# The values of two functions, f and g, are given in a table. One, both, or neither of them may be exponential. Give the exponential models for those that are. begin{array}{|l|l|l|}hline X&-2&-1&0&1&2hline f(x)&0.18&0.9&4.5&22.5&112.5hline g(x)&12&6&3&1.5&0.75hlineend{array} f(x)-? g(x)-? Question
Exponential models The values of two functions, f and g, are given in a table. One, both, or neither of them may be exponential. Give the exponential models for those that are.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{l}\right|}{l}{\left|{l}\right|}\right\rbrace}{h}{l}\in{e}{X}&-{2}&-{1}&{0}&{1}&{2}\backslash{h}{l}\in{e}{f{{\left({x}\right)}}}&{0.18}&{0.9}&{4.5}&{22.5}&{112.5}\backslash{h}{l}\in{e}{g{{\left({x}\right)}}}&{12}&{6}&{3}&{1.5}&{0.75}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
f(x)-?
g(x)-? 2020-12-13
The function f(x) with respect to x is given in the table :
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{l}\right|}{l}{\left|{l}\right|}\right\rbrace}{h}{l}\in{e}{X}&-{2}&-{1}&{0}&{1}&{2}\backslash{h}{l}\in{e}{f{{\left({x}\right)}}}&{0.18}&{0.9}&{4.5}&{22.5}&{112.5}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
From the table it is shown that ,
The multiplier for every next coordinate is 5.
Therefore the common ratio formed is ,
$$\displaystyle{r}={\frac{{{0.9}}}{{{0.18}}}}$$
$$\displaystyle={5}$$
The initial value of the function is $$\displaystyle{a}={4.5}$$
The general form of the exponential form is,
$$\displaystyle{y}={a}\cdot{r}^{{x}}$$
Substitute the values,
$$\displaystyle{y}={4.5}{\left({5}\right)}^{{x}}$$
Thus the exponential function formed is $$\displaystyle{f{{\left({x}\right)}}}={y}={4.5}{\left({5}\right)}^{{x}}$$
The function g(x) with respect to x is given in the table :
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{l}\right|}{l}{\left|{l}\right|}\right\rbrace}{h}{l}\in{e}{X}&-{2}&-{1}&{0}&{1}&{2}\backslash{h}{l}\in{e}{g{{\left({x}\right)}}}&{12}&{6}&{3}&{1.5}&{0.75}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
From the table it is shown that ,
The multiplier for every next coordinate is 0.5.
Therefore the common ratio formed is,
$$\displaystyle{r}={\frac{{{6}}}{{{12}}}}$$
$$\displaystyle={0.5}$$
The initial value of the function is 3.
The general form of the exponential form is,
$$\displaystyle{y}={a}\cdot{r}^{{x}}$$
Substitute the values,
$$\displaystyle{y}={3}{\left({0.5}\right)}^{{x}}$$
Thus the exponential function formed is $$\displaystyle{g{{\left({x}\right)}}}={3}{\left({0.5}\right)}^{{x}}$$

### Relevant Questions The values of two functions, f and g, are given in a table. One, both, or neither of them may be exponential. Give the exponential models for those that are.
f(x)-?
g(x)-??
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{l}\right|}{l}{\left|{l}\right|}\right\rbrace}{h}{l}\in{e}{X}&-{2}&-{1}&{0}&{1}&{2}\backslash{h}{l}\in{e}{f{{\left({x}\right)}}}&{1.125}&{2.25}&{4.5}&{9}&{18}\backslash{h}{l}\in{e}{g{{\left({x}\right)}}}&{16}&{8}&{4}&{2}&{1}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}{x}&{1}&{2}&{3}&{4}&{5}&{6}&{7}&{8}&{9}&{10}\backslash{h}{l}\in{e}{f{{\left({x}\right)}}}&{409.4}&{260.7}&{170.4}&{110.6}&{74}&{44.7}&{32.4}&{19.5}&{12.7}&{8.1}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ The burial cloth of an Egyptian mummy is estimated to contain 560 g of the radioactive materialcarbon-14, which has a half life of 5730 years.
a. Complete the table below. Make sure you justify your answer by showing all the steps.
$$\begin{array}{|l|l|l|}\hline t(\text{in years})&m(\text{amoun of radioactive material})\\\hline0&\\\hline5730\\\hline11460\\\hline17190\\\hline\end{array}$$
b. Find an exponential function that models the amount of carbon-14 in the cloth, y, after t years. Make sure you justify your answer by showing all the steps.
