Applications: Functions

A piecewise function has one rule for each band of inputs, so the first job is always to find the band. On the graph a filled end shows a value that belongs to its band and a hollow end shows one that does not, which settles what happens at exactly 1 hour and at exactly 4 hours. Working backward from a charge, each band is tried in turn, and a solution counts only if it lies in the band whose rule produced it.

1. Find the band before using a rule. A stay of exactly 1 hour satisfies 0 < x ≤ 1, so it belongs to the first band and C(1) = 3. On the graph the first band ends with a filled end at (1, 3), and the second band starts with a hollow end at (1, 4) because x = 1 does not belong to it.
2. A stay of 2.5 hours satisfies 1 < x ≤ 4, so use the second rule: C(2.5) = 2 × 2.5 + 2 = 7.
3. (a) A stay of exactly 1 hour costs $3 and a stay of 2.5 hours costs $7.
4. For a charge of $8, look at each band. The first band always gives 3 and the third always gives 12, so only the second rule can give 8: 2x + 2 = 8, so 2x = 6 and x = 3.
5. (b) Check that the solution lies in the band whose rule gave it: 1 < 3 ≤ 4 is true. She parked for 3 hours. Check: C(3) = 2 × 3 + 2 = 8.

answer(a) $3 for exactly 1 hour and $7 for 2.5 hours; (b) 3 hours

techniquePiecewise Functions · Graphing a Piecewise Function · Function Notation

examsSAT · GCSE Higher · H2

A piecewise function is continuous at a threshold when the two rules give the same value there, so that the graph can be drawn through the threshold without lifting the pen. The first rule fixes the value at the threshold. The second rule has a constant still to be chosen, and asking it to give the same value is one equation in that constant.

1. At the threshold x = 200 the first rule gives B(200) = 0.2 × 200 = 40. A household that uses exactly 200 units pays $40.
2. The bill must not jump when one more unit is used, so the second rule must also give 40 at x = 200: 0.3 × 200 + k = 40.
3. (a) 60 + k = 40, so k = 40 − 60 = −20. The second rule is B(x) = 0.3x − 20.
4. A use of 350 units is more than 200, so use the second rule: B(350) = 0.3 × 350 − 20 = 105 − 20 = 85.
5. (b) The bill is $85. Check by the bands: the first 200 units cost $40 and the other 150 units cost 150 × 0.30 = $45, and 40 + 45 = 85.

answer(a) k = −20; (b) $85

techniqueMaking a Piecewise Function Continuous · Piecewise Functions

When one process follows another, each is a function and the whole is their composite. In fg(x) the function written nearest to x acts first, so fg means g and then f. Swapping the order changes what the fee is taken from, and so it changes the result.

1. The fee is taken first, so g acts first: g(120) = 120 − 5 = 115 dollars are left. Then f changes them: f(115) = 40 × 115 = 4600 pesos.
2. In fg(x) the function nearest to x acts first. Put g(x) into f: fg(x) = f(x − 5) = 40(x − 5) = 40x − 200.
3. (a) fg(x) = 40x − 200, and fg(120) = 4800 − 200 = 4600, so the customer receives 4600 pesos. The −200 is the fee of $5 seen in pesos, because 5 × 40 = 200.
4. In the other order f acts first: f(120) = 40 × 120 = 4800, and then g(4800) = 4800 − 5 = 4795. As a formula, gf(x) = 40x − 5.
5. (b) gf(120) = 4795. The order gf would mean changing all of the money first and then taking a fee of 5 pesos, which leaves the customer 195 pesos better off. The two orders give different functions, so fg ≠ gf.

answer(a) fg(x) = 40x − 200; 4600 pesos; (b) gf(120) = 4795, which would mean changing the money first and then taking a fee of 5 pesos, 195 pesos more for the customer

techniqueComposite Functions · Function Notation

examsGCSE Higher · H2

The inverse function takes every output of f back to its input, so it changes Fahrenheit into Celsius. It undoes the steps of f in the opposite order. The graph of f⁻¹ is the reflection of the graph of f in the line y = x, so the two graphs can only meet each other on that line, at an input that f leaves unchanged.

