# Lesson 5

Changes Over Rational Intervals

## 5.1: Changes Over Intervals (5 minutes)

### Warm-up

The goal of this warm-up is for students to use the properties of exponents to reason about how the value of an exponential function increases over different input intervals. In particular, in order to fully answer the questions, students need to generalize from specific examples to any input increase of the given amount. Students will build on this type of thinking with a simple exponential function when they reason about fractional intervals for more complex functions later in the lesson.

### Student Facing

Consider the exponential function \(h(x)=4^x\). For each question, be prepared to share your reasoning with the class.

- By what factor does \(h\) increase when the exponent \(x\) increases by 1?
- By what factor does \(h\) increase when the exponent \(x\) increases by 2?
- By what factor does \(h\) increase when the exponent \(x\) increases by 0.5?

### Student Response

Student responses to this activity are available at one of our IM Certified Partners

### Activity Synthesis

Invite students to share their reasoning for the factors that align with each of the 3 intervals. If students determined the factors using only specific examples, such as calculating at the growth factor between \(x=0\) and \(x=2\) as 16 since \(g(2)=16\boldcdot g(0)\), ask them to consider how they know that the growth factor is 16 for any interval of size 2 units. Help students translate their verbal reasoning into equations, such as \(4^{x+2}=4^2 \boldcdot 4^x\) or \(\dfrac{4^{x+2}}{4^x}=\dfrac{4^2 \boldcdot 4^x}{4^x}=4^2\).

## 5.2: Machine Depreciation (20 minutes)

### Activity

In a previous lesson, students found the growth factor of an exponential function over a fractional interval. They then rewrote a given exponential function to represent how the quantity would change over a different unit of input. For example, given a quantity expressed as a function of time in weeks, they expressed it as a function of time in days.

That work continues here, but the focus shifts from finding the growth factor for a particular interval to noticing how a function changes over any equal intervals of input. For example, how does the function value change every time the input increases by 0.5—say, from 1 to 1.5, or from 0.25 to 0.75?

To find the factor by which the function \(f\) changes when the input increases by 0.5, students may:

- Use a calculator to divide the numerical values of \(f(0.5)\) and \(f(0)\) (or other equal intervals).
- Find the factor by which \(f\) changes when the input increases by 1 (say from \(f(0)\) to \(f(1)\)), and then take the square root for that factor.

Monitor for students who use each strategy to share during the whole-class discussion.

Making spreadsheet technology available gives students an opportunity to choose appropriate tools strategically (MP5).

### Launch

*Representation: Internalize Comprehension.*Activate or supply background knowledge. Prior to filling in the machine table, provide an alternate scenario with more accessible values. For example, keep the “years since purchase” column identical to that in the activity, but change the other elements. Instead of a machine, use the depreciation of a used motorcycle that originally cost \$1,000, with 0.50 as the decay factor. As in the table in the activity, fill in the outputs on the table for years 0, 1, 2, and 3. Invite students to make calculations for the half year values and to then derive a function from the table.

*Supports accessibility for: Memory; Conceptual processing*

### Student Facing

After purchase, the value of a machine depreciates exponentially. The table shows its value as a function of years since purchase. If a spreadsheet tool is available, consider using it to help you reason about the following questions.

years since purchase | value in dollars |
---|---|

0 | 16,000 |

0.5 | |

1 | 13,600 |

1.5 | |

2 | 11,560 |

3 | 9,826 |

- The value of the machine in dollars is a function \(f\) of time \(t\), the number of years since the machine was purchased. Find an equation defining \(f\) and be prepared to explain your reasoning.
- Find the value of the machine when \(t\) is 0.5 and 1.5. Record the values in the table.
- Observe the values in the table. By what factor did the value of the machine change:
- every one year, say from 1 year to 2 years, or from 0.5 years to 1.5 years?
- every half a year, say from 0 to 0.5 year, or from 1.5 years to 2 years?

- Suppose we know \(f(q)\), the value of the machine \(q\) years since purchase. Explain how we could use \(f(q)\) to find \(f(q+0.5)\), the value of the machine half a year after that point.

### Student Response

Student responses to this activity are available at one of our IM Certified Partners

### Student Facing

#### Are you ready for more?

A bank account is growing exponentially. At the beginning of 2010, the balance, in dollars, was 1,200. At the beginning of 2015, the balance was 1,350. What would the bank account balance be at the beginning of 2018? Give an approximate answer as well as an exact expression.

### Student Response

Student responses to this activity are available at one of our IM Certified Partners

### Anticipated Misconceptions

A linear model for \(f\), based on the values for 0 and 1 year, predicts a value of \$14,800 after 0.5 year, which is very close to the value found with an exponential model. If students find the value for 0.5 year and think that \(f\) may be linear, ask them to calculate what value of the machine the linear model would predict after 2 and 3 years. Then ask them if this agrees with the values provided in the table.

