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Sagot :
To determine which set of ordered pairs could be generated by an exponential function, we need to identify a set where the y-values follow the pattern [tex]\(y = a \cdot r^x\)[/tex], where [tex]\(a\)[/tex] and [tex]\(r\)[/tex] are constants.
Let's analyze each set step-by-step:
### Set 1: [tex]\((1,1),\left(2, \frac{1}{2}\right),\left(3, \frac{1}{3}\right),\left(4, \frac{1}{4}\right)\)[/tex]
To check if this can be generated by an exponential function [tex]\(y = a \cdot r^x\)[/tex]:
- For [tex]\(x = 1\)[/tex], [tex]\(y = 1\)[/tex]: [tex]\(1 = a \cdot r^1\)[/tex], so [tex]\(a = r\)[/tex]
- For [tex]\(x = 2\)[/tex], [tex]\(y = \frac{1}{2}\)[/tex]: [tex]\(\frac{1}{2} = a \cdot r^2 = r \cdot r^2 = r^3\)[/tex], giving us [tex]\(r^3 = \frac{1}{2}\)[/tex]
- For [tex]\(x = 3\)[/tex], [tex]\(y = \frac{1}{3}\)[/tex]: [tex]\(\frac{1}{3} = a \cdot r^3 = r \cdot r^3 = r^4\)[/tex], giving us [tex]\(r^4 = \frac{1}{3}\)[/tex]
- For [tex]\(x = 4\)[/tex], [tex]\(y = \frac{1}{4}\)[/tex]: [tex]\(\frac{1}{4} = a \cdot r^4 = r \cdot r^4 = r^5\)[/tex], giving us [tex]\(r^5 = \frac{1}{4}\)[/tex]
We don't find consistent [tex]\(r\)[/tex] values.
### Set 2: [tex]\((1,1),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{9}\right),\left(4, \frac{1}{16}\right)\)[/tex]
To check if this can be generated by an exponential function [tex]\(y = a \cdot r^x\)[/tex]:
- For [tex]\(x = 1\)[/tex], [tex]\(y = 1\)[/tex]: [tex]\(1 = a \cdot r^1\)[/tex], so [tex]\(a = r\)[/tex]
- For [tex]\(x = 2\)[/tex], [tex]\(y = \frac{1}{4}\)[/tex]: [tex]\(\frac{1}{4} = a \cdot r^2 = r \cdot r^2 = r^3\)[/tex], giving us [tex]\(r^3 = \frac{1}{4}\)[/tex]
- For [tex]\(x = 3\)[/tex], [tex]\(y = \frac{1}{9}\)[/tex]: [tex]\(\frac{1}{9} = a \cdot r^3 = r \cdot r^3 = r^4\)[/tex], giving us [tex]\(r^4 = \frac{1}{9}\)[/tex]
- For [tex]\(x = 4\)[/tex], [tex]\(y = \frac{1}{16}\)[/tex]: [tex]\(\frac{1}{16} = a \cdot r^4 = r \cdot r^4 = r^5\)[/tex], giving us [tex]\(r^5 = \frac{1}{16}\)[/tex]
Again, no consistent [tex]\(r\)[/tex].
### Set 3: [tex]\(\left(1, \frac{1}{2}\right),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{8}\right),\left(4, \frac{1}{16}\right)\)[/tex]
Here, we check the pattern:
- For [tex]\(x = 1\)[/tex], [tex]\(y = \frac{1}{2}\)[/tex]: [tex]\(\frac{1}{2} = a \cdot r^1\)[/tex], so [tex]\(a = \frac{1}{2r}\)[/tex]
- For [tex]\(x = 2\)[/tex], [tex]\(y = \frac{1}{4}\)[/tex]: [tex]\(\frac{1}{4} = a \cdot r^2 = \frac{1}{2r} \cdot r^2 = \frac{r}{2}\)[/tex]
- Since [tex]\(\frac{1}{4} = \frac{r}{2}\)[/tex], solving gives [tex]\(r^2 = \frac{1}{2}\)[/tex], so [tex]\(r = \frac{1}{\sqrt{2}}\)[/tex]
- For [tex]\(x = 3\)[/tex], [tex]\(y = \frac{1}{8}\)[/tex]: [tex]\(\frac{1}{8} = \frac{1}{2} \cdot \left(\frac{1}{\sqrt{2}}^3\right)\)[/tex]
- Check consistency: For [tex]\(x = 4\)[/tex], [tex]\(y = \frac{1}{16}\)[/tex]
### Set 4: [tex]\(\left(1, \frac{1}{2}\right),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{6}\right),\left(4, \frac{1}{8}\right)\)[/tex]
Assuming it could follow [tex]\(y = a \cdot r^x\)[/tex]:
- For [tex]\(x = 1\)[/tex], [tex]\(y = \frac{1}{2}\)[/tex]: [tex]\(\frac{1}{2} = a \cdot r\)[/tex]
- For [tex]\(x = 2\)[/tex], [tex]\(y = \frac{1}{4}\)[/tex]: [tex]\(\frac{1}{4} = a \cdot r^2\)[/tex]
- For [tex]\(x = 3\)[/tex], [tex]\(y = \frac{1}{6}\)[/tex]
- For [tex]\(x = 4\)[/tex], [tex]\(y = \frac{1}{8}\)[/tex]
Inconsistent increments confirm [tex]\(y = a \cdot r^x\)[/tex] doesn't fit.
