Moves the variable from the base to the exponent, which changes growth from additive to multiplicative. Identifies exponential functions from tables and formulas, distinguishes growth from decay by the base, establishes that an exponential eventually outgrows every power function, and introduces compound interest and the number e as the limit of ever-more-frequent compounding.
Subject: Precalculus · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Precalculus · Chapter 4 — Exponential and Logarithmic Functions
§4.1 Exponential Functions, pp. 460-481
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 460-481 — the pages these objectives are drawn from
Warm-up
The difference between adding and multiplying is easy to underestimate.
Discussion prompt
Would you rather be paid a thousand pounds a day for thirty days, or one penny on day one with the amount doubling daily?
Hint: The doubling looks hopeless for the first two weeks. Compute day 30.
Answer:
The flat offer gives 30,000 pounds, and it leads for most of the month — after two weeks the doubling has reached only about 82 pounds.
But day 30 of the doubling is over five million pounds, and the last day alone is worth more than the whole flat offer many times over.
That is the whole chapter in one example. Multiplying by a fixed factor beats adding a fixed amount, eventually and then overwhelmingly — and 'eventually' can be much later than intuition suggests, which is exactly why the intuition is unreliable.
Concept
A linear function adds the same amount for each unit of input. An exponential function multiplies by the same factor instead, which produces growth of an entirely different character.
exponential function — A function of the form a times b to the power x, where b is positive and not 1. The variable is in the exponent, and each unit increase in the input multiplies the output by b.
\[ f(x)=ab^x, \qquad b>0, \; b\ne 1 \]
The base must be positive, or the function would be undefined at many inputs — a negative base raised to a half is not real. And the base cannot be 1, since that would make the function constant and remove everything interesting about it.
Figure (svg): Two tables side by side showing linear growth adding a fixed amount at each step and exponential growth multiplying by a fixed factor, with the values diverging sharply by the last row
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 460-465
Section
Section 1
Concept
An exponential function has a constant base and a variable exponent. A power function has the reverse, and the two behave nothing alike.
The table test is the practical one. Compute the ratio of each output to the previous: if that ratio is constant, the relationship is exponential and the ratio is the base. If the differences are constant instead, it is linear.
Figure (svg): Two tables side by side showing linear growth adding a fixed amount at each step and exponential growth multiplying by a fixed factor, with the values diverging sharply by the last row
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 460-468
Picture it
Both start at 2 and both use the number 3, in different ways.
Figure (svg): Two tables side by side showing linear growth adding a fixed amount at each step and exponential growth multiplying by a fixed factor, with the values diverging sharply by the last row
The linear column adds 3 each row and the exponential multiplies by 3. Five rows in they differ by a factor of nearly thirty, and the gap widens without limit.
Worked example
Check the differences, then the ratios.
\[ \begin{array}{c|cccc} x & 0 & 1 & 2 & 3 \\ \hline y & 5 & 15 & 45 & 135 \end{array} \]
Compute the differences
Why: Consecutive outputs subtracted.
\[ 10, 30, 90 \]
Check whether they are constant
Why: They are not.
Compute the ratios
Why: Consecutive outputs divided.
\[ 3, 3, 3 \]
Read the model
Why: Constant ratio 3, starting at 5.
\[ y = 5(3 ^{x}) \]
Figure (svg): The solution to Worked example identify from a table shown as a ladder of expressions, one row per legal move
\[ y=5\cdot 3^x \]
Verify: test the last entry
Why: At x equal to 3 the model gives 5 times 27, which is 135 — matching the table. The initial value 5 was read straight off the output at zero, since any base to the power zero is 1. Both constants came from the table with no algebra at all.
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 462-465
Sorting
Look at where the variable sits.
Sort into buckets
Sort each rule.
Worked example
Look at where the variable sits.
\[ \text{Classify } f(x)=3^x, \; g(x)=x^3, \; h(x)=3x. \]
Examine the first
Why: Constant base, variable exponent.
Examine the second
Why: Variable base, constant exponent.
Examine the third
Why: No exponent above one.
Compare their growth
Why: At x equal to 10.
\[ 59049, 1000, 30 \]
Figure (svg): The solution to Worked example exponential or power shown as a ladder of expressions, one row per legal move
\[ 3^x \text{ exponential}, \; x^3 \text{ power}, \; 3x \text{ linear} \]
Verify: check the ordering reverses somewhere
Why: At x equal to 2 the three give 9, 8 and 6 — very close. At x equal to 10 they give 59049, 1000 and 30. The exponential started nearly level with the cubic and finished nearly sixty times ahead, which is the divergence this section is about.
