Formal Function Notation: Domain and Range

Split a restaurant bill on your phone. The app runs one calculation - the total divided by the number of people. Put in four and it works; put in zero and it stops dead, because dividing by zero is not something that rule can do. That is not a fault in the app. Every rule in maths comes with a set of inputs it accepts, and a shorter set it has to refuse. It works the other way round too. Solve for the side of a rectangle from its area, and the algebra hands you two answers with one of them negative - and a length cannot be negative. The algebra is not wrong there; it just does not know what the letters stand for. In function questions that final decision has a fixed place - it is the last step of the method, and it is the step people skip.

Two ovals and a handful of arrows carry every word in this topic, so the picture comes before the algebra. One note first. This is functions written out in full, the way Edexcel International GCSE and Cambridge IGCSE ask for it. If your board handles composite and inverse functions without domain and range, the video called Inverse and Composite Functions is the one you want. The oval on the left holds the numbers you are allowed to put in; the oval on the right holds what comes out. Each arrow sends one input to exactly one output, and that is all a function is. Those two sets have names. The left-hand set is the domain, the inputs; the right-hand set is the range, the outputs you actually get. Domain in, range out. Now the writing. If the rule is triple it and subtract four, you write f of x equals three x minus four; some papers write that same rule as f maps x to three x minus four. Either way, f of x is not f multiplied by x - the bracket means feed x in, so f of five gives eleven. Some inputs get thrown out of the domain before you start. Only two things break a rule at this level: a bottom that comes out as zero, and a square root of a negative. The zero bottom is much the commoner. So here is one for you. Take f of x equals one over x minus three, with the whole x minus three on the bottom. Which value has to be excluded from the domain - zero, three, or negative three? Take your pick. I'll wait. The answer is three. Put three in and the bottom becomes zero, and dividing by zero has no answer at all, so three has nothing to be sent to. You write that as: the domain is all values of x except three. The exception is part of the function, not a footnote underneath it.

With the language in place the algebra is short, so here is the standard question type start to finish. Take g of x equals two x plus one, and h of x equals x squared minus five. You are told that g h of x equals one, and that x is greater than or equal to zero. Find the value of x. g h of x means g of h of x, so h goes first and its output feeds into g. The reason is position - h is the one sitting next to the x. Before I do it, have a go yourself. Substitute h into g, set the result equal to one, and solve as far as you can. Pause here and work it through. I'll wait. Substituting means putting x squared minus five into g wherever g has an x. So g of h of x is two lots of x squared minus five, plus one. That opens up to two x squared minus ten plus one, which tidies to two x squared minus nine. Now set that equal to one. Two x squared minus nine equals one gives two x squared equals ten, so x squared equals five, and the square root gives x equals root five or x equals minus root five. And that is where the damage gets done, because the working usually stops right there. The question said x is greater than or equal to zero, so minus root five was never allowed. Cross it out, and write x equals root five. That is the phrase to carry into the exam with you: solve it, then sieve it. Do the algebra, then pour your answers through the restriction and see which one falls out. One thing about writing it down. Root five is about two point two four, but unless the question asks you to round, leave it exact as root five. So the method is four moves: substitute, solve, sieve, then write down the answer that survived.

Inverses run the whole thing backwards, and that changes which values are allowed in the front door. Take f of x equals five over x minus two, with the whole x minus two on the bottom, and x cannot equal two. Find f inverse of x, and state the value excluded from its domain. The method is swap, then rearrange. Write y equals five over x minus two, then swap the two letters, so x equals five over y minus two. Now make y the subject. The swap is not a trick to memorise. An inverse undoes the function, so its input is the original's output, and swapping the letters just says x is now the output of f. Changing the subject of a formula in general is a different topic. Multiply both sides by y minus two, giving x times y minus two equals five. Divide both sides by x, and y minus two equals five over x. Add two, and f inverse of x equals five over x, plus two. Now the domain part. Look at five over x, plus two, and tell me which value of x has to be excluded from the domain of that inverse. Have a think. I'll wait. The answer is zero, because a zero on the bottom breaks it. So f inverse has domain all values of x except zero. And zero is not a random number to land on. Five divided by something can never come out as zero, so zero is the one value f could never produce. The inverse's domain is the original's range. Notice where the two exclusions came from. One was handed to you in the wording; this one you had to find in your own algebra. Both get checked at the end.

Now the exam side of it, where this exact slip turns up written down in an examiner's report. This is one line from a report on a paper where students had to substitute one function into another, set the result equal to a value, and solve - a different pair of functions from ours, same shape of question. Part b saw mixed results; some students substituted h of x into g correctly and went onto rearrange and solve after setting equal to one for three marks. Some students did everything right except they ignored the domain for x and gave plus or minus two as their answer, losing the A mark. Look again at that middle sentence: those students did everything right except for one thing. The substituting, the rearranging and the solving were all correct; one clause in the question went unapplied, and the answer was wrong. So make the restriction visible before you start. When a question says x is greater than or equal to zero, or x cannot equal two, underline it - then at the end write both roots, cross the excluded one out, and write the reason beside it. That last line is the whole difference between a correct method and a correct answer.

Here is the whole thing back in order, quickly, and it fits into five words: solve it, then sieve it. A function is a rule plus the set of inputs it accepts. Domain in, range out - and any value that would put a zero on the bottom leaves the domain before you begin. g h of x means do h first, because h sits next to the x. For an inverse, swap x and y and rearrange, and its domain is the original function's range. Then the step that decides your answer: check every solution against the restriction, and cross out the one the domain does not allow.

Next in the chapter: Differentiating Powers of x, where the same f of x you have just learnt to write gets a brand new operation done to it.

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