Calculus From The Ground Up

A rule adds 3 to its input. Build the table for the inputs 0, 4, and −2.

1. Read the rule as an instruction with a blank in it: take the input, add 3.
2. Put 0 in the blank: 0 + 3 = 3.
3. Put 4 in the blank: 4 + 3 = 7.
4. Put −2 in the blank: −2 + 3 = 1. Adding a positive to a negative moves it toward zero and past it.

Answer. 0 → 3, 4 → 7, −2 → 1.

Notice that the rule never asked whether the input was positive. A function accepts whatever its domain allows, and −2 is allowed here.

A vending machine has given three different snacks for slot B4 this week. Has it broken the promise?

1. Ask what the input is. It is the slot code B4, the same input each time.
2. Ask what the outputs were. Three different snacks.
3. One input, three outputs. The promise is broken.

Answer. Yes. The machine is not computing a function of the slot code.

A working engineer would say the machine is stocked wrong. A mathematician says the same thing differently: the output depends on something the input does not record, so the true input is not the slot code alone.

A taxi charges £3.20 to start, then £1.40 for each mile. Write the rule in words and find the fare for 0, 3, and 10 miles.

1. Identify the input and the output. The input is the number of miles; the output is the fare.
2. Write the rule with a blank: multiply the ⬚ by 1.40, then add 3.20.
3. At 0 miles: 0 × 1.40 + 3.20 = £3.20. The fare is not zero, because the charge to start does not depend on distance.
4. At 3 miles: 4.20 + 3.20 = £7.40. At 10 miles: 14.00 + 3.20 = £17.20.

Answer. £3.20, £7.40, and £17.20.

The two numbers in the rule do different jobs. The 1.40 says how fast the fare grows; the 3.20 says where it starts. Chapter 6 names that pair, and chapter 11 shows the 1.40 is the only one a rate of change can see.

Which of these are functions of the stated input? (a) the price of a fixed-price ticket, given the day. (b) the winning lottery number, given the day. (c) the day of the week, given a date.

1. (a) Each day has one price, and the price is fixed in advance. One input, one settled output.
2. (b) Each day has one winning number once it is drawn, so it is a function of the day. But it cannot be predicted, which is a different complaint from being ill-defined.
3. (c) Each date falls on exactly one weekday. One input, one output.
4. All three qualify. Being unpredictable, or uninteresting, or non-numerical does not disqualify a rule.

Answer. All three are functions.

Part (b) is the one that catches people. A function must be settled, not guessable. If you had asked instead for "a number that might win", the answer would have forked, and that would have failed.

Is {(4, 1), (5, 2), (6, 1), (7, 3)} a function?

1. List the first entries: 4, 5, 6, 7.
2. Check for a repeat. There is none, so no input can possibly have two outputs.
3. The outputs 1 and 1 repeat, but they belong to the different inputs 4 and 6.

Answer. Yes. A repeated output is not a defect.

Does "the square of x" define a function? Does "a square root of x" define one over the real numbers?

1. Squaring: give it 3 and it returns 9; give it −3 and it returns 9. Each single input returns exactly one number.
2. That two inputs share the output 9 is sharing, not splitting.
3. "A square root of x": give it 9 and it returns 3, or −3, without deciding.
4. One input, two candidate outputs, no rule to choose between them.

Answer. Squaring is a function. "A square root of x" is not, unless the principal root is specified, at which point it becomes √(x) and is.

A rule pairs each pupil in a class with a sport they play. Is it a function?

1. Ask whether any pupil could be paired with two sports. A pupil who plays both hockey and tennis has two outputs.
2. Ask whether any pupil could be paired with none. A pupil who plays no sport has no output.
3. Both failures are available, and which one occurs depends on the class rather than on the wording.
4. The rule is only a function if every pupil plays exactly one sport.

Answer. Not in general. It is a function only when each pupil plays exactly one sport.

Rules stated in English very often hide this. "A sport they play" quietly assumes there is exactly one, and the assumption is doing all the work.

Two students disagree. One says {(1, 5), (2, 5), (3, 5)} cannot be a function because the outputs are all identical. Settle it.

