Answer :
We are given an exponential function [tex]\( f \)[/tex] defined by a table and a function [tex]\( g \)[/tex] defined by
[tex]$$
g(x)=-18\left(\frac{1}{3}\right)^x+2.
$$[/tex]
We wish to compare the two functions on the interval [tex]\([-1,2]\)[/tex].
Step 1. Evaluate Function [tex]\( f \)[/tex] on [tex]\([-1,2]\)[/tex]:
From the table, the values of [tex]\( f(x) \)[/tex] are:
[tex]\[
\begin{array}{c|cccc}
x & -1 & 0 & 1 & 2 \\
\hline
f(x) & -22 & -10 & -4 & -1 \\
\end{array}
\][/tex]
The function [tex]\( f \)[/tex] increases from [tex]\(-22\)[/tex] at [tex]\( x=-1 \)[/tex] to [tex]\(-1\)[/tex] at [tex]\( x=2 \)[/tex].
The average rate of increase for [tex]\( f \)[/tex] on this interval is calculated by:
[tex]$$
\text{Average Rate for } f = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{-1 - (-22)}{3} = \frac{21}{3} = 7.
$$[/tex]
Step 2. Evaluate Function [tex]\( g \)[/tex] on [tex]\([-1,2]\)[/tex]:
The function [tex]\( g(x) \)[/tex] is given by
[tex]$$
g(x) = -18\left(\frac{1}{3}\right)^x+2.
$$[/tex]
We compute key values as follows:
- At [tex]\( x=-1 \)[/tex]:
[tex]$$
g(-1) = -18\left(\frac{1}{3}\right)^{-1}+2 = -18\cdot 3+2 = -54+2 = -52.
$$[/tex]
- At [tex]\( x=2 \)[/tex]:
[tex]$$
g(2) = -18\left(\frac{1}{3}\right)^2+2 = -18\cdot \frac{1}{9}+2 = -2+2 = 0.
$$[/tex]
The function [tex]\( g \)[/tex] increases from [tex]\(-52\)[/tex] at [tex]\( x=-1 \)[/tex] to [tex]\( 0 \)[/tex] at [tex]\( x=2 \)[/tex].
The average rate of increase for [tex]\( g \)[/tex] is:
[tex]$$
\text{Average Rate for } g = \frac{g(2) - g(-1)}{2 - (-1)} = \frac{0 - (-52)}{3} = \frac{52}{3} \approx 17.33.
$$[/tex]
Step 3. Compare the Two Functions on [tex]\([-1,2]\)[/tex]:
- Both [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are increasing over the interval.
- The average rate of increase of [tex]\( f \)[/tex] is [tex]\( 7 \)[/tex] while that of [tex]\( g \)[/tex] is approximately [tex]\( 17.33 \)[/tex].
Since [tex]\( g \)[/tex] increases at a faster average rate than [tex]\( f \)[/tex], the correct statement is:
A. Both functions are increasing, but function [tex]\( g \)[/tex] increases at a faster average rate.
[tex]$$
g(x)=-18\left(\frac{1}{3}\right)^x+2.
$$[/tex]
We wish to compare the two functions on the interval [tex]\([-1,2]\)[/tex].
Step 1. Evaluate Function [tex]\( f \)[/tex] on [tex]\([-1,2]\)[/tex]:
From the table, the values of [tex]\( f(x) \)[/tex] are:
[tex]\[
\begin{array}{c|cccc}
x & -1 & 0 & 1 & 2 \\
\hline
f(x) & -22 & -10 & -4 & -1 \\
\end{array}
\][/tex]
The function [tex]\( f \)[/tex] increases from [tex]\(-22\)[/tex] at [tex]\( x=-1 \)[/tex] to [tex]\(-1\)[/tex] at [tex]\( x=2 \)[/tex].
The average rate of increase for [tex]\( f \)[/tex] on this interval is calculated by:
[tex]$$
\text{Average Rate for } f = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{-1 - (-22)}{3} = \frac{21}{3} = 7.
$$[/tex]
Step 2. Evaluate Function [tex]\( g \)[/tex] on [tex]\([-1,2]\)[/tex]:
The function [tex]\( g(x) \)[/tex] is given by
[tex]$$
g(x) = -18\left(\frac{1}{3}\right)^x+2.
$$[/tex]
We compute key values as follows:
- At [tex]\( x=-1 \)[/tex]:
[tex]$$
g(-1) = -18\left(\frac{1}{3}\right)^{-1}+2 = -18\cdot 3+2 = -54+2 = -52.
$$[/tex]
- At [tex]\( x=2 \)[/tex]:
[tex]$$
g(2) = -18\left(\frac{1}{3}\right)^2+2 = -18\cdot \frac{1}{9}+2 = -2+2 = 0.
$$[/tex]
The function [tex]\( g \)[/tex] increases from [tex]\(-52\)[/tex] at [tex]\( x=-1 \)[/tex] to [tex]\( 0 \)[/tex] at [tex]\( x=2 \)[/tex].
The average rate of increase for [tex]\( g \)[/tex] is:
[tex]$$
\text{Average Rate for } g = \frac{g(2) - g(-1)}{2 - (-1)} = \frac{0 - (-52)}{3} = \frac{52}{3} \approx 17.33.
$$[/tex]
Step 3. Compare the Two Functions on [tex]\([-1,2]\)[/tex]:
- Both [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are increasing over the interval.
- The average rate of increase of [tex]\( f \)[/tex] is [tex]\( 7 \)[/tex] while that of [tex]\( g \)[/tex] is approximately [tex]\( 17.33 \)[/tex].
Since [tex]\( g \)[/tex] increases at a faster average rate than [tex]\( f \)[/tex], the correct statement is:
A. Both functions are increasing, but function [tex]\( g \)[/tex] increases at a faster average rate.