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3.2 — Half Interested
In Section 3.1, you saw that exponential relationships grow by multiplying by the same factor for equal-sized increases in the input. This lesson uses that idea in a bank-account situation. A constant yearly percentage means a constant yearly multiplier—not a constant number of dollars in interest.
1. In class: A Solidify Understanding Task
Compound interest over whole years
Suppose an account contains $382.88 now and earns 5% interest each year, compounded annually. Five percent of the current balance is added at each year’s end. Every year the new balance is 105% of the previous balance.
Multiply by 1.05 for each complete year. Retain the unrounded value during calculation and round to cents when displaying money.
\[A(0)=382.88,\quad A(1)=382.88(1.05)=402.024\approx\$402.02.\]After five years, repeat the multiplication five times, or use an exponent:
\[A(5)=382.88(1.05)^5\approx\$488.66.\]| Years from now | Exact calculation before rounding | Balance rounded to cents |
|---|---|---|
| 0 | \(382.88(1.05)^{0}\) | $382.88 |
| 1 | \(382.88(1.05)^{1}\) | $402.02 |
| 2 | \(382.88(1.05)^{2}\) | $422.13 |
| 3 | \(382.88(1.05)^{3}\) | $443.23 |
| 4 | \(382.88(1.05)^{4}\) | $465.39 |
| 5 | \(382.88(1.05)^{5}\) | $488.66 |
Each multiplication contributes one year of growth. Therefore, the compact rule is
This is an exponential function. Its input is elapsed time, its initial value is 382.88, and its constant annual growth factor is 1.05. The exponential equation gives the same balances as repeated year-by-year multiplication because the power counts the number of factors.
Work backward to past balances
To reverse a year of 5% growth, divide by 1.05. You should not subtract 5% of the current balance: that is not the reverse operation of multiplying by 1.05.
One year ago, the balance was \(382.88\div1.05\). Five years ago:
\[A(-5)=\frac{382.88}{(1.05)^5}=382.88(1.05)^{-5}\approx\$300.00.\]Check: growing that starting balance forward for five years returns approximately $382.88.
What does half a year mean?
The same exponential model can assign a balance between whole years. We want a half-year multiplier that produces the full-year multiplier 1.05 after it is used twice. Call that multiplier \(q\).
We use the positive square root because this is a positive growth multiplier. Thus the exponent \( frac12\) means the square-root multiplier:
Simple half-of-the-annual-interest estimate: Half of \(0.05(382.88)\) is \(\$9.572\), giving \(382.88+9.572=\$392.452\approx\$392.45\).
Consistent exponential half-year factor:
\[A(1/2)=382.88(1.05)^{1/2}=382.88\sqrt{1.05}\approx\$392.34.\]The answers differ because dividing the dollar increase in half is additive, whereas an exponential model divides the multiplication into equal factors.
The exponential half-year rule is a mathematical model consistent with continuous interpolation of the exponential function. A real bank’s payment for part of a year depends on its interest and compounding contract; the packet is comparing mathematical models rather than stating a universal banking policy.
The difference grows clearer over longer time spans. Half of the dollar growth over four years is not the correct growth over two years. Because \(1.05^4\) is the four-year factor, the two-year factor must be \(1.05^2\), whose square is \(1.05^4\).
Arithmetic midpoint: \((382.88+465.39)/2\approx\$424.14\), not the two-year balance.
Quarters of a year and three-quarters of a year
For equal quarter-year steps, we need a multiplier \(r\) that becomes 1.05 after four identical multiplications. Therefore \(r^4=1.05\) and \(r=\sqrt[4]{1.05}=1.05^{1/4}\). Three quarter-year steps multiply by this factor three times.
Calculate with full precision, rounding only the final amount:
\[A(3/4)=382.88(1.05)^{3/4}\approx \$397.15.\]Interest earned is approximately \($397.15-382.88=\$14.27\).
Here \(1.05^{3/4}\) may be understood as taking a fourth root to obtain a quarter-year factor, then multiplying by that factor three times. This explains the fractional exponent; detailed rules for operating with rational exponents will be revisited in later sections.
2. Ready: simplify square, cube, fourth, and fifth roots (problems 1–9)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
A radical is a root. The small index tells you the number of equal factors: \(\sqrt[3]{8}=2\) because \(2^3=8\), while \(\sqrt[4]{81}=3\) because \(3^4=81\). Without a written index, \(\sqrt{\phantom x}\) means a square root, and \(\sqrt{25}=5\), the nonnegative principal square root.
To simplify a radical, look for the largest convenient perfect power whose exponent matches its root index. Factor it out of the number under the radical (the radicand).
Choose a perfect square factor, not simply any factor.
Here 8 is a perfect cube: \(2^3=8\).
Because \(2^5=32\), exactly one 2 emerges from the fifth root.
