Operations on Functions Calculator
On the quotient: it is undefined wherever g is zero, and that exclusion stays even when the fraction cancels to something tidy. Try x^2 - 4 with x + 2 — the quotient becomes x − 2, and −2 is still out, because at that value the original was 0 divided by 0. On a graph it is a hole.
This operations on functions calculator adds, subtracts, multiplies and divides two functions, and gives all four results at once.
Type f(x) = x^2 + 1 and g(x) = x - 3, and it works out the values the quotient has to exclude as well.
What the Notation Actually Means
Function operations are written with the operation outside the bracket, which looks stranger than it is.
(f + g)(x) is nothing more elaborate than f(x) + g(x). The same goes for the other three:
(f + g)(x) = f(x) + g(x)
(f − g)(x) = f(x) − g(x)
(f · g)(x) = f(x) × g(x)
(f / g)(x) = f(x) ÷ g(x)
Adding and subtracting functions, then multiplying and dividing them — those four verbs are the whole topic. The notation is doing one job: writing "add these two functions" in four characters instead of a sentence. Once you see that, three of the four operations are just algebra you already do.
The Fourth One Is Different
Dividing functions is where this topic earns its place in the syllabus, and it is not because the division itself is hard.
When you divide one function by another, f(x) ÷ g(x) has no value wherever g(x) = 0. So before you can call the quotient a function at all, you have to find where g is zero and throw those values out of the domain. That step is worth a mark on its own in most mark schemes, and it is the one students skip.
| f(x) | g(x) | (f / g)(x) | Excluded |
|---|---|---|---|
| x² + 1 | x − 3 | (x² + 1)/(x − 3) | x = 3 |
| x² − 4 | x + 2 | x − 2 | x = −2 |
| 3x | x² + 1 | 3x/(x² + 1) | nothing |
The Middle Row Is the Whole Lesson
Look at it again. f(x) = x² − 4 and g(x) = x + 2, so the quotient is (x² − 4)/(x + 2), which cancels neatly to x − 2.
And x = −2 is still excluded.
The cancelling did not undo anything. At x = −2 the original expression was 0 divided by 0, which has no value at all, so the quotient never had a value there. What you have now is a function that agrees with x − 2 everywhere except at that one point — and on a graph it shows as a hole in an otherwise straight line.
Write the answer as x − 2, x ≠ −2. Leaving off the condition is not a rounding error; it is a different function.
Subtraction Loses More Marks Than Division
The sum and difference of functions are where an honest ranking of mistakes would start, and subtraction leads it.
(f − g)(x) means f(x) minus all of g(x). If f(x) = x² + 1 and g(x) = x − 3, then it is (x² + 1) − (x − 3) = x² + 1 − x + 3 = x² − x + 4. That +3 at the end catches people every single time, because the minus has to reach the −3 as well as the x.
Write the bracket first, then remove it. It takes two seconds and it is the difference between a mark and no mark.
A Quick Check on the Product
When you multiply functions, the degree of f · g is always the degree of f plus the degree of g.
Multiply a quadratic by a linear function and you must get a cubic. If your answer comes out quadratic, a term went missing. It costs nothing to glance at, and it catches the commonest multiplication slip before you write it down.
Working at a Single Value
Questions often ask for (f + g)(2) rather than (f + g)(x), and there are two routes to it.
You can combine the functions first and then substitute, or substitute into each function separately and then combine. Both give the same answer, always, because the operation is defined pointwise. The second route is usually faster for one value; the first is better when you need several.
One exception, and it is the obvious one: if g(2) = 0 then (f / g)(2) does not exist, however you approach it.
Order Matters for Two of the Four
(f + g)(x) and (g + f)(x) are the same thing, always. So are the two products. Addition and multiplication of numbers do not care about order, and functions inherit that.
Subtraction and division do care. (f − g)(x) and (g − f)(x) are negatives of each other, and the two quotients are reciprocals — with different excluded values, because one asks where g is zero and the other asks where f is.
It sounds obvious written down. It is worth checking anyway, because a question that asks for (g − f)(x) when you have just worked out (f − g)(x) is a cheap mark to lose.
What This Page Does Not Do
Composing functions — f(g(x)), often written (f ∘ g)(x) — is a different operation and it is not on this page.
It looks similar in a list of formulas and behaves nothing like the other four. Addition and multiplication of functions are commutative; composition is not, and (f ∘ g)(x) rarely equals (g ∘ f)(x). It has its own page, the composition and evaluating functions calculator, which handles it properly rather than as an afterthought here.
How to Use This Operations on Functions Calculator
Put each function in its own box, written as it is given: x^2 + 1 and x - 3. An f(x) = in front is ignored, so you can paste the whole line from your book.
All four results come back together, along with the excluded values. For the domains of the individual functions rather than the combination, there is a domain and range calculator.
Most courses take these four first, then composition, then inverses — three ways of getting a new function out of ones you already have. The inverse function calculator handles the third. If the algebra itself is the sticking point rather than the notation, adding and subtracting polynomials and multiplying polynomials cover the arithmetic behind the sum and the product.
Operations on Functions Calculator FAQ
What does (f + g)(x) mean?
What is the domain of (f / g)?
The fraction cancelled — is the value still excluded?
Why does my subtraction keep coming out wrong?
Is (f + g)(x) the same as (g + f)(x)?
What degree should the product be?
How do I find (f + g)(2)?
Are these the same as composing functions?
Function operations come down to this. Three of the four are ordinary algebra. The fourth needs you to solve g(x) = 0 first and say what comes out of the domain — and that condition survives any cancelling that happens afterwards.
