The domain is handed to you. The range is the part you work out, and it is the part the exam asks for.
Do Now: completing the square, one question5 min
What function, domain and range actually mean8 min
Writing a domain and a range the way the exam marks them4 min
The most common mistake, and the picture that fixes it9 min
The Range Finder6 min
Two worked exam questions10 min
Practice, then the exit ticket13 min
Do Now · 5 min
Write this in completed square form, then answer the question under it.
\(f(x) = x^2 - 4x + 7\)
What is the smallest value \(f(x)\) can take, and which \(x\) gives it?
1 · halve the number in front of \(x\): half of \(-4\) is \(-2\): so the bracket is \((x - 2)^2\)
2 · expand that bracket to see what it actually gives: \((x - 2)^2 = x^2 - 4x + 4\)
3 · compare with what we want: we need \(+7\), we have \(+4\): \(7 - 4 = 3\), so add 3
4 · put it together: \(f(x) = (x - 2)^2 + 3\)
5 · a square can never be negative, so the smallest \((x - 2)^2\) gets is 0: and it hits 0 when \(x - 2 = 0\)
smallest value is 3, at \(x = 2\)
The three words
Function, domain, range.
A function is a rule that takes a number in and gives exactly one number back. One input, one output, every time.
Domain is the set of numbers you are allowed to put in.
Range is the set of numbers that actually come out.
Think of a coffee machine. The buttons that actually work are the domain. The drinks that can actually land in the cup are the range. If the machine has a decaf button but no decaf inside it, decaf is not in the range, whatever the button says.
The domain is almost always given to you. It is the phrase sitting after the formula, like for \(x \ge 1\) or for \(x \in \mathbb{R}\). You are being told which buttons work.
The range is what you have to work out, and it depends on the domain. Change the domain and the range changes with it.
Notation
The same function gets written two ways. Both mean this.
the one you already know
\(f(x) = 2x + 3\) for \(x \ge 0\)
Read it as: f of x equals \(2x + 3\).
the one the exam paper also prints
\(f : x \mapsto 2x + 3\) for \(x \ge 0\)
Read it as: f maps x to \(2x + 3\). The arrow with the bar on it does the same job as the equals sign above.
Same rule, same domain, same answers. If a question hands you the arrow version, rewrite it as \(f(x) = 2x + 3\) on your own paper and carry on.
Cambridge prints \(\ge\) and \(\le\) in a slightly rounder shape, \(\geqslant\) and \(\leqslant\). They mean the same thing. Write whichever you write naturally.
Writing it so it scores
A domain is written in \(x\). A range is written in \(f(x)\).
domain: what goes in
\(x > 1\)
The inputs are called \(x\), so the domain is described in \(x\).
range: what comes out
\(f(x) \ge 3\)
The outputs are called \(f(x)\), so the range is described in \(f(x)\).
This is worth marks on its own. On a real paper, a question asked for a domain and the answer was \(x > 1\). The mark scheme accepted \(x > 1\). It gave zero for the same thing written as \(f^{-1}(x) > 1\).
so it is possible to lose the mark like this: right numbers, right inequality sign, wrong letter, no mark
Before you write a domain or a range down, read your own line back and ask which of the two you are describing.
The most common mistake on this topic
Putting the two ends in is a guess.
Here is the Do Now function again, this time with a domain attached.
\(f(x) = x^2 - 4x + 7\) for \(0 \le x \le 5\)
The tempting move is to put both ends of the domain in and call that the range.
That answer is wrong, and the reason is the thing you found in the Do Now.
from the Do Now: \(f(x) = (x - 2)^2 + 3\), so the lowest point of this curve is at \(x = 2\)
and \(x = 2\) is allowed, because the domain is \(0 \le x \le 5\): \(f(2) = 3\)
3 is smaller than both 7 and 12. The curve goes below both ends on the way across, so the two ends never saw the bottom of the range.
The picture
So where do the top and the bottom actually come from?
The bottom is the vertex. The curve never gets lower than its turning point, and that turning point is inside the domain, so 3 is the smallest output.
bottom of the range \(= 3\)
The top is whichever end climbs higher. From the vertex the curve rises in both directions, so the end that is farther from \(x = 2\) ends up higher.
distance from \(x = 2\): left end \(2 - 0 = 2\) · right end \(5 - 2 = 3\)
3 is farther, so the right end climbs higher: \(f(5) = 12\) beats \(f(0) = 7\)
\(3 \le f(x) \le 12\)
\(f(0) = 7\) is not in the answer. It is inside the range, but it is neither end. An end of the domain does not have to be an end of the range.
The Range Finder
Drag the domain. Watch the range move.
The bar along the bottom is the domain you are allowed to use. The bar up the right side is the range that comes out of it. Move the two ends and watch when putting the ends in gives the right answer, and when it does not.
The pattern. On a straight line the two ends always work, because the line only ever goes one way. On either parabola the ends work right up until the turning point is inside your domain. Try \(6x - x^2\): its turning point is a maximum, so there the ends miss the top of the range instead of the bottom.
The method
How to find a range, every time.
Sketch it over the given domain. A shape, with the two ends marked and any turning point marked. It does not need to be neat.
Find the turning point, then check whether it is inside the domain. For a quadratic, complete the square and read off the vertex. Outside the domain means the two ends are your answer. Inside means the turning point is one end of the range.
Read off the lowest and highest values the curve actually reaches, then write them in \(f(x)\) with the correct inequality signs.