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 465-468
Trap
\[ f(x)=x^2 \text{ and } g(x)=2^x \text{ are essentially the same} \]
Note that both involve a 2 and an x
Why: The same two symbols appear in each, so the functions are treated as similar.
The two are used interchangeably in reasoning about growth.
They are completely different families. One squares its input; the other doubles for each unit of input.
At x equal to 10 they give 100 and 1024. At x equal to 20 they give 400 and over a million. The gap grows without bound.
Ask which of the two is fixed. A fixed exponent means a power function, and a fixed base means an exponential — and only the second one's growth compounds.
Faded example
Outputs 4, 12, 36 and 108 for inputs 0, 1, 2 and 3.
Fill in the blanks
\text3 = 4, \quad \text___ = ___, \quad y = ___\cdot___^x
Why: Each output is 3 times the previous, so the base is 3, and the output at zero is 4, so that is the initial value. Both constants come straight from the table: the base from the ratio and the initial value from the output at input zero.
Discrimination
The test distinguishes the two families.
Sort into buckets
Sort each described table.
Socratic
The definition insists the base is positive and not 1.
Discussion prompt
What goes wrong with a negative base, and what goes wrong with a base of 1?
Hint: Try raising negative 2 to the power one half, and try 1 to any power.
Answer:
A negative base breaks at fractional exponents. Negative 2 to the power one half is the square root of a negative, which is not real — so the function would be undefined at infinitely many inputs and its graph would be a scatter of disconnected points.
A base of 1 makes the function constant, since 1 to any power is 1. That is a perfectly good function but it is a horizontal line, with none of the growth behaviour the family is defined to capture.
So both exclusions remove degenerate cases rather than interesting ones. What is left — positive bases other than 1 — gives exactly the growing and decaying curves the chapter is about.
Section
Section 2
Concept
A base greater than 1 makes the function grow; a base between 0 and 1 makes it decay. Either way the graph approaches the horizontal axis on one side without reaching it.
The asymptote is worth dwelling on. A positive number raised to any power is positive, so the output can become arbitrarily small without ever being zero or negative. That is why an exponential decay model never quite reaches nothing, which matches how radioactive decay and cooling actually behave.
Figure (svg): Two exponential graphs on the same axes: one with base greater than one rising steeply, and one with base between zero and one falling towards the horizontal axis
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 468-473
Picture it
Two bases, one above 1 and one below, on the same axes.
Figure (svg): Two exponential graphs on the same axes: one with base greater than one rising steeply, and one with base between zero and one falling towards the horizontal axis
The two curves are reflections of each other in the vertical axis, because one half to the power x is the same as 2 to the power negative x. Decay is growth run backwards.
Worked example
Read the base, then the initial value.
\[ \text{Describe } f(x)=200(0.85)^x. \]
Read the base
Why: It is 0.85, between 0 and 1.
Read the initial value
Why: The coefficient, at input zero.
\[ \text{starts at } 200 \]
Interpret the base as a rate
Why: Each step keeps 85 percent.
\[ \text{loses } 15 \%\text{ per step} \]
State the asymptote
Why: The horizontal axis.
\[ \text{approaches } 0 \]
Figure (svg): Two exponential graphs on the same axes: one with base greater than one rising steeply, and one with base between zero and one falling towards the horizontal axis
\[ \text{decay, } 15\% \text{ lost per step, asymptote } y=0 \]
Verify: check the percentage reading
Why: After one step the value is 200 times 0.85, which is 170 — a loss of 30, which is 15 percent of 200. The base is what REMAINS each step, so the loss is 1 minus the base. Reading 0.85 as a 85 percent loss rather than a 15 percent one is the standard error here.
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 469-471
Sorting
Compare the base against 1.
Sort into buckets
Sort each function.