For the worksheet, problems 1, 2, 4, 6, and 8 use square roots; 3 and 9 use cube roots; 5 uses a fourth root; and 7 uses a fifth root. Not every radicand contains a perfect-power factor larger than 1.
Check your work — Ready 1–9 answer key
- \(2\sqrt{10}\).
- \(5\sqrt2\).
- \(2\sqrt[3]2\).
- \(6\sqrt2\).
- \(3\).
- \(4\sqrt2\).
- \(2\sqrt[5]5\).
- \(3\sqrt5\).
- \(3\sqrt[3]2\).
3. Set: arithmetic means, geometric means, and interval factors (problems 10–18)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
An arithmetic mean lies halfway between two numbers on an equally spaced additive sequence. Insert several arithmetic means by finding the common difference:
Insert four equally spaced values between 7 and 32. There are five intervals, not four:
\[d=\frac{32-7}{5}=5.\]The full list is \(7,12,17,22,27,32\). The four inserted means are 12, 17, 22, and 27.
A geometric mean fills a position in a multiplicative sequence. When the given endpoints are positive, the common positive ratio \(r\) must satisfy \(r^k= ext{last}/ ext{first}\), where \(k\) is the number of intervals.
Insert two geometric means between 5 and 135. Three equal multiplications lead from 5 to 135:
\[5r^3=135,\qquad r^3=27,\qquad r=3.\]The sequence is \(5,15,45,135\); the means are 15 and 45.
In worksheet problems 16–18, the input values are \(0, frac12,1, frac32,2\). The whole-step factor \(F_w\) advances the input by 1; the column factor \(F_c\) advances it by \( frac12\). Two columns equal one whole step, so \(F_c^2=F_w\).
Suppose \(y(0)=6\) and \(y(1)=24\). Then
\[F_w=\frac{24}{6}=4,\quad F_c=\sqrt4=2.\]Multiply by 2 each column:
| \(x\) | 0 | \( frac12\) | 1 | \( frac32\) | 2 |
|---|---|---|---|---|---|
| \(y\) | 6 | 12 | 24 | 48 | 96 |
You can verify any filled table by skipping every other column: the result must multiply by \(F_w\).
Check your work — Set 10–18 answer key
- 8; difference \(\frac{11-5}{2}=3\).
- a. 11; b. 4; c. −3. Difference \(\frac{-10-18}{4}=-7\).
- a. 9; b. 6; c. 3; d. 0; e. −3. Difference \(\frac{-6-12}{6}=-3\).
- 6, using the positive geometric mean: \(r=\sqrt{\frac{12}{3}}=2\). (A geometric sequence with ratio −2 instead has middle term −6.)
- a. 35; b. 175; ratio \(\sqrt[3]{\frac{875}{7}}=5\).
- a. 12; b. 36; c. 108; d. 324; ratio \(\sqrt[5]{\frac{972}{4}}=3\).
- a. 8; b. 32; c. 64; d. \(F_w=4\); e. \(F_c=2\).
- a. \(4\sqrt2\); b. \(8\sqrt2\); c. 16; d. \(F_w=2\); e. \(F_c=\sqrt2\).
- a. \(5\sqrt3\); b. \(15\sqrt3\); c. 45; d. \(F_w=3\); e. \(F_c=\sqrt3\).
4. Go: evaluating functions and finding inputs (problems 19–24)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
A function rule turns an input into an output. To evaluate \(f(c)\), substitute \(c\) into every occurrence of the input variable. To find the input for a given output, set the formula equal to that output and solve. This homework includes linear, quadratic, and exponential functions, including fractional and negative inputs.
Let \(f(x)=4x+3\). Find \(f(-2)\), find the input where \(f(x)=27\), and find \(f(\tfrac12)\).
\[f(-2)=4(-2)+3=-5.\] \[4x+3=27\Rightarrow4x=24\Rightarrow x=6.\] \[f(1/2)=4(1/2)+3=5.\]Substitute using parentheses to preserve negative signs.
Let \(h(x)=x^2-5x-14\). Evaluate \(h(-3)\), solve \(h(x)=0\), and evaluate \(h(\tfrac12)\).
\[h(-3)=9+15-14=10.\] \[x^2-5x-14=(x-7)(x+2)=0\Rightarrow x=7\text{ or }-2.\] \[h(1/2)=\frac14-\frac52-14=-\frac{65}{4}.\]Unlike a linear equation, the quadratic equation may have two inputs with the same output.
Let \(g(t)=80(1.25)^t\). Find \(g(-2)\), \(g(\tfrac12)\), and the input for which \(g(t)=125\).
\[g(-2)=\frac{80}{(1.25)^2}=51.2.\] \[g(1/2)=80\sqrt{1.25}\approx89.44.\] \[80(1.25)^t=125\Rightarrow (1.25)^t=\frac{125}{80}=1.5625=(1.25)^2\Rightarrow t=2.\]For negative exponents, take the reciprocal; for a half exponent, take the positive square root. This equation was chosen so its input can be recognized without logarithms (a later technique).
