Math  /  Data & Statistics

Questionic 2.2) Einear and Exponential Functions
Name: exponential, or neither. Give a reason given in the table below. For each table, determine if the function 2. \begin{tabular}{|c|c|} \hlinexx & g(x)g(x) \\ \hline 1 & -1 \\ \hline 2 & 0 \\ \hline 3 & 2 \\ \hline 4 & 5 \\ \hline 5 & 9 \\ \hline \end{tabular} 3. 4. \begin{tabular}{|c|c|} \hlinexx & k(x)k(x) \\ \hline 1 & -9 \\ \hline 6 & -2 \\ \hline 11 & 5 \\ \hline 16 & 12 \\ \hline 21 & 19 \\ \hline \end{tabular}

Studdy Solution

STEP 1

1. A function is linear if it has a constant rate of change (difference between consecutive y y -values is constant).
2. A function is exponential if the ratio between consecutive y y -values is constant.
3. If neither condition is met, the function is neither linear nor exponential.

STEP 2

1. Analyze the function g(x) g(x) for linearity or exponentiality.
2. Analyze the function k(x) k(x) for linearity or exponentiality.

STEP 3

Calculate the differences between consecutive g(x) g(x) values:
g(2)g(1)=0(1)=1 g(2) - g(1) = 0 - (-1) = 1 g(3)g(2)=20=2 g(3) - g(2) = 2 - 0 = 2 g(4)g(3)=52=3 g(4) - g(3) = 5 - 2 = 3 g(5)g(4)=95=4 g(5) - g(4) = 9 - 5 = 4
The differences are not constant, so g(x) g(x) is not linear.

STEP 4

Calculate the ratios between consecutive g(x) g(x) values:
g(2)g(1)=01 (undefined) \frac{g(2)}{g(1)} = \frac{0}{-1} \text{ (undefined)} g(3)g(2)=20 (undefined) \frac{g(3)}{g(2)} = \frac{2}{0} \text{ (undefined)} g(4)g(3)=52 \frac{g(4)}{g(3)} = \frac{5}{2} g(5)g(4)=95 \frac{g(5)}{g(4)} = \frac{9}{5}
The ratios are not constant, so g(x) g(x) is not exponential.
Conclusion for g(x) g(x) : Neither linear nor exponential.

STEP 5

Calculate the differences between consecutive k(x) k(x) values:
k(6)k(1)=2(9)=7 k(6) - k(1) = -2 - (-9) = 7 k(11)k(6)=5(2)=7 k(11) - k(6) = 5 - (-2) = 7 k(16)k(11)=125=7 k(16) - k(11) = 12 - 5 = 7 k(21)k(16)=1912=7 k(21) - k(16) = 19 - 12 = 7
The differences are constant, so k(x) k(x) is linear.
Conclusion for k(x) k(x) : Linear.
The function g(x) g(x) is neither linear nor exponential, and the function k(x) k(x) is linear.

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