Worked example
The base is 1 plus the growth rate, or 1 minus the decay rate.
\[ \text{A population of } 5000 \text{ grows } 3\% \text{ a year. Model it.} \]
Identify the initial value
Why: The population now.
\[ a = 5000 \]
Convert the rate to a base
Why: One plus three hundredths.
\[ b = 1.03 \]
Write the model
Why: Initial value times base to the t.
\[ P(t) = 5000(1.03) ^{t} \]
Check after one year
Why: Three percent more.
\[ 5150 \]
Figure (svg): The solution to Worked example build a model from a rate shown as a ladder of expressions, one row per legal move
\[ P(t)=5000(1.03)^t \]
Verify: confirm the base's meaning
Why: After one year the population is 5150, which is 5000 plus 3 percent of 5000. The base 1.03 keeps the original whole and adds the growth, which is why it is 1 plus the rate rather than the rate itself. Using 0.03 as the base would model a 97 percent annual collapse.
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 471-473
Error analysis
A student models 5 percent annual growth.
Annotate
On: \( P(t)=1000(0.05)^t \)
The base is what the quantity is multiplied by, so it must include the original whole. Growth gives 1 plus the rate, and decay gives 1 minus it.
Faded example
A quantity decays by 12 percent each period.
Fill in the blanks
b = 1 - 0.12 = 0.88
Why: A 12 percent loss leaves 88 percent, so the base is 0.88. The base is always what remains, which for decay is 1 minus the rate and for growth is 1 plus it. Checking one step confirms it: 100 becomes 88, a loss of 12.
Prediction
An exponential function has a positive initial value.
Predict first
What is its range?
Correct: All positive numbers, never including zero.
Why: A positive base raised to any power is positive, and multiplying by a positive initial value keeps it positive. The output can become arbitrarily small but never reaches zero, which is why the horizontal axis is an asymptote rather than part of the graph.
Counterexample
A classmate claims a decaying exponential eventually reaches zero.
Discussion prompt
Explain why it does not, and say what it does instead.
Hint: What is a positive number raised to a large power?
Answer:
A positive base raised to any power is positive. One half to the hundredth power is unimaginably small, but it is not zero — and multiplying it by a positive initial value keeps it positive.
So the graph approaches the horizontal axis without ever meeting it. Formally the axis is an asymptote, exactly as in §3.7's rational functions.
This matches the physics it models. Radioactive material never fully decays and a cooling object never quite reaches room temperature, though both get close enough that the difference stops mattering. The model's refusal to reach zero is a feature rather than an artefact, and half-life is defined precisely because 'when will it be gone' has no answer.
Section
Section 3
Concept
Two data points determine an exponential model. Dividing one output by the other eliminates the coefficient and leaves an equation for the base alone.
The division step is the whole technique, and it is the same idea as the scaling shortcut in §3.9: taking a ratio makes the unknown coefficient cancel, leaving fewer unknowns than you started with. Two unknowns and two points reduce to one unknown and one equation.
Figure (svg): Two exponential graphs on the same axes: one with base greater than one rising steeply, and one with base between zero and one falling towards the horizontal axis
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 473-477
Picture it
The base is what distinguishes them, and it is what the ratio of two outputs recovers.
Figure (svg): Two exponential graphs on the same axes: one with base greater than one rising steeply, and one with base between zero and one falling towards the horizontal axis
Any two points on one of these curves determine it completely. The ratio finds the base and either point then fixes the height.
Worked example
Divide, take a root, then substitute.
\[ \text{An exponential passes through } (1,12) \text{ and } (3,108). \text{ Find it.} \]
Divide the outputs
Why: The coefficient cancels.
\[ \frac{108}{12} = 9 \]
Relate to the base
Why: The inputs differ by 2.
\[ b ^{2} = 9 \]
Solve for the base
Why: Positive root, since the base is positive.
\[ b = 3 \]
Substitute a point
Why: Using the first one.
\[ 12 = a(3),\text{ so } a = 4 \]
Figure (svg): The solution to Worked example from two points shown as a ladder of expressions, one row per legal move
\[ y=4\cdot 3^x \]
Verify: check the other point
Why: At x equal to 3 the model gives 4 times 27, which is 108 — the second point, which was not used in finding the coefficient. Checking against the unused point is what confirms both constants rather than just one.
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 474-476
Faded example
An exponential passes through inputs 1 and 4 with outputs 6 and 162.
Fill in the blanks
b^3} = \frac3___ = 27 \;\Longrightarrow\; b = ___
Why: The inputs differ by 3, so the ratio of the outputs is the base cubed. That ratio is 27, whose cube root is 3. Substituting back gives the coefficient: 6 equals a times 3, so a is 2, and the model is 2 times 3 to the x.