Choosing the sign. If the function actually reaches the value, include it, so use \(\le\) or \(\ge\). If it only gets closer and closer without ever arriving, exclude it, so use \(<\) or \(>\). When the domain has two ends, the answer usually needs two ends too.
Worked example 1 · from a real exam paper
\(f(x) = x^2 + 6x - 8\) for \(x \in \mathbb{R}\). Find the range. [3]
\(x \in \mathbb{R}\) means every real number is allowed, so nothing is cut off and the whole parabola counts.
1 · halve the number in front of \(x\): half of 6 is 3: so the bracket is \((x + 3)^2\)
2 · expand it to see what that gives: \((x + 3)^2 = x^2 + 6x + 9\)
3 · we want \(-8\) on the end and we have \(+9\): \(-8 - 9 = -17\), so subtract 17
4 · put it together: \(f(x) = (x + 3)^2 - 17\)
5 · the square is smallest at 0, which happens when \(x = -3\): \(f(-3) = 0 - 17 = -17\)
6 · no end to check this time, because the domain has no ends: the curve rises forever on both sides of \(x = -3\)
\(f(x) \ge -17\)
Three marks: one for the completed square form, one for the value \(-17\), one for writing it as a range in \(f(x)\) with the right sign. Writing \(x \ge -17\) gives away that last mark after doing all the work.
Worked example 2 · no turning point this time
\(f(x) = \dfrac{1}{x}\) for \(x \ge 1\). Find the range.
There is no vertex here, so the question becomes which end is included and which is not.
1 · the smallest input the domain allows is \(x = 1\): \(f(1) = \dfrac{1}{1} = 1\)
2 · make \(x\) bigger and see what happens to the output: \(f(2) = 0.5\), \(f(10) = 0.1\), \(f(1000) = 0.001\)
3 · so the outputs only ever fall, which makes \(f(1) = 1\) the biggest one: nothing in the domain beats it
4 · they fall toward 0, but \(\tfrac{1}{x}\) is never actually 0: no matter how big \(x\) gets, there is always something left
\(0 < f(x) \le 1\)
why ≤ 1
\(x = 1\) is allowed by the domain and it produces exactly 1. The value is reached, so it is included.
why > 0
No allowed \(x\) gives an output of 0. The curve gets as close as you like and stops short, so 0 is left out.
Practice · together · 7 min
State the range of each. Write every answer in \(f(x)\).
Q1 \(f(x) = 5 - 2x\) for \(-1 \le x \le 3\)
a line, so no turning point, and this one falls as \(x\) grows: \(f(-1) = 5 + 2 = 7\), \(f(3) = 5 - 6 = -1\)
the smaller \(x\) gives the bigger output here: so the left end is the top of the range
the \(x^2\) term is negative, so this parabola opens downward and its turning point is a maximum: \(g(x) = -(x - 2)^2 + 16\)
biggest at \(x = 2\), and it falls away forever on both sides: \(g(2) = 16\)
\(g(x) \le 16\)
Q3 \(h(x) = x^2 - 6x + 5\) for \(0 \le x \le 2\)
complete the square: \(h(x) = (x - 3)^2 - 4\), turning point at \(x = 3\)
\(x = 3\) is outside \(0 \le x \le 2\), so the curve only falls across this domain: \(h(0) = 5\), \(h(2) = 4 - 12 + 5 = -3\)
\(-3 \le h(x) \le 5\)
Practice · on your own · 8 min
Your turn. Tap to check when you're done.
Q4 \(p(x) = x^2 - 6x + 5\) for \(0 \le x \le 5\)
same completed square: \(p(x) = (x - 3)^2 - 4\), turning point at \(x = 3\)
\(x = 3\) is inside \(0 \le x \le 5\) this time, so the turning point is the bottom: \(p(3) = -4\)
for the top, the end farther from \(x = 3\) wins: \(3 - 0 = 3\) beats \(5 - 3 = 2\): \(p(0) = 5\), and \(p(5) = 0\) is smaller
\(-4 \le p(x) \le 5\)
Q5 \(q(x) = \dfrac{1}{x - 2}\) for \(x > 2\)
\(x\) is bigger than 2, so \(x - 2\) is a positive number: 1 divided by a positive is positive
\(x - 2\) can be tiny (huge output) or enormous (output near 0): but it never reaches 0
\(q(x) > 0\)
Q6 \(r(x) = \sqrt{x} + 3\) for \(x \ge 0\)
\(\sqrt{x}\) starts at 0 when \(x = 0\), and grows without ever stopping: \(r(0) = 0 + 3 = 3\)
so every output is 3 or above, with no ceiling: \(r(100) = 13\), \(r(10000) = 103\)
\(r(x) \ge 3\)
Q3 and Q4 are the same rule. Only the domain changed, and the answer changed with it. That is the whole lesson in two questions.
Recap and exit ticket
Three things to carry out of here.
Every range question is a sketch question wearing a formula. The algebra finds the turning point. The sketch tells you whether that turning point is inside the domain you were given, and that is what decides the answer.
An end of the domain is not automatically an end of the range. Check the turning point first, then ask which end climbs farther from it.
Completing the square did not finish when quadratics finished. It is now how you find the turning point so you can find the range.
Two checks before you write any range down. Is the letter \(f(x)\) rather than \(x\)? Does your answer have both ends when the domain has both ends?
Next session: composite functions. Two machines in a row, and a rule about which one is allowed to feed the other. The range you learned to find today is what decides it.
Exit ticket · 5 min
Schoology, Paper 1 → 1.2 Functions.
Four questions. One asks for the full method, so show the completed square form.
Press Submit before you close the tab. An attempt left open scores nothing.