3(t + h) - 5(t + h)^2 \\ &= 20 + 3t + 3h - 5t^2 - 10th - 5h^2. \end{aligned}

Subtracting s(t) leaves 3h - 10th - 5h^2, and dividing by h gives

\frac{s(t + h) - s(t)}{h} = 3 - 10t - 5h

meters per second, her average velocity over the interval. At t = 1 it is -7 - 5h:

hintervalaverage rate
1[1, 2]-12
0.5[1, 1.5]-9.5
0.1[1, 1.1]-7.5
0.01[1, 1.01]-7.05
-0.5[0.5, 1]-4.5

Check one entry directly: s(1) = 18 and s(1.5) = 13.25, so the rate on [1, 1.5] is \displaystyle \frac{13.25 - 18}{0.5} = -9.5. The last entry is the interval [0.5, 1]: s(0.5) = 20.25, and \displaystyle \frac{18 - 20.25}{-0.5} = -4.5.

1 2 10 20 t s
h = 1
h = 0.5
Figure 2Bayarmaa’s height in Example 6. Narrower intervals starting at t = 1 give secants that are less steep.

The signs say she is falling on each interval. As h shrinks, from either side, the averages crowd toward -7, a number no single interval gave. Making that estimate of her speed at the instant t = 1 is the next step, below.

Boundaries

  • The width must not be zero. At h = 0 the quotient is \displaystyle \frac{0}{0}, undefined, because one point makes no interval. The simplified form, such as -2x - h + 6, can be evaluated at h = 0, but the number it gives is not the average rate of any interval.
  • Both ends must be in the domain. For \displaystyle \frac{3}{2x + 1} neither end may be \displaystyle -\frac{1}{2}; for \displaystyle \sqrt{3x + 1} both ends need 3x + 1 \ge 0. If the domain has a gap between the ends, the quotient may still compute, but it describes no single stretch of the graph.
  • It is still an average. The quotient sees only the two ends. Two functions that agree at x and at x + h have the same difference quotient there, whatever happens in between.

Where this goes next

Example 6 left a number, -7, that the averages approached without reaching. Estimating a rate at an instant makes that precise from tables of shrinking intervals like the one above, and the simplified difference quotient is what makes those tables quick. In calculus the same idea, carried out exactly, is finding a derivative from its definition: the difference quotient of -x^2 + 6x - 1 becomes -2x + 6 once the width is allowed to vanish.

Exercises

Linear and quadratic rules

Find and simplify the difference quotient \displaystyle \frac{f(x + h) - f(x)}{h} of each function.

  1. p(x) = 9 - 2x
  2. q(x) = x^2 + 5
  3. r(x) = 3x^2 - 4x
  4. k(x) = -2x^2 + x + 7
  5. \displaystyle m(x) = \frac{1}{2}x^2 - 3

Using a difference quotient

  1. Use your answer to exercise 3 to find the average rate of change of r(x) = 3x^2 - 4x on [1, 4]. Check it from r(1) and r(4).
  2. Use your answer to exercise 4 to find the interval of width 2 on which k(x) = -2x^2 + x + 7 has an average rate of change of 0. Check that its ends have equal outputs.
  3. For q(x) = x^2 + 5, use the difference quotient to find the average rate of change on the intervals starting at x = 3 with h = 1, h = 0.1 and h = -0.1.

Higher powers

Find and simplify the difference quotient of each function.

  1. c(x) = x^3
  2. u(x) = x^3 - 2x

Rational and root rules

Find and simplify the difference quotient of each function, and say which values of x and x + h it needs.

  1. \displaystyle n(x) = \frac{1}{x}
  2. \displaystyle w(x) = \frac{4}{x - 3}
  3. \displaystyle v(x) = \frac{x}{x + 2}
  4. \displaystyle R(x) = \sqrt{4x + 1}

The cliff dive

Tiberiu dives from a lower ledge. His height above the water t seconds after he leaves it is T(t) = 15 + 2t - 5t^2 meters, for 0 \le t \le 1.9.

  1. Find and simplify \displaystyle \frac{T(t + h) - T(t)}{h}, and use it to find his average velocity on [1, 1.5], with units.
  2. Ayodele notices that Tiberiu rises for a moment before he falls. For which h > 0 is the average rate of change on [0, h] positive? Check the boundary value.
  3. Find the average rate of change on the intervals starting at t = 1 with h = 0.1, h = 0.01 and h = -0.01. What number do they crowd toward?

Reading a difference quotient

  1. A quadratic function f has f(0) = 5 and difference quotient 4x + 2h - 3. Without finding the rule, find f(1) and f(3).
  2. For f(x) = x^2 - 4x, simplify \displaystyle \frac{f(x) - f(a)}{x - a} where x \ne a. Explain why this is the same average rate as a difference quotient, with a and x as the two ends.