1. Check the demand that actually exists: does any single input have two outputs?
2. Input 1 has only 5. Input 2 has only 5. Input 3 has only 5.
3. No input forks, so the rule is a function.
4. What the student has noticed is that it is not one-to-one, which is a real property with real consequences, but a different one.

Answer. It is a function. The student has confused sharing with splitting.

The property they were reaching for matters in chapter 9: a rule whose outputs are shared cannot be run backwards. So the observation was worth making and the conclusion was wrong.

For g(x) = x² − 3, find g(0), g(2), and g(−2).

1. Read the rule as: square the blank, then subtract 3.
2. g(0) = 0² − 3 = 0 − 3 = −3.
3. g(2) = 2² − 3 = 4 − 3 = 1.
4. g(−2) = (−2)² − 3 = 4 − 3 = 1. The brackets matter: it is the input that is squared, so the minus is squared with it.

Answer. g(0) = −3, g(2) = 1, g(−2) = 1.

Written without brackets, −2² means −(2²) = −4 and the answer would come out as −7. The brackets are not decoration.

For h(t) = 5 − 2t, solve h(t) = −1.

1. This gives the output, so write the rule equal to it: 5 − 2t = −1.
2. Subtract 5 from both sides: −2t = −6.
3. Divide both sides by −2: t = 3.
4. Check by evaluating forwards: h(3) = 5 − 2(3) = 5 − 6 = −1. It agrees.

Answer. t = 3.

Checking an inverse question by running the rule forwards costs one line and catches nearly every sign error.

For f(x) = x², write f(x + h) and simplify f(x + h) − f(x).

1. Put x + h into every blank: f(x + h) = (x + h)².
2. Expand: (x + h)² = x² + 2xh + h². The middle term is the one that is always forgotten.
3. Subtract f(x), which is x²: (x² + 2xh + h²) − x².
4. The x² terms cancel, leaving 2xh + h².

Answer. f(x + h) = x² + 2xh + h², and f(x + h) − f(x) = 2xh + h².

Every term that survives has an h in it. That is not a coincidence, and chapter 13 depends on it.

A table gives f(0) = 1, f(1) = 3, f(2) = 5. Write the rule as a formula, in words, and as three points.

1. Look at what the outputs do between rows: 1 to 3 to 5, rising by 2 each time the input rises by 1.
2. A constant rise per step means the rule has the shape 2x plus something fixed.
3. Use one row to pin the fixed part: at x = 0 the output is 1, and 2(0) = 0, so the fixed part is 1.
4. Check the rule against a row not used to build it: f(2) = 2(2) + 1 = 5. It agrees.

Answer. f(x) = 2x + 1; "double the input, then add one"; the points (0, 1), (1, 3), (2, 5).

Three representations of one rule, and the fourth is the graph beside them. The table was the given form and the formula was recovered, which is the direction textbooks drill least and applications need most.

State the domain of q(x) = 1/(x − 3).

1. Scan for the three causes. There is a denominator containing x, and no even root.
2. Set the denominator to zero and solve: x − 3 = 0, so x = 3.
3. That input, and only that input, must go.

Answer. All real numbers except 3, written (−∞, 3) and (3, ∞).

The numerator plays no part. A common slip is to answer "x is not 0", which is where the denominator would vanish if it read x rather than x − 3.

State the real domain and range of r(x) = √(x + 2).

1. The cause here is the even root, so demand that the whole expression under it is not negative: x + 2 at least 0.
2. Solve: x at least −2. The condition lands on x + 2, not on x alone.
3. For the range, the square root returns non-negative values only, and as x runs upward from −2 the inside runs upward from 0.
4. So the outputs start at √(0) = 0 and rise without bound.

Answer. Domain [−2, ∞). Range [0, ∞).

State the domain of f(x) = √(x)/(x - 4).

1. Two causes are present at once, so check both rather than stopping at the first.
2. The square root demands x at least 0.
3. The denominator forbids x − 4 = 0, so x = 4 must go.
4. Combine: start from [0, ∞) and remove the single point 4.

Answer. [0, 4) and (4, ∞).

The endpoint 0 survives because the root accepts it and the denominator there is −4, which is harmless. Each cause has to be checked against the whole expression, not against the letter x.

Find the range of f(x) = 1/x.