1. The function f multiplies by 1.8 and then adds 32. The inverse undoes these steps in the opposite order: first subtract 32, then divide by 1.8.
2. In symbols, let y = 1.8x + 32. Then y − 32 = 1.8x, so x = y − 321.8. Writing the input as x, f⁻¹(x) = x − 321.8.
3. (a) f⁻¹(95) = 95 − 321.8 = 631.8 = 35, so 95^°F is 35^°C. Check: f(35) = 1.8 × 35 + 32 = 63 + 32 = 95.
4. A temperature that reads the same on both scales is an input that f leaves unchanged: f(x) = x, so 1.8x + 32 = x.
5. Subtract x and 32 from both sides: 0.8x = −32, so x = −320.8 = −40.
6. (b) The two scales agree at −40 degrees. On the graph this is the point (−40, −40), where y = f(x) meets the line y = x, and the graph of f⁻¹ passes through the same point. Check: 1.8 × (−40) + 32 = −72 + 32 = −40.

answer(a) f⁻¹(x) = x − 321.8; 35^°C; (b) −40 degrees

techniqueInverse Functions

examsSAT · GCSE Higher · H2

A change made inside the bracket, to the input, moves a graph sideways, and a change made outside, to the output, moves or stretches it up and down. Neither change needs the formula to be worked out again: it is enough to follow one known point, here the turning point, through the change.

1. First find the peak of a normal day. f(x) = 5x(8 − x) is zero at x = 0 and at x = 8, so the axis of symmetry is x = 4, and f(4) = 5 × 4 × 4 = 80. The pool is fullest at 12 noon, with 80 people.
2. The graph of y = f(x − 2) is the graph of y = f(x) moved 2 units to the right, because the new function needs an input that is 2 larger to give the same output. The turning point (4, 80) moves to (6, 80).
3. (a) The greatest number is still 80 people, and it is reached at x = 6, which is 2 p.m. Check: f(6 − 2) = f(4) = 80.
4. On the rainy day every output is multiplied by 0.6, so the function is 0.6f(x). This is a vertical stretch with scale factor 0.6: every point stays above the same value of x and moves to 0.6 of its height.
5. (b) The turning point (4, 80) moves to (4, 0.6 × 80) = (4, 48). The greatest number is 48 people, still at 12 noon.

answer(a) 80 people at x = 6, which is 2 p.m.; (b) 0.6f(x); 48 people, at 12 noon

techniqueTransforming Graphs · Stretching a Graph Vertically

examsSAT · GCSE Higher · H2

The formula for R accepts any number, but the situation does not. The domain is the set of prices the situation allows, and each limit in the story is an inequality in x. The range is the set of takings those prices can produce. For a quadratic the greatest or least output may be at the turning point and not at an end of the domain, so all three places are checked.

1. The number of buyers cannot be more than the 400 seats: 600 − 20x ≤ 400, so 200 ≤ 20x and x ≥ 10.
2. The number of buyers cannot be negative: 600 − 20x ≥ 0, so 20x ≤ 600 and x ≤ 30.
3. (a) Both limits must hold, so the domain is 10 ≤ x ≤ 30.
4. For the range, find the turning point first. R(x) = x(600 − 20x) is zero at x = 0 and at x = 30, so the axis of symmetry is x = 15, which is inside the domain. R(15) = 15 × 300 = 4500. The coefficient of x² is negative, so this is the greatest value.
5. Now the two ends of the domain: R(10) = 10 × 400 = 4000 and R(30) = 30 × 0 = 0. The least value is 0.
6. (b) The range is 0 ≤ R ≤ 4500. The takings run from nothing, at a price of $30, up to $4500, at a price of $15.

answer(a) 10 ≤ x ≤ 30; (b) 0 ≤ R ≤ 4500

techniqueDomain and Range

examsSAT · H2

Doubling every hour multiplies by 2 once for each hour, which is the exponential function 500 × 2ⁿ. Finding the count at a given hour is a substitution. Finding the hour for a given count puts the unknown in the index, and a logarithm is the tool that brings an index down to the line.