Worked example
The exponent is the difference of the inputs, whatever it is.
\[ \text{An exponential passes through } (2,50) \text{ and } (7,1600). \text{ Find the base.} \]
Divide the outputs
Why: The coefficient cancels.
\[ \frac{1600}{50} = 32 \]
Find the difference of inputs
Why: Seven minus two.
\[ 5 \]
Relate to the base
Why: The base to that power.
\[ b ^{5} = 32 \]
Take the fifth root
Why: Thirty two is two to the fifth.
\[ b = 2 \]
Figure (svg): The solution to Worked example non-consecutive inputs shown as a ladder of expressions, one row per legal move
\[ b = 2 \]
Verify: find the coefficient and check
Why: Substituting the first point: 50 equals a times 4, so a is 12.5. At x equal to 7 the model gives 12.5 times 128, which is 1600 — matching. The exponent in the base equation is always the difference of the inputs, so non-consecutive points work exactly as well as consecutive ones.
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 476-477
Trap
\[ b^5=32 \;\Longrightarrow\; b = \frac{32}{5} = 6.4 \]
Undo the exponent by dividing
Why: The 5 is applied to the base, so it is divided out.
The base is reported as 6.4.
An exponent is undone by a root, not a division. The fifth root of 32 is 2, not 6.4.
Checking: 2 to the fifth is 32, while 6.4 to the fifth is over ten thousand. The two are not close.
Match the operation to its inverse. Multiplication is undone by division and exponentiation by a root, and mixing the two is the commonest error in setting up exponential models.
Prediction
Two points on an exponential are used, and their outputs are divided.
Predict first
What happens to the coefficient?
Correct: It cancels, leaving an equation in the base alone.
Why: Both outputs contain the coefficient as a factor, so dividing removes it entirely. This is the same cancellation trick used for the scaling questions in §3.9, and it is what reduces two unknowns to one — which is why two points are exactly enough.
Ranking
For building an exponential model from two points.
Put in order
Why: Dividing first removes the coefficient so the base can be found alone. Once the base is known, either point determines the coefficient. Checking with the point not used in the last step is what verifies both constants rather than only confirming the arithmetic of one substitution.
Explain it to yourself
An exponential model has two unknown constants.
Discussion prompt
Explain why exactly two data points determine it, and what would happen with only one.
Hint: How many unknowns and how many equations?
Answer:
The model has two unknowns: the coefficient and the base. Two data points give two equations, which is exactly enough to determine two unknowns.
With one point there are infinitely many exponentials through it — pick any base at all and the coefficient adjusts to fit. So one point constrains the model without determining it.
The situation matches §2.1's lines, which also needed two facts. The difference is only in how the two are used: for a line you subtract to find the slope, and here you divide to find the base, because the model is multiplicative rather than additive.
Section
Section 4
Concept
Any exponential with base above 1 eventually exceeds any polynomial, no matter how large the polynomial's degree or coefficients. A higher degree only delays the crossover.
The reason is that each unit step multiplies an exponential by a fixed factor, while it multiplies a polynomial by a factor that approaches 1. So the exponential keeps gaining ground proportionally and the polynomial's growth rate flattens relative to it.
Figure (svg): An exponential and a high-degree polynomial plotted together, with the polynomial ahead near the origin and the exponential overtaking it decisively further out
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 477-479
Picture it
A cubic leads for a while and then loses decisively.
Figure (svg): An exponential and a high-degree polynomial plotted together, with the polynomial ahead near the origin and the exponential overtaking it decisively further out
Before the crossover the cubic looks like the faster grower, and a plot restricted to that region would be badly misleading. This is why the eventual behaviour has to be reasoned about rather than observed.
Worked example
Compare the two at a series of inputs.
\[ \text{Where does } 2^x \text{ overtake } x^3? \]
Compare at a small input
Why: At x equal to 5.
\[ 32\text{ against } 125 \]
Compare further out
Why: At x equal to 9.
\[ 512\text{ against } 729 \]
Compare again
Why: At x equal to 10.
\[ 1024\text{ against } 1000 \]
Locate the crossover
Why: Between 9 and 10.
\[ \text{just below } 10 \]
Figure (svg): An exponential and a high-degree polynomial plotted together, with the polynomial ahead near the origin and the exponential overtaking it decisively further out
\[ \text{crossover near } x\approx 9.94 \]
Verify: check it never reverses
Why: At x equal to 20 the exponential is over a million and the cubic is 8000. At x equal to 30 the exponential exceeds a billion and the cubic is 27000. The gap widens permanently, which is what 'eventually and decisively' means — the crossover happens once and is never undone.