Let \(m(x)=3(4^x)\). Evaluate \(m(-1)\), solve \(m(x)=12\), and evaluate \(m(\tfrac12)\).
\[m(-1)=3/4,\qquad 3(4^x)=12\Rightarrow4^x=4\Rightarrow x=1,\qquad m(1/2)=3\sqrt4=6.\]A factor outside the power is not part of the exponent: \(3(4^x)\neq (12)^x\).
For packet problem 21(b), \(210(1.08)^t=420\) leads to \((1.08)^t=2\). The exact input need not be a simple fraction or integer. Without logarithms, estimate using a table or graph of powers of 1.08 until the output is close to 420; a calculator can refine the estimate. Do not confuse an approximate numerical input with an exact algebraic solution.
The story says the bank will pay a “half-year portion” of interest but does not define the bank’s actual partial-year interest convention. Carlos’s half-of-annual-interest amount and Clarita’s exponential half-year interpolation represent different mathematical models. The latter preserves a constant multiplicative growth rate across equal intervals; it does not by itself establish what a real bank is legally required to pay.
Section 3.2, Go problem 21(b): The equation \(210(1.08)^t=420\) requires a non-obvious time value. As the lesson has not yet introduced logarithms, a numerical estimate from tables or technology is reasonable unless the instructor supplies another method.
Check your work — Go 19–24 answer key
- a. −13; b. \(x=14\); c. −6.
- a. \(\frac2{81}\); b. \(x=4\); c. \(2\sqrt3\).
- a. \(210(1.08)^{12}\approx528.82\); b. \(t=\frac{\ln2}{\ln1.08}\approx9.01\); c. \(210\sqrt{1.08}\approx218.24\).
- a. 14; b. \(x=-3,2\); c. \(-\frac{21}{4}\).
- a. 44; b. \(x=\frac95\); c. \(\frac{13}{2}\).
- a. \(\frac1{625}\); b. \(x=0\); c. 5.
5. Additional learning resources
These independently published references are optional. They provide further explanation and practice; some continue into topics beyond today’s work.
| Resource | What it helps with | Scope / what to skip |
|---|---|---|
| Paul’s Online Math Notes — Radicals | Root indices, simplifying radicals, and worked examples supporting Ready 1–9. | Some extra material: use numerical radicals and the early simplification examples; skip complicated variables and radical equations. |
| Paul’s Online Math Notes — Rational Exponents | The meaning of fractional and negative powers for the bank-account task. | Some extra material: focus on definitions and numerical examples; defer extended exponent manipulations to Sections 3.3–3.4. |
| Khan Academy — Exponential functions | Optional interactive study of exponential growth and interpretation. | Significantly broader: use only the basic exponential-growth material; not all topics in the unit belong to 3.2. |
6. Teacher’s review (teachers and parents)
Main task, problems 7–13: The statement about a bank paying a “half-year portion” of interest lacks a specified bank convention. The subsequent preference for \(A(t)=382.88(1.05)^t\) is mathematically appropriate for an exponential interpolation model, but is not automatically a statement about contractual bank interest. Suggested revision: “Assume the account follows the exponential model \(A(t)=382.88(1.05)^t\) for every real number of years \(t\). Compare this with paying half the nominal annual interest.”
Ready introductory explanation: The packet suggests that square roots of non-square numbers are necessarily irrational; this is acceptable for positive integers that are not perfect squares, but false for arbitrary non-square rational numbers (for example \(\sqrt{1/4}=1/2\)). State the integer condition explicitly.
Go 21(b): Finding an input from \(210(1.08)^t=420\) is not readily accomplished exactly with the algebra introduced at this point. Clarify that a numerical approximation or graph is acceptable, unless logarithms have been taught elsewhere.
7. Coverage of the Section 3.2 packet
| Packet component | Problems | Where covered |
|---|---|---|
| Main task | 1–4: annual growth / exponent model | In class |
| Main task | 5–6: backward growth / negative powers | In class |
| Main task | 7–12: half-year interpolation and comparison | In class, graph, clarifications |
| Main task | 13: three-quarter-year interest | In class |
| Ready | 1–9: simplify roots of varying indices | Section 2 |
| Set | 10–12: arithmetic means; 13–15: geometric means; 16–18: whole-step and column growth factors | Section 3 |
| Go | 19–24: evaluating and solving for function inputs | Section 4, four original worked Go examples |
Independent supplemental material · Secondary Math II · Module 3 · Section 3.2. The original mathematics curriculum is © Mathematics Vision Project and is licensed under CC BY 4.0.
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