Answer :
To complete the table, we need to fill in the missing values for [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], [tex]\(d\)[/tex], [tex]\(e\)[/tex], and [tex]\(f\)[/tex] based on the exponential functions given.
Given:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline $x$ & $2^x$ & $3 \cdot 2^x$ & $2^{3 x}$ \\ \hline -2 & $\frac{1}{4}$ & $\frac{3}{4}$ & $\frac{1}{64}$ \\ \hline -1 & $\frac{1}{2}$ & $\frac{3}{2}$ & $\frac{1}{8}$ \\ \hline 0 & $a$ & $b$ & $c$ \\ \hline 1 & $d$ & $e$ & $f$ \\ \hline 2 & 4 & 12 & 64 \\ \hline \end{tabular} \][/tex]
We know the following:
- When [tex]\(x = 0\)[/tex]:
- [tex]\(2^x = 2^0 = 1\)[/tex], so [tex]\(a = 1\)[/tex].
- [tex]\(3 \cdot 2^x = 3 \cdot 2^0 = 3 \cdot 1 = 3\)[/tex], so [tex]\(b = 3\)[/tex].
- [tex]\(2^{3x} = 2^{3 \cdot 0} = 2^0 = 1\)[/tex], so [tex]\(c = 1\)[/tex].
- When [tex]\(x = 1\)[/tex]:
- [tex]\(2^x = 2^1 = 2\)[/tex], so [tex]\(d = 2\)[/tex].
- [tex]\(3 \cdot 2^x = 3 \cdot 2^1 = 3 \cdot 2 = 6\)[/tex], so [tex]\(e = 6\)[/tex].
- [tex]\(2^{3x} = 2^{3 \cdot 1} = 2^3 = 8\)[/tex], so [tex]\(f = 8\)[/tex].
The completed table is:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline $x$ & $2^x$ & $3 \cdot 2^x$ & $2^{3 x}$ \\ \hline -2 & $\frac{1}{4}$ & $\frac{3}{4}$ & $\frac{1}{64}$ \\ \hline -1 & $\frac{1}{2}$ & $\frac{3}{2}$ & $\frac{1}{8}$ \\ \hline 0 & 1 & 3 & 1 \\ \hline 1 & 2 & 6 & 8 \\ \hline 2 & 4 & 12 & 64 \\ \hline \end{tabular} \][/tex]
Given:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline $x$ & $2^x$ & $3 \cdot 2^x$ & $2^{3 x}$ \\ \hline -2 & $\frac{1}{4}$ & $\frac{3}{4}$ & $\frac{1}{64}$ \\ \hline -1 & $\frac{1}{2}$ & $\frac{3}{2}$ & $\frac{1}{8}$ \\ \hline 0 & $a$ & $b$ & $c$ \\ \hline 1 & $d$ & $e$ & $f$ \\ \hline 2 & 4 & 12 & 64 \\ \hline \end{tabular} \][/tex]
We know the following:
- When [tex]\(x = 0\)[/tex]:
- [tex]\(2^x = 2^0 = 1\)[/tex], so [tex]\(a = 1\)[/tex].
- [tex]\(3 \cdot 2^x = 3 \cdot 2^0 = 3 \cdot 1 = 3\)[/tex], so [tex]\(b = 3\)[/tex].
- [tex]\(2^{3x} = 2^{3 \cdot 0} = 2^0 = 1\)[/tex], so [tex]\(c = 1\)[/tex].
- When [tex]\(x = 1\)[/tex]:
- [tex]\(2^x = 2^1 = 2\)[/tex], so [tex]\(d = 2\)[/tex].
- [tex]\(3 \cdot 2^x = 3 \cdot 2^1 = 3 \cdot 2 = 6\)[/tex], so [tex]\(e = 6\)[/tex].
- [tex]\(2^{3x} = 2^{3 \cdot 1} = 2^3 = 8\)[/tex], so [tex]\(f = 8\)[/tex].
The completed table is:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline $x$ & $2^x$ & $3 \cdot 2^x$ & $2^{3 x}$ \\ \hline -2 & $\frac{1}{4}$ & $\frac{3}{4}$ & $\frac{1}{64}$ \\ \hline -1 & $\frac{1}{2}$ & $\frac{3}{2}$ & $\frac{1}{8}$ \\ \hline 0 & 1 & 3 & 1 \\ \hline 1 & 2 & 6 & 8 \\ \hline 2 & 4 & 12 & 64 \\ \hline \end{tabular} \][/tex]
