Answer :
To determine which exponential model best represents the given data, we consider four different exponential models:
1. [tex]\( f(x) = 5(1.2)^x \)[/tex]
2. [tex]\( f(x) = 5(0.2)^x \)[/tex]
3. [tex]\( f(x) = 2(5)^x \)[/tex]
4. [tex]\( f(x) = 2(0.5)^x \)[/tex]
The table of data points is:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -2 & 128 \\ \hline -1 & 27 \\ \hline 0 & 5 \\ \hline 1 & 1 \\ \hline 2 & 0.1 \\ \hline \end{array} \][/tex]
We need to evaluate the sum of the squared deviations between the observed data points and the values predicted by each model. This can be done by calculating the deviation for each data point, squaring it, and then summing these values for all data points.
### Calculating the Sum of Squared Deviations for Each Model
#### Model 1: [tex]\( f(x) = 5(1.2)^x \)[/tex]
For each [tex]\( x \)[/tex]:
- [tex]\( f(-2) = 5(1.2)^{-2} \approx 3.47 \)[/tex]
- [tex]\( f(-1) = 5(1.2)^{-1} \approx 4.17 \)[/tex]
- [tex]\( f(0) = 5(1.2)^0 = 5 \)[/tex]
- [tex]\( f(1) = 5(1.2)^1 = 6 \)[/tex]
- [tex]\( f(2) = 5(1.2)^2 = 7.2 \)[/tex]
Sum of Squared Deviations:
[tex]\[ (128 - 3.47)^2 + (27 - 4.17)^2 + (5 - 5)^2 + (1 - 6)^2 + (0.1 - 7.2)^2 \][/tex]
#### Model 2: [tex]\( f(x) = 5(0.2)^x \)[/tex]
For each [tex]\( x \)[/tex]:
- [tex]\( f(-2) = 5(0.2)^{-2} = 125 \)[/tex]
- [tex]\( f(-1) = 5(0.2)^{-1} = 25 \)[/tex]
- [tex]\( f(0) = 5(0.2)^0 = 5 \)[/tex]
- [tex]\( f(1) = 5(0.2)^1 = 1 \)[/tex]
- [tex]\( f(2) = 5(0.2)^2 = 0.2 \)[/tex]
Sum of Squared Deviations:
[tex]\[ (128 - 125)^2 + (27 - 25)^2 + (5 - 5)^2 + (1 - 1)^2 + (0.1 - 0.2)^2 \][/tex]
#### Model 3: [tex]\( f(x) = 2(5)^x \)[/tex]
For each [tex]\( x \)[/tex]:
- [tex]\( f(-2) = 2(5)^{-2} = 0.08 \)[/tex]
- [tex]\( f(-1) = 2(5)^{-1} = 0.4 \)[/tex]
- [tex]\( f(0) = 2(5)^0 = 2 \)[/tex]
- [tex]\( f(1) = 2(5)^1 = 10 \)[/tex]
- [tex]\( f(2) = 2(5)^2 = 50 \)[/tex]
Sum of Squared Deviations:
[tex]\[ (128 - 0.08)^2 + (27 - 0.4)^2 + (5 - 2)^2 + (1 - 10)^2 + (0.1 - 50)^2 \][/tex]
#### Model 4: [tex]\( f(x) = 2(0.5)^x \)[/tex]
For each [tex]\( x \)[/tex]:
- [tex]\( f(-2) = 2(0.5)^{-2} = 8 \)[/tex]
- [tex]\( f(-1) = 2(0.5)^{-1} = 4 \)[/tex]
- [tex]\( f(0) = 2(0.5)^0 = 2 \)[/tex]
- [tex]\( f(1) = 2(0.5)^1 = 1 \)[/tex]
- [tex]\( f(2) = 2(0.5)^2 = 0.5 \)[/tex]
Sum of Squared Deviations:
[tex]\[ (128 - 8)^2 + (27 - 4)^2 + (5 - 2)^2 + (1 - 1)^2 + (0.1 - 0.5)^2 \][/tex]
### Identifying the Best Model
By evaluating each model using the sum of squared deviations, we find the model that fits the data most closely. This model will have the smallest total sum of squared deviations.
The results from our computations indicate that the sum of squared deviations for the second model [tex]\( f(x) = 5(0.2)^x \)[/tex] is the smallest.
Thus, the best model to represent the data is:
[tex]\[ f(x) = 5(0.2)^x \][/tex]
With a sum of squared deviations of approximately 13.01.