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 478-478
Prediction
Compare an exponential with base 1.01 against x to the hundredth power.
Predict first
Which is eventually larger?
Correct: The exponential, though very much later.
Why: Every exponential with base above 1 eventually beats every polynomial. A base barely above 1 and a degree of 100 push the crossover extraordinarily far out, but it happens all the same, and afterwards the exponential pulls away without bound. Degree and coefficients affect only when, never whether.
Worked example
The conclusion is unchanged; only the crossover moves.
\[ \text{Does } 2^x \text{ also overtake } x^{10}? \]
Compare at a moderate input
Why: At x equal to 20.
Note who leads
Why: The polynomial, by a huge margin.
Compare much further out
Why: At x equal to 60.
Conclude
Why: The exponential passes it near there.
Figure (svg): The solution to Worked example a higher degree only delays it shown as a ladder of expressions, one row per legal move
\[ \text{Yes: the crossover is near } x\approx 59. \]
Verify: notice what changed and what did not
Why: Raising the degree from 3 to 10 moved the crossover from about 10 to about 60 — a substantial delay. But it still happened, and after it the exponential pulls away just as decisively. No degree, however large, prevents it; the crossover simply moves further out.
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 478-479
Error analysis
A student compares two functions on a plot from 0 to 5.
Annotate
On: \( \text{on } [0,5]: \; x^3 \text{ is above } 2^x \;\Longrightarrow\; x^3 \text{ grows faster} \)
Extending the window is the fix, and knowing roughly where to extend it to requires the reasoning rather than the picture. This is a case where the graph actively misleads unless the range is chosen with the answer already in mind.
Sorting
Compare each pair at very large inputs.
Sort into buckets
Sort each pair by which function wins eventually.
Faded example
Compare 2 to the power 20 against 20 cubed.
Fill in the blanks
2^8000 \approx 1\,048\,576, \qquad 20^3 = exponential \;\Longrightarrow\; \text______ \text___
Why: Twenty cubed is 8000 while 2 to the twentieth is over a million — the exponential leads by a factor of over 130. At x equal to 5 the cubic led comfortably, so the reversal has already happened and the gap is widening rapidly.
Socratic
The result holds for every base above 1 and every degree.
Discussion prompt
Explain why, in terms of what each function does at each step.
Hint: By what factor does each grow when the input increases by one?
Answer:
An exponential is multiplied by its fixed base at every unit step, forever. Going from x to x plus 1 multiplies it by b, whatever x is.
A polynomial's growth factor per step approaches 1. Going from 100 to 101 multiplies the cube by about 1.03; going from 1000 to 1001 multiplies it by about 1.003. The larger the input, the less it gains proportionally.
So the exponential keeps its full advantage at every step while the polynomial's advantage evaporates. Eventually the exponential is multiplying by b while the polynomial multiplies by nearly nothing, and from there the outcome is inevitable.
Section
Section 5
Concept
Interest compounded n times a year multiplies by one plus the rate over n, that many times per year. Increasing n raises the total, but towards a limit rather than without bound.
\[ A=P\Bigl(1+\frac{r}{n}\Bigr)^{nt}, \qquad A=Pe^{rt} \]
The diminishing returns are worth seeing. At 100 percent annual interest, compounding once doubles your money, twice gives 2.25, monthly gives about 2.61, and continuously gives e, about 2.718. Going from yearly to monthly gains a lot; going from monthly to continuously gains very little.
Figure (svg): A card showing the compound interest formula with each symbol labelled, and beneath it the continuous compounding formula that arises as the compounding frequency grows
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 479-481
Picture it
Four symbols, and what happens when one of them grows without bound.
Figure (svg): A card showing the compound interest formula with each symbol labelled, and beneath it the continuous compounding formula that arises as the compounding frequency grows
The number e is not a convention. It is the value the compounding formula approaches, which is why it turns up whenever growth is continuous rather than stepped.