c. If the burial cloth is estimated to contain 49.5% of the original amount of carbon-14, how long ago was the mummy buried. Give exact answer. Make sure you justify your answer by showing all the steps. The burial cloth of an Egyptian mummy is estimated to contain 560 g of the radioactive materialcarbon-14, which has a half life of 5730 years.
a. Complete the table below. Make sure you justify your answer by showing all the steps.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{l}\right|}{l}{\left|{l}\right|}\right\rbrace}{h}{l}\in{e}{t}{\left(\text{in years}\right)}&{m}{\left(\text{amoun of radioactive material}\right)}\backslash{h}{l}\in{e}{0}&\backslash{h}{l}\in{e}{5730}\backslash{h}{l}\in{e}{11460}\backslash{h}{l}\in{e}{17190}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
b. Find an exponential function that models the amount of carbon-14 in the cloth, y, after t years. Make sure you justify your answer by showing all the steps.
c. If the burial cloth is estimated to contain 49.5% of the original amount of carbon-14, how long ago was the mummy buried. Give exact answer. Make sure you justify your answer by showing all the steps. The following table lists the reported number of cases of infants born in the United States with HIV in recent years because their mother was infected.
Source:
Centers for Disease Control and Prevention.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{Year}&\text{amp, Cases}\backslash{h}{l}\in{e}{1995}&{a}\mp,\ {295}\backslash{h}{l}\in{e}{1997}&{a}\mp,\ {166}\backslash{h}{l}\in{e}{1999}&{a}\mp,\ {109}\backslash{h}{l}\in{e}{2001}&{a}\mp,\ {115}\backslash{h}{l}\in{e}{2003}&{a}\mp,\ {94}\backslash{h}{l}\in{e}{2005}&{a}\mp,\ {107}\backslash{h}{l}\in{e}{2007}&{a}\mp,\ {79}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
a) Plot the data on a graphing calculator, letting $$\displaystyle{t}={0}$$ correspond to the year 1995.
b) Using the regression feature on your calculator, find a quadratic, a cubic, and an exponential function that models this data.
c) Plot the three functions with the data on the same coordinate axes. Which function or functions best capture the behavior of the data over the years plotted?
d) Find the number of cases predicted by all three functions for 20152015. Which of these are realistic? Explain. Several models have been proposed to explain the diversification of life during geological periods. According to Benton (1997), The diversification of marine families in the past 600 million years (Myr) appears to have followed two or three logistic curves, with equilibrium levels that lasted for up to 200 Myr. In contrast, continental organisms clearly show an exponential pattern of diversification, and although it is not clear whether the empirical diversification patterns are real or are artifacts of a poor fossil record, the latter explanation seems unlikely. In this problem, we will investigate three models fordiversification. They are analogous to models for populationgrowth, however, the quantities involved have a differentinterpretation. We denote by N(t) the diversification function,which counts the number of taxa as a function of time, and by rthe intrinsic rate of diversification.
(a) (Exponential Model) This model is described by $$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{e}}}{N}\ {\left({8.86}\right)}.$$ Solve (8.86) with the initial condition N(0) at time 0, and show that $$\displaystyle{r}_{{{e}}}$$ can be estimated from $$\displaystyle{r}_{{{e}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ {\left({8.87}\right)}$$
(b) (Logistic Growth) This model is described by $$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{l}}}{N}\ {\left({1}\ -\ {\frac{{{N}}}{{{K}}}}\right)}\ {\left({8.88}\right)}$$ where K is the equilibrium value. Solve (8.88) with the initial condition N(0) at time 0, and show that $$\displaystyle{r}_{{{l}}}$$ can be estimated from $$\displaystyle{r}_{{{l}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ +\ {\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}\ {\left({8.89}\right)}$$ for $$\displaystyle{N}{\left({t}\right)}\ {<}\ {K}.$$
(c) Assume that $$\displaystyle{N}{\left({0}\right)}={1}$$ and $$\displaystyle{N}{\left({10}\right)}={1000}.$$ Estimate $$\displaystyle{r}_{{{e}}}$$ and $$\displaystyle{r}_{{{l}}}$$ for both $$\displaystyle{K}={1001}$$ and $$\displaystyle{K}={10000}.$$
(d) Use your answer in (c) to explain the following quote from Stanley (1979): There must be a general tendency for calculated values of $$\displaystyle{\left[{r}\right]}$$ to represent underestimates of exponential rates,because some radiation will have followed distinctly sigmoid paths during the interval evaluated.