1. Ask which outputs are actually produced. Try to hit the output 5: solve 1/x = 5, giving x = 1/5, which is allowed.
2. The same works for any non-zero target y, because x = 1/y is then an allowed input.
3. Now try to hit 0: solve 1/x = 0. Nothing divided into 1 gives nothing.
4. So every output except 0 is produced, and 0 is not.

Answer. (−∞, 0) and (0, ∞).

Notice the method. To find a range, set the rule equal to a general output and ask which targets can be solved for. That is a chapter 3 solving question, asked once for every possible answer at the same time.

Plot y = 2x − 1 for x = −2, −1, 0, 1, 2.

1. Evaluate at each input in turn: 2(−2) − 1 = −5, then −3, −1, 1, 3.
2. Write each as a pair, input first: (−2, −5), (−1, −3), (0, −1), (1, 1), (2, 3).
3. Plot each point, going across by the first number and up by the second. Negative outputs go below the horizontal axis.
4. The five points lie on a straight line, which is what the family in chapter 6 predicts.

Answer. (−2, −5), (−1, −3), (0, −1), (1, 1), (2, 3).

Read the behaviour of g(x) = x² from its graph.

1. Falling for negative inputs, rising for positive inputs.
2. Flat at exactly one place, the low point at (0, 0).
3. It crosses both axes at that same point and never goes below the horizontal axis.
4. Far out in either direction it climbs without bound, and symmetrically, because (−x)² = x².

Answer. A single lowest point at the origin, symmetric about the vertical axis, unbounded above and bounded below by 0.

That last sentence is the range from chapter 4, read off the picture instead of argued from the formula.

A graph passes through (1, 4) and (1, 7). What can you conclude without seeing the rest of it?

1. Both points share the first coordinate 1, so both describe the same input.
2. They give that input the outputs 4 and 7.
3. The vertical line at x = 1 therefore meets the graph at least twice.
4. That is the fork chapter 2 ruled out, and nothing elsewhere on the graph can undo it.

Answer. It cannot be the graph of a function.

One pair of points settles it. Proving a graph is a function needs every vertical line; proving it is not needs only one.

From the graph of y = x² - 4, state the intercepts and the lowest point.

1. The vertical intercept is the output at x = 0, which is 0 − 4 = −4, giving the point (0, −4).
2. The horizontal intercepts are the inputs giving output 0: x² − 4 = 0, so x² = 4 and x is 2 or −2.
3. That is two crossings, at (−2, 0) and (2, 0). Two inputs sharing the output 0 is permitted.
4. The lowest point is where the parent x² was flat, moved down 4: (0, −4).

Answer. Intercepts (0, −4), (−2, 0) and (2, 0); lowest point (0, −4).

Here the vertical intercept and the lowest point happen to coincide. That is a feature of this rule, not a general fact.

A table gives outputs 5, 8, 13, 20, 29 for inputs 0, 1, 2, 3, 4. Which family?

1. First differences: 8 − 5 = 3, then 5, 7, 9. Not constant, so not linear.
2. Second differences: 5 − 3 = 2, then 2, 2. Constant, so quadratic.
3. Check the ratios in case it is exponential instead: 8/5 = 1.6, 13/8 = 1.625. Not constant, which confirms it.
4. The constant second difference is 2, and for x² the second difference is also 2, so the rule is x² plus something linear. At x = 0 the output is 5, and the pattern fits x² + 2x + 5.

Answer. Quadratic. The rule is x² + 2x + 5.

Check one row that was not used to build it: at x = 3, 9 + 6 + 5 = 20. It agrees.

For y = x², compare the output change from x = 0 to 1 with the change from x = 3 to 4.

1. From 0 to 1: the outputs go 0 to 1, a change of 1.
2. From 3 to 4: the outputs go 9 to 16, a change of 7.
3. Both input steps had size 1, yet the output changes differ by a factor of seven.

Answer. A change of 1, against a change of 7.

This is exactly why a single number cannot describe how fast x² grows. Chapter 11 answers by measuring over a stated interval; chapter 13 answers by shrinking the interval to nothing.

A table gives outputs 4, 12, 36, 108 for inputs 0, 1, 2, 3. Identify the family and the rule.