1. [tex]\( f(x) = 5(1.2)^x \)[/tex]
2. [tex]\( f(x) = 5(0.2)^x \)[/tex]
3. [tex]\( f(x) = 2(5)^x \)[/tex]
4. [tex]\( f(x) = 2(0.5)^x \)[/tex]
The table of data points is:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -2 & 128 \\ \hline -1 & 27 \\ \hline 0 & 5 \\ \hline 1 & 1 \\ \hline 2 & 0.1 \\ \hline \end{array} \][/tex]
We need to evaluate the sum of the squared deviations between the observed data points and the values predicted by each model. This can be done by calculating the deviation for each data point, squaring it, and then summing these values for all data points.
### Calculating the Sum of Squared Deviations for Each Model
#### Model 1: [tex]\( f(x) = 5(1.2)^x \)[/tex]
For each [tex]\( x \)[/tex]:
- [tex]\( f(-2) = 5(1.2)^{-2} \approx 3.47 \)[/tex]
- [tex]\( f(-1) = 5(1.2)^{-1} \approx 4.17 \)[/tex]
- [tex]\( f(0) = 5(1.2)^0 = 5 \)[/tex]
- [tex]\( f(1) = 5(1.2)^1 = 6 \)[/tex]
- [tex]\( f(2) = 5(1.2)^2 = 7.2 \)[/tex]
Sum of Squared Deviations:
[tex]\[ (128 - 3.47)^2 + (27 - 4.17)^2 + (5 - 5)^2 + (1 - 6)^2 + (0.1 - 7.2)^2 \][/tex]
#### Model 2: [tex]\( f(x) = 5(0.2)^x \)[/tex]
For each [tex]\( x \)[/tex]:
- [tex]\( f(-2) = 5(0.2)^{-2} = 125 \)[/tex]
- [tex]\( f(-1) = 5(0.2)^{-1} = 25 \)[/tex]
- [tex]\( f(0) = 5(0.2)^0 = 5 \)[/tex]
- [tex]\( f(1) = 5(0.2)^1 = 1 \)[/tex]
- [tex]\( f(2) = 5(0.2)^2 = 0.2 \)[/tex]
Sum of Squared Deviations:
[tex]\[ (128 - 125)^2 + (27 - 25)^2 + (5 - 5)^2 + (1 - 1)^2 + (0.1 - 0.2)^2 \][/tex]
#### Model 3: [tex]\( f(x) = 2(5)^x \)[/tex]
For each [tex]\( x \)[/tex]:
- [tex]\( f(-2) = 2(5)^{-2} = 0.08 \)[/tex]
- [tex]\( f(-1) = 2(5)^{-1} = 0.4 \)[/tex]
- [tex]\( f(0) = 2(5)^0 = 2 \)[/tex]
- [tex]\( f(1) = 2(5)^1 = 10 \)[/tex]
- [tex]\( f(2) = 2(5)^2 = 50 \)[/tex]
Sum of Squared Deviations:
[tex]\[ (128 - 0.08)^2 + (27 - 0.4)^2 + (5 - 2)^2 + (1 - 10)^2 + (0.1 - 50)^2 \][/tex]
#### Model 4: [tex]\( f(x) = 2(0.5)^x \)[/tex]
For each [tex]\( x \)[/tex]:
- [tex]\( f(-2) = 2(0.5)^{-2} = 8 \)[/tex]
- [tex]\( f(-1) = 2(0.5)^{-1} = 4 \)[/tex]
- [tex]\( f(0) = 2(0.5)^0 = 2 \)[/tex]
- [tex]\( f(1) = 2(0.5)^1 = 1 \)[/tex]
- [tex]\( f(2) = 2(0.5)^2 = 0.5 \)[/tex]
Sum of Squared Deviations:
[tex]\[ (128 - 8)^2 + (27 - 4)^2 + (5 - 2)^2 + (1 - 1)^2 + (0.1 - 0.5)^2 \][/tex]
### Identifying the Best Model
By evaluating each model using the sum of squared deviations, we find the model that fits the data most closely. This model will have the smallest total sum of squared deviations.
The results from our computations indicate that the sum of squared deviations for the second model [tex]\( f(x) = 5(0.2)^x \)[/tex] is the smallest.
Thus, the best model to represent the data is:
[tex]\[ f(x) = 5(0.2)^x \][/tex]
With a sum of squared deviations of approximately 13.01.