(e) Explain why the exponential model is a good approximation to the logistic model when $$\displaystyle\frac{{N}}{{K}}$$ is small compared with 1. The table gives the midyear population of Japan, in thousands, from 1960 to 2010.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{Year}&\text{Population}\backslash{h}{l}\in{e}{1960}&{94.092}\backslash{h}{l}\in{e}{1965}&{98.883}\backslash{h}{l}\in{e}{1970}&{104.345}\backslash{h}{l}\in{e}{1975}&{111.573}\backslash{h}{l}\in{e}{1980}&{116.807}\backslash{h}{l}\in{e}{1985}&{120.754}\backslash{h}{l}\in{e}{1990}&{123.537}\backslash{h}{l}\in{e}{1995}&{125.327}\backslash{h}{l}\in{e}{2000}&{126.776}\backslash{h}{l}\in{e}{2005}&{127.715}\backslash{h}{l}\in{e}{2010}&{127.579}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
Use a calculator to fit both an exponential function and a logistic function to these data. Graph the data points and both functions, and comment on the accuracy of the models. [Hint: Subtract 94,000 from each of the population figures. Then, after obtaining a model from your calculator, add 94,000 to get your final model. It might be helpful to choose $$\displaystyle{t}={0}$$ to correspond to 1960 or 1980.] The annual sales S (in millions of dollars) for the Perrigo Company from 2004 through 2010 are shown in the table. $$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{Year}&{2004}&{2005}&{2006}&{2007}&{2008}&{2009}&{2010}\backslash{h}{l}\in{e}\text{Sales, S}&{898.2}&{1024.1}&{1366.8}&{1447.4}&{1822.1}&{2006.9}&{2268.9}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ a) Use a graphing utility to create a scatter plot of the data. Let t represent the year, with $$\displaystyle{t}={4}$$ corresponding to 2004. b) Use the regression feature of the graphing utility to find an exponential model for the data. Use the Inverse Property $$\displaystyle{b}={e}^{{{\ln{\ }}{b}}}$$ to rewrite the model as an exponential model in base e. c) Use the regression feature of the graphing utility to find a logarithmic model for the data. d) Use the exponential model in base e and the logarithmic model to predict sales in 2011. It is projected that sales in 2011 will be \$2740 million. Do the predictions from the two models agree with this projection? Explain. $$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{Year}&\text{Population}\backslash{h}{l}\in{e}{1960}&{3581}\backslash{h}{l}\in{e}{1965}&{3723}\backslash{h}{l}\in{e}{1970}&{3877}\backslash{h}{l}\in{e}{1975}&{4007}\backslash{h}{l}\in{e}{1980}&{4086}\backslash{h}{l}\in{e}{1985}&{4152}\backslash{h}{l}\in{e}{1990}&{4242}\backslash{h}{l}\in{e}{1995}&{4359}\backslash{h}{l}\in{e}{2000}&{4492}\backslash{h}{l}\in{e}{2005}&{4625}\backslash{h}{l}\in{e}{2010}&{4891}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
Use a calculator to fit both an exponential function and a logistic function to these data. Graph the data points and both functions, and comment on the accuracy of the models. [Hint: Subtract 3500 from each of the population figures. Then, after obtaining a model from your calculator, add 3500 to get your final model. It might be helpful to choose $$\displaystyle{t}={0}$$ to correspond to 1960.] The following table shows the approximate average household income in the United States in 1990, 1995, and 2003. ($$\displaystyle{t}={0}$$ represents 1990.)
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{t(Year)}&{0}&{5}&{13}\backslash{h}{l}\in{e}\text{H(Household Income in}\ \{1},{000}{)}&{30}&{35}&{43}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
a) Linear: $$\displaystyle{H}{\left({t}\right)}={m}{b}\ +\ {b}$$
b) Quadratic: $$\displaystyle{H}{\left({t}\right)}={a}{t}^{{{2}}}\ +\ {b}{t}\ +\ {c}$$
c) Exponential: $$\displaystyle{H}{\left({t}\right)}={A}{b}^{{{t}}}$$