Reasoning

  1. Show that the difference quotient of any quadratic f(x) = ax^2 + bx + c is 2ax + ah + b. Check it against Example 2 and exercise 3.
  2. Tiberiu puts h = 0 into \displaystyle \frac{f(x + h) - f(x)}{h} for Example 2’s f before simplifying and gets \displaystyle \frac{0}{0}. Bayarmaa simplifies first and gets -2x + 6. Explain why Tiberiu’s result is undefined, and why Bayarmaa’s number is not an average rate of change of any interval.
  3. Explain why, for any polynomial f, every term of f(x + h) - f(x) contains a factor of h. Why does the argument not show this for \displaystyle f(x) = \frac{1}{x}, even though the difference quotient still simplifies?
Answers
  1. -2. p(x + h) - p(x) is (9 - 2x - 2h) - (9 - 2x), which is -2h, and \displaystyle \frac{-2h}{h} = -2.
  2. 2x + h. q(x + h) - q(x) is (x^2 + 2xh + h^2 + 5) - (x^2 + 5), which is 2xh + h^2; divide by h.
  3. 6x + 3h - 4. r(x + h) - r(x) is 3(x + h)^2 - 4(x + h) - (3x^2 - 4x), which is 6xh + 3h^2 - 4h; divide by h.
  4. -4x - 2h + 1. k(x + h) is -2(x + h)^2 + (x + h) + 7, and subtracting k(x) leaves -4xh - 2h^2 + h; divide by h.
  5. \displaystyle x + \frac{1}{2}h. m(x + h) - m(x) is \displaystyle \frac{1}{2}(x + h)^2 - \frac{1}{2}x^2, which is \displaystyle xh + \frac{1}{2}h^2; divide by h.
  6. 11. With x = 1 and h = 3: 6(1) + 3(3) - 4 = 11. Check: r(1) = -1, r(4) = 32, and \displaystyle \frac{32 - (-1)}{3} = 11.
  7. \displaystyle \left[-\frac{3}{4}, \frac{5}{4}\right]. With h = 2 the quotient is -4x - 4 + 1 = -4x - 3, which is 0 at \displaystyle x = -\frac{3}{4}. Check: \displaystyle k\left(-\frac{3}{4}\right) = k\left(\frac{5}{4}\right) = \frac{41}{8}. The interval is centered on \displaystyle \frac{1}{4}, the vertex input of k.
  8. 7, 6.1 and 5.9. The quotient 2x + h at x = 3 is 6 + h. The last is the interval [2.9, 3].
  9. 3x^2 + 3xh + h^2. (x + h)^3 - x^3 is 3x^2h + 3xh^2 + h^3; divide by h.
  10. 3x^2 + 3xh + h^2 - 2. The cube part is exercise 9, and -2(x + h) + 2x = -2h adds -2.
  11. \displaystyle \frac{-1}{x(x + h)}. \displaystyle \frac{1}{x + h} - \frac{1}{x} is \displaystyle \frac{x - (x + h)}{x(x + h)}, which is \displaystyle \frac{-h}{x(x + h)}; divide by h. Needs x \ne 0 and x + h \ne 0.
  12. \displaystyle \frac{-4}{(x + h - 3)(x - 3)}. The numerator over the common denominator is 4(x - 3) - 4(x + h - 3) = -4h. Needs x \ne 3 and x + h \ne 3.
  13. \displaystyle \frac{2}{(x + h + 2)(x + 2)}. The numerator over the common denominator is (x + h)(x + 2) - x(x + h + 2), which is 2h. Needs x \ne -2 and x + h \ne -2.
  14. \displaystyle \frac{4}{\sqrt{4x + 4h + 1} + \sqrt{4x + 1}}. Multiplying by the conjugate turns the numerator into (4x + 4h + 1) - (4x + 1), which is 4h. Needs 4x + 1 \ge 0 and 4x + 4h + 1 \ge 0.
  15. 2 - 10t - 5h, and -10.5 meters per second. T(t + h) - T(t) is 2h - 10th - 5h^2. With t = 1, h = 0.5: 2 - 10 - 2.5 = -10.5, matching T(1) = 12, T(1.5) = 6.75 and \displaystyle \frac{6.75 - 12}{0.5} = -10.5.
  16. 0 < h < 0.4. At t = 0 the quotient is 2 - 5h, positive when h < 0.4. At h = 0.4 it is 0: T(0.4) = 15 + 0.8 - 0.8 = 15 = T(0).
  17. -8.5, -8.05 and -7.95 meters per second, crowding toward -8. At t = 1 the quotient is -8 - 5h.
  18. f(1) = 4 and f(3) = 14. From the quotient, f(x + h) = f(x) + h(4x + 2h - 3). With x = 0, h = 1: 5 + 1(-1) = 4. With x = 0, h = 3: 5 + 3(3) = 14.
  19. x + a - 4. f(x) - f(a) is (x^2 - a^2) - 4(x - a), which factors as (x - a)(x + a - 4); cancel x - a. It is change in output over change in input between a and x, the difference quotient with a for the start and h = x - a.
  20. f(x + h) is a(x + h)^2 + b(x + h) + c; subtracting ax^2 + bx + c leaves 2axh + ah^2 + bh, and dividing by h gives 2ax + ah + b. Example 2 has a = -1, b = 6: -2x - h + 6. Exercise 3 has a = 3, b = -4: 6x + 3h - 4.
  21. With h = 0 both ends are x: the top is f(x) - f(x) = 0 and the bottom is 0, and division by 0 is undefined. Simplifying cancelled h using h \ne 0, so -2x - h + 6 equals the quotient only for h \ne 0; its value at h = 0 belongs to no interval. It is the number the averages approach as the width shrinks.
  22. f(x + h) - f(x) is a polynomial in h, and at h = 0 it equals f(x) - f(x) = 0, so its constant term is 0 and every remaining term carries h. For \displaystyle \frac{1}{x} the difference \displaystyle \frac{1}{x + h} - \frac{1}{x} is not a polynomial in h, so the argument does not apply; its factor of h appears only in the numerator after combining over a common denominator.
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0