1. First differences: 8, 24, 72. Not constant, so not linear.
2. Second differences: 16, 48. Not constant either, so not quadratic.
3. Ratios: 12/4 = 3, 36/12 = 3, 108/36 = 3. Constant, so exponential.
4. The constant factor is 3 and the output at x = 0 is 4, so the rule is 4 × 3ˣ.

Answer. Exponential, with the rule 4·3ˣ.

Check a row not used to build it: at x = 2, 4 × 9 = 36. It agrees. The value at x = 0 is always the multiplier, because 3 to the power 0 is 1.

Two rules are proposed for the same data: 2x + 1 and 2ˣ. Test both against the outputs 1, 2, 4, 8 at inputs 0, 1, 2, 3.

1. Test 2x + 1: at x = 0 it gives 1, which matches. At x = 1 it gives 3, and the data says 2.
2. One disagreement is enough to reject it.
3. Test 2ˣ: 1, 2, 4, 8. All four match.
4. Confirm by mechanism rather than by luck: the ratios of the data are 2, 2, 2, which is what an exponential rule with base 2 must produce.

Answer. The data is 2ˣ. The linear rule fails at the second point.

Matching one point proves nothing, which is why the first test was abandoned as soon as a second point disagreed. Any rule can be made to pass through a single point.

Describe the change from x² to (x + 4)² − 1.

1. Find the parent: it is x², with the whole thing squared, so nothing has been stretched.
2. The + 4 is inside the bracket, so it moves horizontally and it moves against its sign: four to the left.
3. The − 1 is outside, so it moves vertically and does what it says: one down.
4. Check with a point. The parent has its low point at (0, 0). The new rule has its low point where the bracket is zero, at x = −4, with height −1.

Answer. Left 4, and down 1. The low point moves from (0, 0) to (−4, −1).

Write a square function shifted right 3 and up 5.

1. Right 3 is horizontal, so it goes inside the bracket, against its sign: (x − 3).
2. Up 5 is vertical, so it goes outside: + 5.
3. Assemble: (x − 3)² + 5.
4. Check: at x = 3 the bracket is 0 and the height is 5, so the low point sits at (3, 5), which is three right and five up from the origin.

Answer. (x − 3)² + 5.

The graph of y = f(x) has a low point at (1, 2). Where is the low point of y = f(x + 3) - 4?

1. Deal with the inside first: + 3 inside the bracket moves the picture three to the left, against its sign.
2. The low point moves from x = 1 to x = 1 − 3 = −2.
3. Now the outside: − 4 lowers every height by 4.
4. The height moves from 2 to 2 − 4 = −2.

Answer. (−2, −2).

Notice that this was answered without knowing what f is. A transformation acts on the picture, so any named feature of the parent can be tracked without the formula.

Write y = x² - 6x + 11 in the shifted form, and read off its low point.

1. Aim for the shape (x − h)² + k, because that form states the low point directly.
2. Take half the coefficient of x, which is −3, and square it to get 9.
3. Write x² − 6x + 9 as (x − 3)², so the rule is (x − 3)² + 11 − 9.
4. That is (x − 3)² + 2, a parent parabola moved right 3 and up 2.

Answer. (x − 3)² + 2, with its low point at (3, 2).

This is completing the square, arriving as a chapter 7 question rather than an algebra exercise. Its purpose here is to expose the shift that was hidden by the expansion.

Let f(x) = x² and g(x) = x − 3. Find f(g(5)) and g(f(5)).

1. For f(g(5)), start inside: g(5) = 5 − 3 = 2.
2. Hand 2 to f: f(2) = 2² = 4.
3. For g(f(5)), start inside again, but now the inner rule is f: f(5) = 25.
4. Hand 25 to g: g(25) = 25 − 3 = 22.

Answer. f(g(5)) = 4 and g(f(5)) = 22.

The gap between 4 and 22 is a useful shock. Squaring after subtracting is a very different instruction from subtracting after squaring.

For the same f and g, write formulas for f(g(x)) and g(f(x)).

1. f(g(x)): the input to f is the whole of g(x), which is x − 3. So square that: (x − 3)².
2. g(f(x)): the input to g is x², and g subtracts 3 from whatever it receives: x² − 3.
3. Check at x = 5 against the previous example: (5 − 3)² = 4, and 5² − 3 = 22. Both agree.

Answer. f(g(x)) = (x − 3)² and g(f(x)) = x² − 3.