1. (a) 2⁶ = 64, so after 6 hours N = 500 × 64 = 32 000 bacteria.
2. For a million, solve 500 × 2ⁿ = 1 000 000. Divide both sides by 500: 2ⁿ = 2000.
3. The unknown is an index, so take logarithms to base 10 of both sides and use the power law: n log₁₀ 2 = log₁₀ 2000, so n = log₁₀ 2000log₁₀ 2.
4. A calculator gives log₁₀ 2 = 0.3010. Since 2000 = 2 × 1000, log₁₀ 2000 = log₁₀ 2 + log₁₀ 1000 = 0.3010 + 3 = 3.3010. So n = 3.30100.3010 ≈ 10.97.
5. (b) The number passes a million a little before n = 11, so the first whole hour is 11. Check: 2¹⁰ = 1024 gives 512 000, which is under a million, and 2¹¹ = 2048 gives 1 024 000, which is over.

answer(a) 32 000 bacteria; (b) after 11 hours

techniqueSolving Exponential Equations · Exponential Growth · Growth and Decay Problems

examsO-Level · SAT

Exponential decay multiplies by the same factor below 1 in every equal interval of time. Here the factor is 0.5 for every 4 hours, and the index x4 counts the half-lives that have passed. A whole number of half-lives needs only halving. An amount that falls between two halvings puts the unknown in the index, and logarithms bring it down.

1. (a) 12 hours is 12 ÷ 4 = 3 half-lives, so the amount is halved three times: A = 320 × 0.5³ = 320 × 18 = 40 mg.
2. For the next dose solve 320 × 0.5^x/4 = 100. Divide both sides by 320: 0.5^x/4 = 100320. Halving is dividing by 2, so turn both sides over: 2^x/4 = 320100 = 3.2.
3. Take logarithms to base 10 of both sides and use the power law: x4 log₁₀ 2 = log₁₀ 3.2.
4. A calculator gives log₁₀ 2 = 0.3010. Since 3.2 = 2⁵10, log₁₀ 3.2 = 5 × 0.3010 − 1 = 0.5050. So x4 = 0.50500.3010 ≈ 1.678, and x ≈ 4 × 1.678 ≈ 6.71.
5. (b) The next dose is due after about 6.7 hours. Check: the amount is 160 mg after one half-life, at 4 hours, and 80 mg after two, at 8 hours. 100 mg lies between these amounts, and 6.7 hours lies between these times.

answer(a) 40 mg; (b) about 6.7 hours

techniqueExponential Decay · Solving Exponential Equations · Growth and Decay Problems

examsO-Level · SAT · GCSE Higher

Adding the same amount every year gives a linear function, and multiplying by the same factor every year gives an exponential function. The exponential starts more slowly here, because 8% of $1000 is only $80, but its yearly gain grows while the linear gain stays at $100. The equation A = B has n both in an index and in a product, so it cannot be solved by algebra, and the two are compared year by year.

1. Aisha's savings rise by the same amount each year, so A is a linear function. After 5 years A = 1000 + 100 × 5 = 1500.
2. Ben's savings are multiplied by 1.08 each year, so B is an exponential function. Work up the powers: 1.08² = 1.1664, 1.08³ ≈ 1.25971, 1.08⁴ ≈ 1.36049 and 1.08⁵ ≈ 1.46933.
3. (a) After 5 years Aisha has $1500 and Ben has 1000 × 1.46933 = $1469.33, so Ben is $30.67 behind.
4. Carry on year by year. At n = 6, A = 1600 and B = 1000 × 1.08⁶ ≈ 1586.87, so Ben is still behind.
5. At n = 7, A = 1700 and B = 1000 × 1.08⁷ ≈ 1713.82, so Ben is ahead.
6. (b) Ben first has more than Aisha after 7 years. He stays ahead from then on, because 8% of a balance above $1700 is more than $136 a year, which is more than Aisha's $100, and it keeps growing.

answer(a) Aisha has $1500 and Ben has $1469.33; (b) after 7 years

techniqueExponential Growth · Growth and Decay Problems

examsO-Level · SAT · GCSE Higher

A decibel scale is a logarithmic scale: it records the index of the power of 10 and not the intensity itself. Multiplying the intensity therefore adds to the level, which is the product law of logarithms, log(ab) = log a + log b, at work.

