Difference Quotient and Average Rate of Change Calculator
This difference quotient calculator works out [f(x+h) − f(x)] ÷ h and simplifies it, showing each step. On the second tab it is an average rate of change calculator, taking the two ends of an interval and giving you a number.
Type x² + 3x and you get 2x + h + 3.
What the Difference Quotient Is
It measures how much a function changes between x and x + h, divided by the gap between them:
[ f(x + h) − f(x) ] ÷ h
That is rise over run, with both points taken off a curve rather than off a straight line. It is also the last piece of algebra before the derivative in calculus, which is why it turns up in precalculus at all — but the derivative itself is a separate job and is not what this page does.
Factor First, Divide Second
The whole task is algebra, and the order of it decides whether you get the answer.
Take f(x) = x². Expand f(x + h) in full, subtract f(x), and look at what is left:
(x² + 2xh + h²) − x² = 2xh + h² → h(2x + h) → 2x + h
Never divide by h before the top is tidy. Take the h out as a factor of the whole numerator, then cancel it in one move. Dividing term by term partway through is where the marks go, because it is far too easy to lose a piece on the way.
The other habit worth having is expanding f(x + h) properly. It is not f(x) + f(h), and (x + h)² is not x² + h². Write the bracket out in full every time, even when you are sure.
Why the h Is Bound to Come Out
This turns a hopeful step into a certain one, and it is worth thirty seconds.
Put h = 0 into the top. You get f(x) − f(x), which is zero — whatever f happens to be. Anything that vanishes at h = 0 must have h as a factor. So the cancelling is not luck; it has to work.
And that gives you a free check. If your numerator has a term with no h in it, you have made an error, and you can find it before wasting the rest of the question on it.
When h Stays Below the Line
You will read, in more than one place, that every h below the line has to disappear and that one left there means you slipped. That is true for polynomials and false for fractions, and believing it costs marks.
Work out f(x) = 1/x and the answer is:
−1 ÷ [ (x + h)·x ]
There is the h, sitting below the line, and nothing is wrong. That h never came from the formula. It came from f itself, which had an x on the bottom to begin with, so putting x + h into it produced an x + h down there. The h that cancelled was a different one entirely.
The rule worth carrying is narrower and true: for a polynomial, the h must cancel completely. If one is left underneath after x² + 3x, go back and look. For a fraction or a root, expect it and leave it alone. This page tells you which shape you gave it and whether the h belongs down there.
One more shape catches people out, and it is not really a fraction at all. x²/3 has a number underneath, not an x, so nothing about it survives the cancelling — the answer is (2x + h)/3, and the 3 is a coefficient wearing a denominator’s clothes. Treat a constant on the bottom as part of the multiplier and the whole thing behaves like the polynomial it is.
A Square Root Needs the Conjugate
Subtracting two square roots gets you nowhere by expanding, because there is nothing to expand. The move is to multiply top and bottom by the same two roots added:
√(x + h) − √x → multiply by √(x + h) + √x
The top becomes a difference of squares, the roots vanish from it, and what is left is (x + h) − x, which is just h. That h cancels, and the answer is 1 ÷ [√(x + h) + √x]. The roots end up below the line and stay there, which is the same situation as the fraction above.
Average Rate of Change: The Same Formula with Numbers
Write the interval as starting at a and running a distance h, so that b = a + h. Then:
[ f(b) − f(a) ] ÷ (b − a) is [ f(a + h) − f(a) ] ÷ h
Letter for letter, the same thing. The gap between your two ends is the h. Put numbers in and you get a rate; leave the letters and you get an expression. That is the only difference, and it is why both tabs are on one page.
For f(x) = 2x² across [1, 4]: f(4) = 32, f(1) = 2, so the rate is 30 ÷ 3 = 10. Across that stretch, f rose by 10 for each step of one in x. It says nothing about any single point inside the stretch, only about the stretch as a whole.
Two things stop a rate existing. If f has no value at one end — 1/x at x = 0 — there is no rate to find. And if both ends are the same number, the gap is zero and nothing can be divided by it. Naming either is the answer, not a failure.
Using the Difference Quotient Calculator
Pick a tab, then type the function the way you write it: ^ for powers, / for division, brackets wherever you would put them. Fractions and decimals both work and nothing is rounded.
Polynomials, fractions of polynomials and √(ax + b) are all worked out in full. A constant function is fine too — f(x) = 5 gives 0, since a constant never changes. Anything outside that is refused by name rather than half-answered, and every answer is put back through the definition on real numbers before you see it.