IDNLearn.com provides a user-friendly platform for finding and sharing knowledge. Join our Q&A platform to get accurate and thorough answers to all your pressing questions.
Sagot :
To determine the relationship between the functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] given in the table, we should carefully analyze their values for different [tex]\( x \)[/tex] values. Here's the table for reference:
[tex]\[ \begin{tabular}{|c|c|c|} \hline $x$ & $f(x)=2^x$ & $g(x)=\left(\frac{1}{2}\right)^x$ \\ \hline 2 & 4 & \frac{1}{4} \\ \hline 1 & 2 & \frac{1}{2} \\ \hline 0 & 1 & 1 \\ \hline -1 & \frac{1}{2} & 2 \\ \hline -2 & \frac{1}{4} & 4 \\ \hline \end{tabular} \][/tex]
Let's compare the values of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] at each given [tex]\( x \)[/tex]:
1. When [tex]\( x = 2 \)[/tex]:
- [tex]\( f(2) = 2^2 = 4 \)[/tex]
- [tex]\( g(2) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)[/tex]
2. When [tex]\( x = 1 \)[/tex]:
- [tex]\( f(1) = 2^1 = 2 \)[/tex]
- [tex]\( g(1) = \left(\frac{1}{2}\right)^1 = \frac{1}{2} \)[/tex]
3. When [tex]\( x = 0 \)[/tex]:
- [tex]\( f(0) = 2^0 = 1 \)[/tex]
- [tex]\( g(0) = \left(\frac{1}{2}\right)^0 = 1 \)[/tex]
4. When [tex]\( x = -1 \)[/tex]:
- [tex]\( f(-1) = 2^{-1} = \frac{1}{2} \)[/tex]
- [tex]\( g(-1) = \left(\frac{1}{2}\right)^{-1} = 2 \)[/tex]
5. When [tex]\( x = -2 \)[/tex]:
- [tex]\( f(-2) = 2^{-2} = \frac{1}{4} \)[/tex]
- [tex]\( g(-2) = \left(\frac{1}{2}\right)^{-2} = 4 \)[/tex]
By examining these values, we can notice the following pattern:
- For positive [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] yields values greater than 1, while [tex]\( g(x) \)[/tex] yields values less than 1.
- Both functions yield 1 when [tex]\( x = 0 \)[/tex].
- For negative [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] yields values less than 1, while [tex]\( g(x) \)[/tex] yields values greater than 1.
- Essentially, [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are inversely related and symmetrical about the [tex]\( y \)[/tex]-axis.
Therefore, the relationship between the functions can be concluded as:
The functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are reflections over the [tex]\( y \)[/tex] axis.
[tex]\[ \begin{tabular}{|c|c|c|} \hline $x$ & $f(x)=2^x$ & $g(x)=\left(\frac{1}{2}\right)^x$ \\ \hline 2 & 4 & \frac{1}{4} \\ \hline 1 & 2 & \frac{1}{2} \\ \hline 0 & 1 & 1 \\ \hline -1 & \frac{1}{2} & 2 \\ \hline -2 & \frac{1}{4} & 4 \\ \hline \end{tabular} \][/tex]
Let's compare the values of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] at each given [tex]\( x \)[/tex]:
1. When [tex]\( x = 2 \)[/tex]:
- [tex]\( f(2) = 2^2 = 4 \)[/tex]
- [tex]\( g(2) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)[/tex]
2. When [tex]\( x = 1 \)[/tex]:
- [tex]\( f(1) = 2^1 = 2 \)[/tex]
- [tex]\( g(1) = \left(\frac{1}{2}\right)^1 = \frac{1}{2} \)[/tex]
3. When [tex]\( x = 0 \)[/tex]:
- [tex]\( f(0) = 2^0 = 1 \)[/tex]
- [tex]\( g(0) = \left(\frac{1}{2}\right)^0 = 1 \)[/tex]
4. When [tex]\( x = -1 \)[/tex]:
- [tex]\( f(-1) = 2^{-1} = \frac{1}{2} \)[/tex]
- [tex]\( g(-1) = \left(\frac{1}{2}\right)^{-1} = 2 \)[/tex]
5. When [tex]\( x = -2 \)[/tex]:
- [tex]\( f(-2) = 2^{-2} = \frac{1}{4} \)[/tex]
- [tex]\( g(-2) = \left(\frac{1}{2}\right)^{-2} = 4 \)[/tex]
By examining these values, we can notice the following pattern:
- For positive [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] yields values greater than 1, while [tex]\( g(x) \)[/tex] yields values less than 1.
- Both functions yield 1 when [tex]\( x = 0 \)[/tex].
- For negative [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] yields values less than 1, while [tex]\( g(x) \)[/tex] yields values greater than 1.
- Essentially, [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are inversely related and symmetrical about the [tex]\( y \)[/tex]-axis.
Therefore, the relationship between the functions can be concluded as:
The functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are reflections over the [tex]\( y \)[/tex] axis.
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.