Chapter 7 met (x − 3)² as a shift of the parent x². It is the same expression, now arrived at from a different direction.

Let f(x) = √(x) and g(x) = x − 4. What inputs are allowed for f(g(x))?

1. First gate: g accepts every real number, so it refuses nothing.
2. Second gate: f is a square root, so it needs a non-negative input. Its input here is g(x) = x − 4.
3. Demand x − 4 at least 0, giving x at least 4.
4. Sanity check with x = 1: g(1) = −3, and f cannot take −3. Correctly refused.

Answer. x at least 4, that is [4, ∞).

Neither rule alone bars x = 1. The chain does. This is why the domain of a composite must be worked out from the chain, not inherited from either link.

Write h(x) = √(3x - 5) as f(g(x)) with two simpler rules, then find its domain.

1. Ask what is done last. The square root is applied to everything else, so f is the root.
2. Ask what is done first. The expression 3x − 5 is built before the root sees it, so g(x) = 3x − 5.
3. Check: f(g(x)) = √(3x − 5), which is h.
4. For the domain, g accepts everything but f needs a non-negative input, so demand 3x − 5 at least 0, giving x at least 5/3.

Answer. f(x) = √(x) and g(x) = 3x − 5, with domain [5/3, ∞).

Decomposing like this, outer rule and inner rule, is the exact preparation the chain rule needs in Book 3. There the question is always "what is the outer function, and what is the inner one?"

Find the inverse of f(x) = 3x − 5.

1. Write it as y = 3x − 5.
2. Swap x and y: x = 3y − 5. The input and output have exchanged roles, which is the whole idea.
3. Solve for y: add 5 to get x + 5 = 3y, then divide by 3.
4. So y = (x + 5)/3.

Answer. f⁻¹(x) = (x + 5)/3.

Verify that inverse by composition, in both directions.

1. f⁻¹(f(x)): the inner rule gives 3x − 5, and the outer adds 5 then divides by 3.
2. That is ((3x − 5) + 5)/3 = 3x/3 = x.
3. f(f⁻¹(x)): the inner rule gives (x + 5)/3, and the outer triples it then subtracts 5.
4. That is 3·(x + 5)/3 − 5 = (x + 5) − 5 = x.

Answer. Both compositions return x, so the inverse is correct.

Chapter 8 question 11 predicted this: a pair of rules composing to x in both directions is precisely a function and its inverse.

Find the inverse of f(x) = (2x + 1)/5.

1. Write y = (2x + 1)/5.
2. Swap the letters: x = (2y + 1)/5.
3. Multiply both sides by 5: 5x = 2y + 1.
4. Subtract 1 and divide by 2: y = (5x − 1)/2.

Answer. f⁻¹(x) = (5x − 1)/2.

Check with one value rather than trusting the algebra. f(2) = 5/5 = 1, and f⁻¹(1) = (5 − 1)/2 = 2. It returns.

Does f(x) = x² - 4x have an inverse over all real numbers? If not, restrict it.

1. Test one-to-one by looking for two inputs sharing an output. Try f(0) = 0 and f(4) = 16 − 16 = 0.
2. Two inputs give 0, so the horizontal line at height 0 meets the graph twice and there is no inverse.
3. Find the turning point, because that is where the two halves meet: writing it as (x − 2)² − 4 shows the low point at x = 2.
4. Restrict to x at least 2, keeping the rising half only. On that half no output is shared.

Answer. No inverse over all real numbers. Restricted to [2, ∞) it has one.

The turning point is always where to cut, because it is the only place a smooth curve turns from falling to rising and so starts repeating outputs. Chapter 13 gives a way to find it without completing the square.

A tank holds 80 litres at 9am and 50 litres at 3pm. Find ΔV and Δt, and say what the signs mean.

1. Volume: new minus old, so ΔV = 50 − 80 = −30 litres.
2. Time: 3pm minus 9am, so Δt = 6 hours.
3. The negative ΔV says the tank lost 30 litres. It is not an error to be dropped.
4. Δt is positive because time ran forwards, which it does in every question in this book.

Answer. ΔV = −30 litres and Δt = 6 hours.

Δt is almost always positive, so the sign of a rate almost always comes from the top of the fraction. That is a useful thing to know when checking an answer.

