Each standard function is pinned down by its picture as much as by its rule

0:00Seven functions with names. Identity. Constant. Polynomial. Rational. Modulus. Signum. Greatest integer. Each gets a rule and each gets a picture, and the picture is not decoration. It is the second half of the specification. Because there is a question you cannot answer from the rule alone. Is this function covered by one expression, or is it stated in pieces? And if it is stated in pieces, what happens where the pieces meet?

0:35One expression draws one unbroken shape. Pieces carry their joins in the picture, and you can see them. So the sorting that matters is not by what the formulas look like. It is by how many expressions it took to say the thing. Sort them that way and they fall four and three. Four are covered by a single expression that applies wherever it is legal: identity, constant, squaring, and one polynomial divided by another.

1:08Three are stated by cases: modulus, signum, and the greatest integer. Why must that sorting be read off the writing, not the values? Take the identity function, which sends every input to itself. Write it in three cases instead. Below nought, send the input to itself. At nought, send it to nought. Above nought, send it to itself. Three cases where there was one, and it agrees with the original at every input tested.

1:41Same values. Different specification. So you have to read the writing. Start with the simplest. The identity function takes any real number and returns it unchanged. One expression, no cases. Take an input, go up to the graph, come back to the vertical axis, and you arrive at the number you started with. That is what the line through the origin is saying. Its domain is all the real numbers and so is its range, because every real number is the output of exactly one input — itself.

2:21Run it over forty-one sample inputs and you get forty-one different outputs. Nothing repeats. Hold on to that. Pick a number and stay there. The constant function sends every input to one fixed number. Take three. Nought goes to three. Seventeen goes to three. Minus a thousand goes to three. Its graph is a horizontal line, and every input arrives at the same height. The domain is all the real numbers, exactly as before.

2:57But run the same forty-one inputs and count the different outputs. One. Forty-one inputs, one output. The range has a single member. That looks like a violation and it is not. Nothing forbids many inputs from landing on the same place; what is forbidden is one input landing on two. And no input here lands on two — every one lands on three and nothing else. This is the most extreme function that still obeys the rule, and it obeys it comfortably.

3:34Those two look unrelated. They are one shape with different settings. Multiply the input by something, then add something. Any rule of that form draws a straight line, and both of the functions you have just seen are of that form. The identity is the multiplier one with nothing added. The constant three is the multiplier nought with three added. Check it: multiplier one, nothing added, and it agrees with the identity at every input tested. Multiplier nought, three added, and it agrees with the constant.

4:13And those two settings do not agree with each other, which is what tells you the form is doing real work. The multiplier controls the tilt. The added number controls the height, and it is where the line crosses the upright axis. Two named functions, one form, and neither is a special case worth memorising separately. Third of the four, and the one with an entry condition. A polynomial function is a sum of powers of the input, each with a number in front.

4:49The condition is on the powers. Every exponent must be a whole number, from nought upwards. One that qualifies: the input cubed, minus the input squared, plus two. Exponents three, two and nought. All whole. A second: the input to the fourth, plus the square root of two times the input. The square root of two is not even a fraction, and it still qualifies. This is the part people get wrong.

5:20The condition says nothing about the numbers in front. Only the exponents are constrained. Now one that fails: the input raised to the power two thirds, plus twice the input. Exactly one exponent is the problem, and it is the two thirds. Change the number in front of it to anything you like and it still fails. Change the exponent to two and it passes immediately. And a negative whole number is refused too. The condition is from nought upwards, not merely whole.

5:57The squaring rule is a polynomial, so draw one properly. Nine inputs, minus four through to four. Square each. Sixteen. Nine. Four. One. Nought. One. Four. Nine. Sixteen. Plot and join, and you get a narrow curve resting on the origin. How many different outputs among those nine? Five. Not nine, because minus four and four both go to sixteen, and so on inwards. Now the range. Not one of those outputs is below nought, and sweeping far more widely turns up no negative output either.

6:36So the range is the squares. Nothing below nought is in it. It is not all the real numbers. The bottom half of the vertical axis is never reached, and the curve resting on the origin is that fact drawn. Cube instead, and the shape changes completely. Nought to nought. One to one. Minus one to minus one. Two to eight. Minus two to minus eight. Three to twenty-seven. Minus three to minus twenty-seven.

7:10Two things, and the second is a trap. First, negative inputs now give negative outputs. Not one comes out positive. Squaring folded the negatives up onto the positives. Cubing does not, which is why this is an S through the origin and the other was a bowl. Second, look at the last two values. On axes reaching eight in each direction, two of those seven outputs are off the page.

7:41Twenty-seven and minus twenty-seven are not in the frame. The curve leaves through the top and comes back in through the bottom. Students read the visible part as the whole function. The frame is a choice about what to draw, never a statement about where the function stops. Fourth of the four, and the first that gives something up. Take one polynomial and divide it by another. That is a rational function — a single expression, with a condition attached.

8:16Wherever the bottom is nought there is no value, because there is no such division. The simplest one divides one by the input. So the input nought has to go, and everything else stays. Take nine inputs: minus two, minus one and a half, minus one, minus a half, a quarter, a half, one, one and a half, and two. The outputs are minus a half, minus two thirds, minus one, minus two, four, two, one, two thirds, and a half.

8:50Seven of those nine can be written down exactly in two decimal places. Two cannot: the two thirds, with a sign. Write the table in two decimals and it says minus nought point six seven, and that is not the value. It is within a hundredth of it — close enough to write down, and not close enough to be true. A table that rounds does you a small favour and tells you a small lie.