### Activity Synthesis

Invite previously identified students to share how they found the decay factor for the half-year intervals in the order listed in the Activity Narrative. Students may notice that the numerical values found from dividing two function values or from estimating square roots may not be identical, raising the question of whether the decay factor is really equal over equal intervals. One advantage to finding an expression for the factor of change, namely \(0.85^{(0.5)}\), is that it allows us to see that the decay factor is the same whenever the input is increased by 0.5.

If not mentioned in students’ explanation, clarify that if we reason about the decay factor by writing expressions, we can see that they are indeed equal. For example, we know that the decay factor over one year is 0.85, so the change from, say, \(f(0)\) to \(f(0.5)\) could be written as: \(\dfrac{16,\!000 \boldcdot (0.85)^{(0.5)}}{16,\!000}\), which equals \(0.85^{(0.5)}\). The change from \(f(1.5)\) to \(f(2)\) could be written as: \(\dfrac{16,\!000 \boldcdot (0.85)^{2}}{16,\!000 \boldcdot (0.85)^{(1.5)}}\), which also equals \(0.85^{(0.5)}\). Using exact expressions here is important as the quotients of the approximate values in the table are close to \(0.85^{(0.5)}\) but differ slightly because of rounding: for example \(\frac{14,751.27}{16,000}\) and \(\frac{13,600}{14,571.27}\) are very close to one another and to \((0.85)^{(0.5)}\) but all three quantities are slightly different.

Make sure students see that the process of finding the factor of decay every half a year can be generalized, that is, that the factor is the same no matter which half-year you examine: \(\dfrac {f(t + 0.5)} {f(t)} = \dfrac {16,\!000 \cdot (0.85)^{(t+0.5)}}{16,\!000 \cdot (0.85)^t}=\dfrac {16,\!000 \cdot (0.85)^t \cdot (0.85)^{(0.5)}}{16,\!000 \cdot (0.85)^t}=(0.85)^{(0.5)}\approx0.92\).

*Speaking: MLR8 Discussion Supports.*As students share how they found the decay factor for the one-year and half-year intervals, press for details by asking how they know that the decay factor is always 0.85 for each one-year interval or approximately 0.92 for each half-year interval. Show concepts multi-modally by displaying the table that shows the value of the machine, in dollars, as a function of years since purchase. Draw arrows labeled 0.85 between each one-year interval and arrows labeled 0.92 between each half-year interval to highlight the fact that the decay factor is the same whenever the input is increased by 1 or 0.5. This will help students justify why the value of an exponential function changes by a constant factor over equal intervals.

*Design Principle(s): Support sense-making; Maximize meta-awareness*

## 5.3: Fever Medicine (10 minutes)

### Activity

In the previous activity, students use the fact that an exponential function changes by the same factor over equal intervals in order to fill out a table of values, and they explain why this is always true using an equation for the function. In this activity, students continue to use the equal factors over equal intervals property of exponential functions in a less scaffolded situation. They are given the coordinates of points on a graph and need to use this information to figure out the value of the given function for a different input.

### Launch

*Writing, Listening, Conversing: MLR1 Stronger and Clearer Each Time.*Use this routine to help students improve their written responses to the first question. At the appropriate time, invite students to meet with 2–3 partners to share and receive feedback on their work. Display feedback prompts that will help students strengthen their ideas and clarify their language. For example, "Your explanation tells me . . .", "Can you say more about why you . . . ?", and "A detail (or word) you could add is _____, because . . . ." Invite students to go back and revise or refine their written responses based on the feedback from peers.

*Design Principle(s): Optimize output (for justification); Cultivate conversation*

*Representation: Internalize Comprehension.*Use color coding and annotations to highlight connections between representations in a problem. For example, invite students to label the smaller \(\frac14\) intervals on the horizontal axis, annotate the points on their graph with the information given in the text, and highlight the length of the intervals they are comparing.

*Supports accessibility for: Visual-spatial processing*

### Student Facing

The graph shows the amount of medicine in a child’s body \(h\) hours after taking the medicine. The amount of medicine decays exponentially.

- After \(\frac{1}{4}\) hour there are about 7 mg of medicine left. After \(\frac{3}{4}\) hour there are about 3.5 mg of medicine left. About how many mg of medicine are left after 1\(\frac{3}{4}\) hours? Explain how you know.
- How does the decay rate from \(\frac{1}{4}\) hour to \(\frac{1}{2}\) hour compare to the decay rate from \(\frac{1}{2}\) hour to \(\frac{3}{4}\) hour? Explain how you know.

### Student Response

Student responses to this activity are available at one of our IM Certified Partners

### Anticipated Misconceptions

Students may be uncomfortable with the fact that the vertical axis is not labeled. Remind them that the problem gives them the information they need about how much medicine is left after \(\frac{1}{4}\) hour and \(\frac{3}{4}\) hour. The graph is mainly intended to help visualize the situation.