A car covers 150 miles in 3 hours. Find its rate. Then a second car covers 150 miles in 2 hours.

1. Rate means per one hour, so divide the distance by the time: 150 ÷ 3 = 50.
2. The units come along with the arithmetic: miles divided by hours gives miles per hour.
3. Second car: 150 ÷ 2 = 75 miles per hour.
4. The same distance in less time gives a bigger rate, because the divisor shrank while the top stayed the same.

Answer. 50 mph and 75 mph.

Notice which number moved. Dividing by a smaller number gives a bigger answer, and that single fact is why the derivative in chapter 13 needs a limit rather than just setting the interval to zero.

A line passes through (2, 5) and (6, 17). Find Δy, Δx, and the rate of change.

1. Δy is the change in the second coordinate: 17 − 5 = 12.
2. Δx is the change in the first: 6 − 2 = 4.
3. Divide: Δy/Δx = 12/4 = 3.
4. Check the order. Taking both differences the other way gives −12 and −4, and (−12)/(−4) is still 3, so the answer does not depend on which point you call first, provided you are consistent.

Answer. Δy = 12, Δx = 4, and the rate of change is 3.

That last step matters. Being consistent is the whole requirement; there is no rule about which point must come first.

A tank drains from 80 litres to 50 litres over 6 hours. Find the rate of change, and say what its sign means.

1. ΔV = 50 − 80 = −30 litres, from example 1.
2. Δt = 6 hours.
3. Divide: −30/6 = −5 litres per hour.
4. The negative sign says the volume is falling. The tank loses 5 litres in each hour.

Answer. −5 litres per hour.

A negative rate does not mean time ran backwards. Δt was positive throughout; the minus came from the top of the fraction, where the quantity genuinely decreased.

Find the average rate of change of f(x) = x² from x = 2 to x = 5.

1. Evaluate at both ends: f(2) = 4 and f(5) = 25.
2. Output change: 25 − 4 = 21.
3. Input change: 5 − 2 = 3.
4. Divide: 21/3 = 7.

Answer. 7.

The curve is not climbing at 7 anywhere near x = 2, and is climbing faster than 7 by x = 5. Seven is the steady rate that would have covered the same ground in the same interval.

Find the average rate of change of g(t) = 60t from t = 1 to t = 4, and interpret the units.

1. g(1) = 60 and g(4) = 240.
2. Output change 180, input change 3, so the rate is 60.
3. If t is hours and g is miles, the answer is 60 miles per hour.
4. Any other interval gives the same answer, because the rule is linear and chapter 6 showed its differences are constant.

Answer. 60 distance-units per time-unit.

For a linear rule, the average rate over every interval is the same number: the m in mx + c. This is the one family where a single number really does describe the whole rule.

For f(x) = x², calculate [f(2 + h) - f(2)]/h and simplify.

1. f(2 + h) = (2 + h)² = 4 + 4h + h². The middle term is the one that matters.
2. Subtract f(2) = 4: the difference is 4h + h².
3. Divide by h: (4h + h²)/h = 4 + h, valid for every h except 0, where the division is not allowed.
4. Read the result. At h = 1 the average rate is 5; at h = 0.1 it is 4.1; at h = 0.01 it is 4.01.

Answer. 4 + h.

The answers are crowding around 4 as h shrinks, without ever being 4. That crowding is a limit, and 4 is the derivative of x² at x = 2. Book 2 makes the crowding precise; Book 3 makes it routine.

A tank holds V(t) = t² litres after t minutes. Find the average filling rate over the first 3 minutes, and over the third minute alone.

1. Over the first 3 minutes: V(0) = 0 and V(3) = 9, so the rate is (9 − 0)/3 = 3 litres per minute.
2. Over the third minute alone the interval is from t = 2 to t = 3.
3. V(2) = 4 and V(3) = 9, so the rate is (9 − 4)/1 = 5 litres per minute.
4. The second figure is larger because the tank fills faster as time goes on, and the first has been diluted by the slow early minutes.

Answer. 3 litres per minute overall; 5 litres per minute during the third.

Both are correct answers to different questions. Whenever an average rate is quoted, the interval is part of the answer, and an average rate with no interval attached means nothing.