Lesson 3

In an earlier unit, students learned the meaning of rational exponents and how to evaluate them. This warm-up reviews that work, preparing students to explore exponential functions with rational input values by focusing on two basic fractional exponents, \(\frac12\) and \(\frac13\), and some of their properties.

Briefly survey the class on whether each expression is a solution to the given equation. If there is a disagreement about whether an expression is a solution, ask students with contrasting views to share their reasoning so agreement can be reached. For an expression that everyone agrees is a solution, invite 2–3 students to share their reasoning.

Make sure students recall that a number raised to an exponent of \(\frac{1}{2}\), as in \(x^{\frac{1}{2}}\), is \(\sqrt{x}\). Highlight that squaring each number, \(x^{\frac{1}{2}}\) and \(\sqrt{x}\), results in \(x\). \(\displaystyle (x^{\frac{1}{2}})^2 = x^1 = x\) \(\displaystyle (\sqrt{x})^2 = \sqrt{x} \boldcdot \sqrt{x} = x\)

Likewise, a number raised to an exponent of \(\frac13\), as in \(p^{\frac13}\), is defined to be \(\sqrt[3]{p}\). Cubing each number, \(p^{\frac13}\) and \(\sqrt[3]{p}\), results in \(p\). \(\displaystyle (p^{\frac{1}{3}})^3 = p^1 = p\) \(\displaystyle (\sqrt[3]p)^3 = \sqrt[3]{p} \boldcdot \sqrt[3]{p} \boldcdot \sqrt[3]{p} = p\)

In this activity, students construct an exponential function to model the population growth in Florida. Then, they interpret the function when evaluated at certain rational numbers. They do so quantitatively, in terms of the population, and abstractly, in terms of an expression with a fractional exponent (MP2).

As students work, identify those who express the population of a given year as an exponential expression and those who use a calculator to estimate a numerical value. Both approaches are important: the first, because it gives an opportunity to interpret a fractional power, and the second, because it gives a concrete sense of the population growth.

Provide access to devices that can run Desmos or other graphing technology.

If students struggle with graphing the function, ask them to think about what horizontal and vertical dimensions to set for the graphing window so the time in decades and the population in thousands can be shown meaningfully. For example, graphing an input range from 0 to 20 would mean assuming the population of Florida grew the same percent for 200 years, which is unlikely.

Focus the discussion on the meaning of \(f(0.5)\) and \(f(1.8)\) in this context and on how to reason about these values. Display a table for all to see, like the one shown here, to fill in during the discussion:

Select previously identified students to share their responses for the population of Florida with the inputs 0.5 and 1.8, written as expressions (\(54,\!500 \boldcdot (1.6)^{0.5}\) and \(54,\!500 \boldcdot (1.6)^{1.8}\)) and as numerical values (about 68,938 and 127,003). Help students see that:

• We can think of the population in 1860, 2 decades after 1840, as \(54,\!500 \boldcdot (1.6)\boldcdot (1.6)\) because the 60% increase is applied twice. Although we can’t write the same kind of expanded expression for the population 0.5 decade after 1840, we can use properties of exponents to reason about its value.
• Here are three different ways to write the growth factor for the population 5 years after 1840: \((1.6)^{0.5}\) or \({1.6}^{\frac12}\) is \(\sqrt{1.6}\).

Conclude by asking students to write three different ways to express the growth factor for the population 18 years after 1840 (\((1.6)^{1.8}\) or \({1.6}^{\frac{18}{10}}\) is \(\sqrt[18]{(1.6)^{10}}\)).

In the previous activity, students interpreted a description of an exponential function, wrote an equation, and produced a graph. In this activity, they work the other way around. Given a graph of an exponential relationship, they produce an equation and reason about the context. Students then express the change in the function in terms of its decay rate, recalling the connection between decay factor and decay rate.

Begin the discussion by inviting students to share how they derived an expression representing the function \(f\) and the meaning of 12 and \(\frac{1}{16}\) (or 0.0625) in the expression \(12 \boldcdot \left(\frac{1}{16}\right)^t\) (or \(12 \boldcdot (0.0625)^t\)).

Next, focus the discussion on the interpretation of \(f(\frac13)\) in this context and how its value could be estimated graphically and reasoned algebraically. Highlight:

• \(\frac13\) an hour is the same as 20 minutes.
• Using the equation, \(f(\frac13)=12 \boldcdot (0.0625)^{\frac13}\), which is approximately 4.8 milligrams of medicine.

Conclude the discussion by asking students to identify the decay factor for 20 minutes and what it means. (\((0.0625)^{\frac13}\), or about 39.7% of the drug remains in the body after 20 minutes.)