1. (a) For one machine II₀ = 10⁸. A logarithm to base 10 asks what power of 10 gives the number, so log₁₀ 10⁸ = 8 and L = 10 × 8 = 80 decibels.
2. With two machines the intensity is 2 × 10⁸ I₀, so L = 10 log₁₀(2 × 10⁸).
3. The logarithm of a product is the sum of the logarithms. A calculator gives log₁₀ 2 = 0.3010, so log₁₀(2 × 10⁸) = log₁₀ 2 + log₁₀ 10⁸ = 0.3010 + 8 = 8.3010.
4. (b) L = 10 × 8.3010 = 83.01, which is 83 decibels to the nearest decibel.
5. Doubling the intensity adds 10 log₁₀ 2 ≈ 3 decibels, whatever the level was before. A rise of 10 decibels is a whole extra power of 10, which is ten times the intensity, so a rise of 3 decibels for twice the intensity is in keeping with the scale.

answer(a) 80 decibels; (b) 83 decibels

techniqueThe Laws of Logarithms · Exponentials and Logarithms

examsO-Level

Exponential data make a curve, and it is hard to tell one curve from another by eye. Taking logarithms of y = a × b^x turns it into the equation of a straight line in x and log y. A straight line is easy to recognize, and its gradient and intercept give b and a.

1. If y = a × b^x, take logarithms to base 10 of both sides: log₁₀ y = log₁₀ a + x log₁₀ b. This has the form Y = c + mx, so a plot of log₁₀ y against x is a straight line with gradient log₁₀ b and intercept log₁₀ a.
2. The values of log₁₀ y rise by 0.5 every week, so the five points lie on a straight line. From (0, 1.0) to (4, 3.0) the gradient is 3.0 − 1.04 − 0 = 24 = 0.5.
3. (a) The gradient is 0.5. The line meets the vertical axis at 1.0, so the intercept is 1.
4. The intercept is log₁₀ a = 1, so a = 10¹ = 10. The gradient is log₁₀ b = 0.5, so b = 10^0.5 = √10. Since 3.16² = 9.9856 and 3.17² = 10.0489, √10 ≈ 3.16.
5. (b) The model is y = 10 × 3.16^x: the channel started with 10 subscribers, and the number is multiplied by about 3.16 every week. Check with week 2: 10 × 3.16² ≈ 99.9, which is close to the 100 in the table.

answer(a) gradient 0.5, intercept 1; (b) y = 10 × 3.16^x

techniqueStraightening Growth with Logarithms · Exponentials and Logarithms

examsO-Level · H2

A logistic model grows like an exponential while there is plenty of room and then levels off at the largest population the lake can feed, which is the carrying capacity. In the formula the term 8 × 2^−x is large at first and fades to nothing, and what is left when it has gone is the carrying capacity. Halfway up, the equation for x is an exponential equation.

1. At the start x = 0 and 2⁰ = 1, so P = 9001 + 8 = 9009 = 100.
2. As x grows, 2^−x = 12^x becomes smaller and smaller: by x = 10 it is 11024. The bottom of the fraction closes on 1, so P closes on 9001 = 900 and never passes it.
3. (a) The lake was stocked with 100 fish, and its carrying capacity is 900 fish. On the graph the carrying capacity is the horizontal asymptote P = 900.
4. Half of the carrying capacity is 450. Put P = 450: 9001 + 8 × 2^−x = 450, so 1 + 8 × 2^−x = 2 and 8 × 2^−x = 1.
5. Multiply both sides by 2^x: 8 = 2^x. Since 8 = 2³, x = 3.
6. (b) The population grows fastest after 3 years, when there are 450 fish. Check: 2⁻³ = 18, so P = 9001 + 1 = 450. Before this the curve bends upward like an exponential, and after it the curve bends over toward the asymptote.

answer(a) 100 fish were put in, and the carrying capacity is 900 fish; (b) after 3 years

techniqueLogistic Growth and Carrying Capacity · Solving Exponential Equations

examsO-Level · SAT

A replay shows the same heights at different times. Replacing x by ax inside the bracket does exactly that: the replay shows a moment of the real dive when ax equals the real time, so every time on the graph is divided by a and every height stays the same. This is a horizontal stretch with scale factor 1a, and a value of a less than 1 slows the dive down.