Worked example
Substitute carefully; the exponent is n times t.
\[ \text{Invest } 2000 \text{ at } 6\% \text{ compounded quarterly for } 5 \text{ years.} \]
Identify the four values
Why: Principal, rate, frequency, time.
\[ P = 2000, r = 0.06, n = 4, t = 5 \]
Compute the per-period rate
Why: Rate divided by frequency.
\[ \frac{0.06}{4} = 0.015 \]
Compute the number of periods
Why: Frequency times years.
\[ 4 \times 5 = 20 \]
Evaluate
Why: Two thousand times 1.015 to the twentieth.
\[ \text{about } 2694 \]
Figure (svg): A card showing the compound interest formula with each symbol labelled, and beneath it the continuous compounding formula that arises as the compounding frequency grows
\[ A=2000(1.015)^{20}\approx 2694 \]
Verify: compare with simple interest
Why: Simple interest would give 6 percent of 2000 for 5 years, which is 600, for a total of 2600. Compounding gained an extra 94 by earning interest on interest. That extra is the whole point of compounding, and it grows dramatically over longer periods.
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 480-480
Faded example
Five thousand at 4 percent compounded monthly for 3 years.
Fill in the blanks
A = 5000\Bigl(1+\frac1236}\Bigr)^___\cdot 3}, \quad \text___ = ___
Why: Monthly means twelve compoundings a year, so the rate is divided by 12 and the exponent is 12 times 3, which is 36. The two 12s must match — dividing the rate by 12 while using an exponent of 3 would model three years of monthly-rate growth compounded only annually.
Worked example
The simpler formula, and the comparison that shows how little is gained.
\[ \text{The same } 2000 \text{ at } 6\% \text{ compounded continuously for } 5 \text{ years.} \]
Use the continuous formula
Why: Principal times e to the rate times time.
\[ A = 2000 e ^{0.3} \]
Compute the exponent
Why: Rate times years.
\[ 0.06 \times 5 = 0.3 \]
Evaluate the exponential
Why: e to the power 0.3.
\[ \text{about } 1.3499 \]
Multiply
Why: By the principal.
\[ \text{about } 2700 \]
Figure (svg): The solution to Worked example continuous compounding shown as a ladder of expressions, one row per legal move
\[ A=2000e^{0.3}\approx 2700 \]
Verify: compare with quarterly
Why: Quarterly gave 2694 and continuous gives 2700 — a difference of 6 pounds over five years, on 2000. Compounding continuously rather than quarterly is worth almost nothing, which is the diminishing returns made concrete. Most of the benefit of compounding is captured by compounding a few times a year.
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 480-481
Trap
\[ A=2000(1+0.06)^{20} \]
Substitute the rate and the number of periods
Why: The annual rate is used with the total number of compounding periods.
The result is about 6414, more than tripling the investment.
The rate must be divided by the frequency. Each quarter earns a quarter of the annual rate, which is 1.5 percent, not the full 6 percent.
The correct base is 1.015, giving about 2694 rather than 6414. The error applies the annual rate four times a year.
The rate and the exponent must be consistent. If the exponent counts quarters, the rate must be per quarter — divide one and multiply the other by the same n.
Prediction
The compounding frequency is increased from monthly to daily.
Predict first
What happens to the final amount?
Correct: It rises slightly, with diminishing returns.
Why: More frequent compounding always gives more, but the gains shrink rapidly. Going from annual to monthly is worth a noticeable amount; from monthly to daily is worth very little, and from daily to continuous almost nothing. The whole sequence converges to the continuous value, which caps how much frequency can ever be worth.
Sorting
Discrete compounding uses one formula and continuous another.
Sort into buckets
Sort each situation.
Real world
The number e was not chosen; it was discovered.
Discussion prompt
Why does e appear in the continuous formula, rather than some rounder number?
Hint: What is the value of one plus one over n, all to the n, as n grows?
Answer:
Take 100 percent annual interest and push the compounding frequency up. The multiplier is one plus one over n, all raised to the n, and computing it for larger and larger n gives 2, then 2.25, then about 2.61, then about 2.7169.
Those values converge, and their limit is about 2.71828. That number is e, and it is defined by this limit rather than being selected for convenience.
So e is as natural as pi, and for a similar reason: it is what a particular process converges to. It turns up everywhere continuous growth appears — population, radioactivity, cooling, charge on a capacitor — because all of them are the same limiting process in different clothing.
Comparison
Fill the blanks from memory. Two families, and confusing them is the source of most errors in this chapter.
Comparison matrix
| linear | exponential | |
|---|---|---|
| each step | adds a fixed amount | multiplies by a fixed factor |
| what is constant | the difference between outputs | the ratio between outputs |
| where the variable is | the base, to the first power | the exponent |
| eventual growth | beaten by any higher power | beats every power |
| value at input zero | the intercept | the coefficient a |
The second row is the practical test on a table. Compute the differences and the ratios, and whichever is constant names the family.