9:20Now draw it, because the picture says what the table cannot. The graph comes in two separate pieces, one in the top right and one in the bottom left, and neither touches either axis. The vertical axis is untouched because nought is not an input. The horizontal axis is untouched too, and that is a different fact. Sweep a wide range of inputs and ask which are sent to nought.

9:48None. Not one. To get nought out you would have to divide one by something enormous and land exactly on nothing. So nought is missing from the range as well as the domain, for two entirely different reasons. And the branches never cross, because the output always carries the sign of the input. And you cannot repair the missing input by declaring the function on the whole line anyway. Declare it everywhere and the arithmetic still refuses at nought. The clause covers the input and there is still no answer to give.

10:27A domain is a promise about where the rule works. A wider promise does not make the rule work in more places. And notice what the break is not. It is not a case boundary — there is only one expression here. It is the one place that expression is not legal. Now the three stated in cases. The first has exactly one boundary. The modulus function. At or above nought, return the input. Below nought, return the negative of the input.

11:03Two clauses, one boundary, at nought. Read the second clause carefully; the usual story about it is wrong. It does not say delete the minus sign. It says take the negative of the input — which comes out positive precisely because the input was negative. At minus five, the negative of minus five is five. The clause computes; it does not edit. That distinction survives when the input stops being a number and starts being an expression, and the deletion story does not.

11:40Now the boundary. The first clause owns nought — but ask what would happen if you gave it to the other. The first gives nought. The second gives the negative of nought, which is also nought. They agree, so the choice changes no value. That is not automatic. Square the input up to two and triple it from two upwards, and the boundary input two is sent to four by one clause and six by the other — two places, so not a function at all.

12:12The modulus boundary answers cleanly. And the picture shows the join. Two straight arms meeting at the origin, one arriving with slope minus one, the other leaving with slope plus one. Both sides carry the same value to the meeting point, so there is no gap — but the direction changes. That is a corner, and it is the only one anywhere in the sweep. Second of the three. The signum function reports the sign and throws away everything else.

12:48Above nought, one. Below nought, minus one. At nought, nought. Three clauses, and the range has exactly three members however widely you sweep. Minus one, nought, and one. So it is defined at nought, and its value there is nought. Not a gap, not undefined. The picture is two horizontal rays, one at height one and one at minus one, with a single point at the origin between them. Neither ray reaches the vertical axis, and the drawing marks that with a hollow circle at each inner end.

13:25It arrives at nought from one side carrying minus one and from the other carrying one. That is a jump, not a corner, and it is the only input in the sweep where it happens. Now a second formula for the same function. Take the modulus of the input and divide by the input. Above nought that is a positive over itself, which is one. Below nought, a positive over a negative, which is minus one.

13:56It agrees with the three-clause version at every input tested — except one. At nought it divides by nought and refuses. The two descriptions part company at exactly one input, which is why that formula always arrives with the value at nought supplied by hand. Last of the three, and the one that generates the most wrong answers. The greatest integer function returns the largest whole number that does not exceed the input.

14:30Read that slowly. Largest. Whole number. Does not exceed. At two point seven, the whole numbers that do not exceed it are two, one, nought and downwards. The largest is two. Now minus one point five. The whole numbers that do not exceed it are minus two, minus three, and downwards. Minus one does exceed it, so minus one is not a candidate. The largest candidate is minus two. Minus one point five goes to minus two.

15:04Rounding would have said minus one, because minus one is nearer. This function never rounds. Chopping off the decimal part would also have said minus one, and that is the more dangerous mistake, because it works on the positives and fails silently on the negatives. Two checks on that. Across a wide sweep, the value never exceeds the input, and the next whole number up always does. That pair is the definition, and every value it produces satisfies both.

15:41Draw it, and the case structure becomes impossible to miss. From nought up to but not including one, the answer is nought. From one up to but not including two, one. From minus one up to but not including nought, minus one. Each stretch is one clause, the value is constant across it, and the picture is a staircase. Over a window from minus three to four there are seven treads, at heights minus three, minus two, minus one, nought, one, two and three.

16:15Each tread includes its left end and excludes its right one. At one, the answer is one. At one point nine nine nine, still one. At two, the answer is two. The value belongs to the step it starts, and the drawing marks the far end with a hollow circle. Every one of those breaks is a jump, and not one is a corner. Now count the treads you can actually see drawn.

16:43Six, not seven. Because the tread at height nought runs along the horizontal axis itself, and it is not drawn as a separate segment. The axis is already there. Count the drawn segments, find six, and conclude the function skips a stretch between nought and one. It does not. That step is doing exactly what the other six do. It is hiding on a line drawn for another reason. Put all seven back on one board.

17:17Exactly one of the seven refuses an input anywhere in the sweep, and it is the reciprocal. None of the seven sends any input to two places. They are all functions; the question was only ever how they were written. Two of them jump somewhere: the signum and the greatest integer. Exactly one has a corner: the modulus. Add those up. All three of the case-defined functions show their boundary in the picture, as a corner or as a jump.

17:51And not one of the four one-expression functions shows any such feature. That is the whole claim, measured rather than asserted. How a function is specified is visible in its graph. So when you meet a new function, the useful question is not what its formula looks like. It is whether one expression covers it — and if not, what happens where the pieces meet. Declare the domain. State the rule. Then draw it, and let the drawing tell you whether you have one thing or several stitched together.

18:27Because the picture is not an illustration of the specification. It is the rest of it.