### Activity Synthesis

Invite some students to share how they calculated the amount of medicine left after \(1\frac{3}{4}\) hours. Make sure students see that:

- The interval between \(\frac{3}{4}\) and \(1\frac{3}{4}\) can be seen as two half-hour intervals. So if we know the growth factor of another half-hour interval, say \(\frac{1}{4}\) to \(\frac{3}{4}\), we can find the factor from \(\frac{3}{4}\) to \(1\frac{3}{4}\) by squaring that growth factor.
- The graph can be useful to check how reasonable the estimate of 0.9 mg is for the \(y\)-coordinate of \(C\). This is \(\frac{1}{4}\) of the \(y\)-coordinate value for \(B\) and this makes sense looking at the graph.

Consider asking students how they could find the decay factor for \(\frac{1}{4}\) hour. Since there are two quarter hours in \(\frac{1}{2}\) hour, the decay factor for \(\frac{1}{2}\) hour is the square of the decay factor for \(\frac{1}{4}\) hour.

## Lesson Synthesis

### Lesson Synthesis

The goal of this discussion is to make sure students see that the value of an exponential function changes by a constant factor over equal intervals, for both whole-number intervals and fractional ones.

Present an example like this one: The function \(f(h) = 200 \boldcdot \left(\frac {1}{16}\right)^h\) describes the amount of medicine \(f(h)\), in milligrams, in a body \(h\) hours after administering 200 mg of the medicine. Ask students,

- “How is the amount of medicine changing every hour?” (It is decaying by a factor of \(\frac{1}{16}\) each hour. After one hour, only \(\frac{1}{16}\) of the initial amount remains.)
- “By what factor does the medicine decay in the first 15 minutes?” (The medicine decays by a factor of \(\frac12\) because \(\frac {f\left(\frac14\right)}{f(0)} =\frac {200 \boldcdot \left( \frac{1}{16}\right)^\frac14}{200\boldcdot \left(\frac{1}{16}\right)^0} = \left(\frac{1}{16}\right)^\frac14 =\sqrt [4]\frac{1}{16}=\frac12\).)
- “By what factor does the medicine decay in between minute 30 and 45?” (This is also \(\frac12\), because \(\frac {f\left(\frac34\right)}{f\left(\frac24\right)} = \frac {200 \boldcdot \left(\frac{1}{16}\right)^\frac34}{200 \boldcdot \left(\frac{1}{16}\right)^\frac24} = \left(\frac{1}{16}\right)^\frac14 = \sqrt [4]\frac{1}{16} = \frac12\).)
- “Suppose that at some point after taking the medicine there are 32 mg of medicine in a patient’s body. How many milligrams are in the body 45 minutes later? How do we know?” (4 mg. In 45 minutes, the medicine will have decayed by a factor of \(\frac12\) three times, so we can multiply 32 by \(\left(\frac12\right)^3\) or \(\frac{1}{8}\) to get 4.)

## 5.4: Cool-down - Cost of A Haircut (5 minutes)

### Cool-Down

Cool-downs for this lesson are available at one of our IM Certified Partners

## Student Lesson Summary

### Student Facing

Earlier we learned that, for an exponential function, every time the input increases by a certain amount the output changes by a certain factor.

For example, the population of a country, in millions, can be modeled by the exponential function \(f(c) = 5 \boldcdot 16^c\), where \(c\) is time in centuries since 1900. By this model, the growth factor for any one century after the initial measurement is 16.

What about the growth factor for any one decade (one tenth of a century)? Let’s start by finding the growth factors between 1910 and 1920 (\(c\) between 0.1 and 0.2) and between 1960 and 1970 (\(c\) between 0.6 and 0.7). To do that we can calculate the quotients of the function at those input values.

- from 1910 to 1920: \(\dfrac {f(0.2)}{f(0.1)}= \dfrac {5 \boldcdot 16^{(0.2)}}{5 \boldcdot 16^{(0.1)}}\), which equals \(16^{(0.1)}\) (or \(\sqrt [10]{16}\))
- from 1960 to 1970: \(\dfrac {f(0.7)}{f(0.6)} = \dfrac {5 \boldcdot 16^{(0.7)}}{5 \boldcdot 16^{(0.6)}}\), which equals \(16^{(0.1)}\) (or \(\sqrt [10]{16}\))

Now we can generalize about the growth factor for *any* one decade using the population \(x\) centuries after 1900, \(f(x)\), and the population one decade (one tenth of a century) after that point, \(f(x + 0.1)\).

- from \(x\) to \((x + 0.1)\): \(\dfrac {f(x + 0.1)}{f(x)} = \dfrac {5 \boldcdot 16^{(x+0.1)}}{5 \boldcdot 16^{x}}\), which also equals \(16^{(0.1)}\) (or \(\sqrt [10]{16}\))

This is consistent with what we know about how exponential functions change over whole-number intervals: they always increase or decrease by equal factors over equal intervals. This is true even when the intervals are fractional.