1. First find the two moments of the real dive. She enters the water when f(x) = 0: 10 + 5x − 5x² = 0, so x² − x − 2 = 0 and (x − 2)(x + 1) = 0. The root x = −1 is before she left the platform, so it is rejected, and she enters the water at x = 2. The highest point is on the axis of symmetry, halfway between the roots, at x = −1 + 22 = 0.5, where f(0.5) = 10 + 2.5 − 1.25 = 11.25 m.
2. In the replay the height is f(0.25x), so at time x the replay shows the real moment 0.25x. A real moment appears when 0.25x equals it, which is when x is that moment divided by 0.25, or 4 times it. The graph is stretched sideways with scale factor 4, and every height stays the same.
3. (a) The highest point: 0.25x = 0.5, so x = 2 seconds into the replay. Entering the water: 0.25x = 2, so x = 8 seconds into the replay. Check: f(0.25 × 8) = f(2) = 10 + 10 − 20 = 0.
4. The second replay must show the real moment 2 seconds at x = 5, so a × 5 = 2 and a = 25 = 0.4. The graph is stretched sideways with scale factor 10.4 = 2.5, and 2 × 2.5 = 5 seconds, as the slot requires.
5. (b) Replace every x in f by 0.4x: f(0.4x) = 10 + 5(0.4x) − 5(0.4x)² = 10 + 2x − 0.8x² meters. Check: at x = 5 the height is 10 + 10 − 20 = 0, and the highest point is at x = 0.50.4 = 1.25, where the height is 10 + 2.5 − 1.25 = 11.25 m, the same as in the real dive.

answer(a) at her highest point 2 seconds into the replay, and into the water 8 seconds into it; (b) a = 0.4, and her height is 10 + 2x − 0.8x² meters

techniqueStretching a Graph Horizontally · Transforming Graphs

examsH2

The time for prices to double is the power to which 1.05 must be raised to give 2, which is a logarithm to base 1.05. A calculator has no key for that base. The change of base rule, log_b x = log₁₀ xlog₁₀ b, turns it into a quotient of two common logarithms, which the calculator does have.

1. Prices have doubled when P × 1.05ⁿ = 2P. Divide both sides by P: 1.05ⁿ = 2. So n is the power of 1.05 that gives 2, which is n = log_1.05 2. The price P has cancelled, so the answer is the same for every item.
2. By the change of base rule, log_1.05 2 = log₁₀ 2log₁₀ 1.05. A calculator gives log₁₀ 2 = 0.30103 and log₁₀ 1.05 = 0.021189, so n = 0.301030.021189 = 14.207.
3. (a) Prices double after 14.2 years. Check: 1.05¹⁴ ≈ 1.980 and 1.05¹⁵ ≈ 2.079, so the doubling falls between the 14th year and the 15th.
4. At 2% a year the yearly factor is 1.02, and the doubling time is log_1.02 2 = log₁₀ 2log₁₀ 1.02 = 0.301030.0086002 = 35.003 years. Check: 1.02³⁵ ≈ 2.000.
5. (b) 35.003 − 14.207 = 20.796, so prices would take about 20.8 more years to double. Subtracting the unrounded times and rounding at the end keeps the first decimal place right.

answer(a) 14.2 years; (b) about 20.8 more years, since at 2% prices take about 35.0 years to double

techniqueThe Change of Base Rule · Solving Exponential Equations

examsO-Level

The graph of M = log₁₀ A passes through (1, 0), (10, 1), (100, 2) and (1000, 3): each tenfold swing is one step up. It rises steeply at first and then more and more slowly, so small and very large earthquakes fit on one short range of magnitudes. A number added outside the logarithm moves the whole graph up without changing its shape.

1. On the graph of M = log₁₀ A, each tenfold swing adds 1: log₁₀ 1000 = 3 and log₁₀ 10 000 = 4. A swing of 5000 micrometers lies between these, so its magnitude is between 3 and 4.
2. (a) log₁₀ 5000 = log₁₀ 5 + log₁₀ 1000 = 0.699 + 3 = 3.699, so the magnitude is 3.7 to 1 decimal place.
3. In M = log₁₀ A + 1 the 1 is added outside the logarithm, to the output. Every point (A, M) of the first graph moves to (A, M + 1): the graph is translated 1 unit up, in the direction of the M-axis.
4. The magnitude belongs to the earthquake, not to the station, so it is still 3.699 at 300 km. Solve log₁₀ A + 1 = 3.699: log₁₀ A = 2.699, so A = 10^2.699 ≈ 500. Exactly: log₁₀ A = log₁₀ 5000 − log₁₀ 10 = log₁₀ 500.
5. (b) The new graph is the graph of M = log₁₀ A translated 1 unit up, and the same earthquake swings the needle by 500 micrometers at the 300 km station, a tenth of the swing at 100 km. Check: log₁₀ 500 + 1 = 2.699 + 1 = 3.699.