Pattern
Five steps, depending on what you were given.
Step 2 is where the errors are. A rate must be converted to a base by adding to or subtracting from 1, and an exponent must be undone by a root rather than a division.
OpenStax Algebra and Trigonometry 2e, §6.1 Exponential Functions §6.1
Check
The base includes the original whole.
Check your understanding
A quantity grows by 7 percent each period. What is the base of its exponential model?
Answer: A
Why: The base is what the quantity is multiplied by each period, which is the original whole plus the growth: 1 plus 0.07. Checking one step confirms it: 100 becomes 107.
Check
Ratio, not difference.
Check your understanding
A table has outputs 8, 20, 50 and 125. What kind of function is it?
Answer: A
Why: The differences are 12, 30 and 75, which are not constant, but the ratios are all 2.5. A constant ratio means exponential, and the ratio is the base.
Check
Eventually, and always.
Check your understanding
Which is larger for sufficiently large x: 1.1 to the power x, or x to the power 50?
Answer: A
Why: Every exponential with base above 1 eventually exceeds every polynomial. A base as small as 1.1 and a degree as large as 50 push the crossover extremely far out, but it happens, and afterwards the exponential pulls away without bound.
Real world
Exponential growth is notoriously hard to judge by intuition, which has practical consequences.
Discussion prompt
Why do people consistently underestimate exponential growth, and where does that matter?
Hint: What does the curve look like during its early stages?
Answer:
Early on an exponential looks almost flat, and its increments are small. Intuition extrapolates linearly from what it has seen, which produces a forecast far below the truth.
The doubling in the warm-up is the standard demonstration: after two weeks it is worth 82 pounds and looks like a bad deal, and by day 30 it is worth millions. Nothing about the early behaviour signals what is coming.
It matters wherever a quantity compounds — epidemics, compound debt, viral spread, and the growth of savings. In each case the useful discipline is to reason about the doubling time rather than about the current rate of increase, since the doubling time is constant while the increments are not.
Commit first
State your confidence along with your answer.
Predict first
An exponential decay model's outputs approach what value as the input grows?
Correct: Zero, approached but never reached.
Why: A positive base raised to any power is positive, so the output stays positive however large the input becomes. The horizontal axis is an asymptote rather than part of the graph, which is why questions about decay ask for half-lives rather than for the time until nothing is left.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate the difference between a power function and an exponential function, and why it matters so much.
Hint: Ask which of the base and the exponent is fixed in each.
Answer:
In a power function the exponent is fixed and the base varies: x squared, x cubed. In an exponential the base is fixed and the exponent varies: 2 to the x, 3 to the x.
It matters because their growth is of a different order. Each unit step multiplies an exponential by its base, forever; a polynomial's growth factor per step fades towards 1. So the exponential eventually wins against every power, regardless of degree.
A good explanation includes the practical warning: the crossover can be far out, so a plot over a small window can show the polynomial ahead and mislead completely. Reasoning about the step behaviour is more reliable than looking.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second causes wrong answers that are wrong by enormous factors rather than by a little, since using the rate as the base models near-total collapse instead of modest growth. The fourth is the one whose ideas recur most, since e appears throughout the rest of the chapter.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Make two tables side by side from the same starting value: one adding a fixed amount each step and one multiplying by a fixed factor. Run both to six rows and note how far apart they finish. Then sketch one growth and one decay exponential on the same axes, marking the shared point at input zero and the horizontal asymptote, and write beside each what its base tells you about the percentage change per step.
If your two tables started close and finished far apart, you have drawn the fact the rest of the chapter depends on.
Recap
Five things, and the first is the distinction the whole chapter rests on.
| if you remember one thing | it should be this |
|---|---|
| about the definition | exponentials multiply where linear functions add |
| about the base | it is 1 plus the growth rate, or 1 minus the decay rate |
| about growth order | an exponential beats every power eventually, whatever its degree |
| about e | it is the limit of ever-more-frequent compounding, not a convention |
Section 4.2 graphs these functions and applies Chapter 1's transformations to them, where the horizontal asymptote is the feature that moves and the one that is most often forgotten.
OpenStax, Precalculus, §4.1 Exponential Functions §4.1, pp. 460-481 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Precalculus — $55/session, free consultation.