### Conclusion
Set 3: [tex]\(\left(1, \frac{1}{2}\right),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{8}\right),\left(4, \frac{1}{16}\right)\)[/tex] fits the exponential function pattern.
Let's analyze each set step-by-step:
### Set 1: [tex]\((1,1),\left(2, \frac{1}{2}\right),\left(3, \frac{1}{3}\right),\left(4, \frac{1}{4}\right)\)[/tex]
To check if this can be generated by an exponential function [tex]\(y = a \cdot r^x\)[/tex]:
- For [tex]\(x = 1\)[/tex], [tex]\(y = 1\)[/tex]: [tex]\(1 = a \cdot r^1\)[/tex], so [tex]\(a = r\)[/tex]
- For [tex]\(x = 2\)[/tex], [tex]\(y = \frac{1}{2}\)[/tex]: [tex]\(\frac{1}{2} = a \cdot r^2 = r \cdot r^2 = r^3\)[/tex], giving us [tex]\(r^3 = \frac{1}{2}\)[/tex]
- For [tex]\(x = 3\)[/tex], [tex]\(y = \frac{1}{3}\)[/tex]: [tex]\(\frac{1}{3} = a \cdot r^3 = r \cdot r^3 = r^4\)[/tex], giving us [tex]\(r^4 = \frac{1}{3}\)[/tex]
- For [tex]\(x = 4\)[/tex], [tex]\(y = \frac{1}{4}\)[/tex]: [tex]\(\frac{1}{4} = a \cdot r^4 = r \cdot r^4 = r^5\)[/tex], giving us [tex]\(r^5 = \frac{1}{4}\)[/tex]
We don't find consistent [tex]\(r\)[/tex] values.
### Set 2: [tex]\((1,1),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{9}\right),\left(4, \frac{1}{16}\right)\)[/tex]
To check if this can be generated by an exponential function [tex]\(y = a \cdot r^x\)[/tex]:
- For [tex]\(x = 1\)[/tex], [tex]\(y = 1\)[/tex]: [tex]\(1 = a \cdot r^1\)[/tex], so [tex]\(a = r\)[/tex]
- For [tex]\(x = 2\)[/tex], [tex]\(y = \frac{1}{4}\)[/tex]: [tex]\(\frac{1}{4} = a \cdot r^2 = r \cdot r^2 = r^3\)[/tex], giving us [tex]\(r^3 = \frac{1}{4}\)[/tex]
- For [tex]\(x = 3\)[/tex], [tex]\(y = \frac{1}{9}\)[/tex]: [tex]\(\frac{1}{9} = a \cdot r^3 = r \cdot r^3 = r^4\)[/tex], giving us [tex]\(r^4 = \frac{1}{9}\)[/tex]
- For [tex]\(x = 4\)[/tex], [tex]\(y = \frac{1}{16}\)[/tex]: [tex]\(\frac{1}{16} = a \cdot r^4 = r \cdot r^4 = r^5\)[/tex], giving us [tex]\(r^5 = \frac{1}{16}\)[/tex]
Again, no consistent [tex]\(r\)[/tex].
### Set 3: [tex]\(\left(1, \frac{1}{2}\right),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{8}\right),\left(4, \frac{1}{16}\right)\)[/tex]
Here, we check the pattern:
- For [tex]\(x = 1\)[/tex], [tex]\(y = \frac{1}{2}\)[/tex]: [tex]\(\frac{1}{2} = a \cdot r^1\)[/tex], so [tex]\(a = \frac{1}{2r}\)[/tex]
- For [tex]\(x = 2\)[/tex], [tex]\(y = \frac{1}{4}\)[/tex]: [tex]\(\frac{1}{4} = a \cdot r^2 = \frac{1}{2r} \cdot r^2 = \frac{r}{2}\)[/tex]
- Since [tex]\(\frac{1}{4} = \frac{r}{2}\)[/tex], solving gives [tex]\(r^2 = \frac{1}{2}\)[/tex], so [tex]\(r = \frac{1}{\sqrt{2}}\)[/tex]
- For [tex]\(x = 3\)[/tex], [tex]\(y = \frac{1}{8}\)[/tex]: [tex]\(\frac{1}{8} = \frac{1}{2} \cdot \left(\frac{1}{\sqrt{2}}^3\right)\)[/tex]
- Check consistency: For [tex]\(x = 4\)[/tex], [tex]\(y = \frac{1}{16}\)[/tex]
### Set 4: [tex]\(\left(1, \frac{1}{2}\right),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{6}\right),\left(4, \frac{1}{8}\right)\)[/tex]
Assuming it could follow [tex]\(y = a \cdot r^x\)[/tex]:
- For [tex]\(x = 1\)[/tex], [tex]\(y = \frac{1}{2}\)[/tex]: [tex]\(\frac{1}{2} = a \cdot r\)[/tex]
- For [tex]\(x = 2\)[/tex], [tex]\(y = \frac{1}{4}\)[/tex]: [tex]\(\frac{1}{4} = a \cdot r^2\)[/tex]
- For [tex]\(x = 3\)[/tex], [tex]\(y = \frac{1}{6}\)[/tex]
- For [tex]\(x = 4\)[/tex], [tex]\(y = \frac{1}{8}\)[/tex]
Inconsistent increments confirm [tex]\(y = a \cdot r^x\)[/tex] doesn't fit.
### Conclusion
Set 3: [tex]\(\left(1, \frac{1}{2}\right),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{8}\right),\left(4, \frac{1}{16}\right)\)[/tex] fits the exponential function pattern.
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