answer(a) 3.7; (b) a translation of 1 unit up, in the direction of the M-axis; a swing of 500 micrometers

techniqueLogarithmic Graphs · Transforming Graphs · The Laws of Logarithms

examsO-Level · SAT

A power law B = a × Mⁿ makes a curve, and one power is hard to tell from another by eye. Taking logarithms of both sides gives log B = log a + n log M, which is a straight line when log B is plotted against log M. Its gradient is the power n itself, and its intercept is log a.

1. Take logarithms to base 10 of B = a × Mⁿ. The logarithm of a product is a sum, and the power law brings n down: log₁₀ B = log₁₀ a + n log₁₀ M. This has the form Y = c + mX, with X = log₁₀ M and Y = log₁₀ B.
2. From (0, 1.845) to each of the other points the gradient is the same: 2.748 − 1.8451.204 = 0.9031.204 = 0.75, 3.276 − 1.8451.908 = 1.4311.908 = 0.75 and 3.942 − 1.8452.796 = 2.0972.796 = 0.75. So the four points lie on one straight line.
3. (a) The gradient is 0.75. The mammal of mass 1 kg has log₁₀ M = 0, so its point (0, 1.845) is on the vertical axis, and the intercept is 1.845.
4. The gradient is the power itself, so n = 0.75. The intercept is log₁₀ a, so a = 10^1.845 = 69.98, which is 70 to 2 significant figures: the resting energy use of a mammal of 1 kg. The model is B = 70 × M^0.75.
5. (b) For the bear, 256 = 4⁴, so 256^0.25 = 4 and 256^0.75 = 4³ = 64. Then B = 70 × 64 = 4480 kilocalories per day. Check on the line: log₁₀ 256 = 2.408 and 1.845 + 0.75 × 2.408 = 3.651, and 10^3.651 ≈ 4480.

answer(a) the gradient from the first point to each of the others is 0.75, so the points lie on a straight line with gradient 0.75 and intercept 1.845; (b) n = 0.75 and a = 70, so B = 70 × M^0.75; about 4480 kilocalories per day

techniqueStraightening a Power Law · The Laws of Logarithms

examsO-Level · H2

A logarithmic model y = a + b ln x keeps rising for as long as x grows, but each extra week adds less than the one before, because ln x rises more and more slowly. Two data points fix the two constants. At x = 1 the logarithm is 0, which gives a at once, and the second point then gives b.

1. At x = 1, ln 1 = 0, so the model gives y = a. Her speed after 1 week is 25 words per minute, so a = 25.
2. At x = 5: 25 + b ln 5 = 45, so b ln 5 = 20 and b = 20ln 5 = 201.6094 = 12.427.
3. (a) a = 25 and b = 12.4 to 3 significant figures, so the model is y = 25 + 12.4 ln x.
4. After 12 weeks: y = 25 + 12.427 ln 12 = 25 + 12.427 × 2.4849 = 25 + 30.88 = 55.88, which is 56 words per minute to the nearest word per minute.
5. For 60 words per minute: 25 + 12.427 ln x = 60, so ln x = 3512.427 = 2.8165. The inverse of ln is e to the power, so x = e^2.8165 ≈ 16.7.
6. (b) The model predicts 56 words per minute after 12 weeks, and it first gives at least 60 words per minute after 17 weeks. Check: at x = 16, y = 25 + 12.427 × 2.7726 = 59.45, and at x = 17, y = 25 + 12.427 × 2.8332 = 60.21. The first 4 weeks of practice added 20 words per minute, and the 7 weeks after that add fewer than 11.

answer(a) a = 25 and b = 12.4; (b) 56 words per minute after 12 weeks; the model first gives at least 60 words per minute after 17 weeks

techniqueFitting a Logarithmic Model · The Natural Logarithm

examsO-Level