Answer :
Certainly! Let's evaluate the expression \( \frac{2x^2 + 4x - 6}{x - 1} \) for each value of \( x \) listed in the table step by step:
1. For \( x = 0 \):
[tex]\[ \frac{2 \cdot 0^2 + 4 \cdot 0 - 6}{0 - 1} = \frac{0 + 0 - 6}{-1} = \frac{-6}{-1} = 6.0 \][/tex]
2. For \( x = 0.5 \):
[tex]\[ \frac{2 \cdot 0.5^2 + 4 \cdot 0.5 - 6}{0.5 - 1} = \frac{2 \cdot 0.25 + 2 - 6}{-0.5} = \frac{0.5 + 2 - 6}{-0.5} = \frac{-3.5}{-0.5} = 7.0 \][/tex]
3. For \( x = 0.9 \):
[tex]\[ \frac{2 \cdot 0.9^2 + 4 \cdot 0.9 - 6}{0.9 - 1} = \frac{2 \cdot 0.81 + 3.6 - 6}{-0.1} = \frac{1.62 + 3.6 - 6}{-0.1} = \frac{-0.78}{-0.1} \approx 7.8 \][/tex]
(\(7.8\) is rounded from \(7.799999999999995\))
4. For \( x = 0.99 \):
[tex]\[ \frac{2 \cdot 0.99^2 + 4 \cdot 0.99 - 6}{0.99 - 1} = \frac{2 \cdot 0.9801 + 3.96 - 6}{-0.01} = \frac{1.9602 + 3.96 - 6}{-0.01} = \frac{-0.0798}{-0.01} \approx 7.98 \][/tex]
(\(7.98\) is approximately \(7.980000000000047\))
5. For \( x = 0.999 \):
[tex]\[ \frac{2 \cdot 0.999^2 + 4 \cdot 0.999 - 6}{0.999 - 1} = \frac{2 \cdot 0.998001 + 3.996 - 6}{-0.001} = \frac{1.996002 + 3.996 - 6}{-0.001} = \frac{-0.007998}{-0.001} \approx 8.0 \][/tex]
(\(8.0\) is approximately \(7.99799999999972\))
6. For \( x = 1.001 \):
[tex]\[ \frac{2 \cdot 1.001^2 + 4 \cdot 1.001 - 6}{1.001 - 1} = \frac{2 \cdot 1.002001 + 4.004 - 6}{0.001} = \frac{2.004002 + 4.004 - 6}{0.001} = \frac{0.008002}{0.001} \approx 8.0 \][/tex]
(\(8.0\) is approximately \(8.00200000000028\))
7. For \( x = 1.01 \):
[tex]\[ \frac{2 \cdot 1.01^2 + 4 \cdot 1.01 - 6}{1.01 - 1} = \frac{2 \cdot 1.0201 + 4.04 - 6}{0.01} = \frac{2.0402 + 4.04 - 6}{0.01} = \frac{0.0802}{0.01} \approx 8.0 \][/tex]
(\(8.0\) is approximately \(8.019999999999953\))
8. For \( x = 1.1 \):
[tex]\[ \frac{2 \cdot 1.1^2 + 4 \cdot 1.1 - 6}{1.1 - 1} = \frac{2 \cdot 1.21 + 4.4 - 6}{0.1} = \frac{2.42 + 4.4 - 6}{0.1} = \frac{0.82}{0.1} \approx 8.2 \][/tex]
(\(8.2\) is rounded from \(8.199999999999996\))
9. For \( x = 1.5 \):
[tex]\[ \frac{2 \cdot 1.5^2 + 4 \cdot 1.5 - 6}{1.5 - 1} = \frac{2 \cdot 2.25 + 6 - 6}{0.5} = \frac{4.5 + 6 - 6}{0.5} = \frac{4.5}{0.5} = 9.0 \][/tex]
10. For \( x = 2 \):
[tex]\[ \frac{2 \cdot 2^2 + 4 \cdot 2 - 6}{2 - 1} = \frac{2 \cdot 4 + 8 - 6}{1} = \frac{8 + 8 - 6}{1} = \frac{10}{1} = 10.0 \][/tex]
Now, we can fill in the table with these evaluations:
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$\frac{2 x^2+4 x-6}{x-1}$[/tex] \\
\hline
0 & 6.0 \\
0.5 & 7.0 \\
0.9 & 7.8 \\
0.99 & 7.98 \\
0.999 & 8.0 \\
1.001 & 8.0 \\
1.01 & 8.0 \\
1.1 & 8.2 \\
1.5 & 9.0 \\
2 & 10.0 \\
\hline
\end{tabular}
And that's the completed solution for each [tex]\( x \)[/tex] in the given table.
1. For \( x = 0 \):
[tex]\[ \frac{2 \cdot 0^2 + 4 \cdot 0 - 6}{0 - 1} = \frac{0 + 0 - 6}{-1} = \frac{-6}{-1} = 6.0 \][/tex]
2. For \( x = 0.5 \):
[tex]\[ \frac{2 \cdot 0.5^2 + 4 \cdot 0.5 - 6}{0.5 - 1} = \frac{2 \cdot 0.25 + 2 - 6}{-0.5} = \frac{0.5 + 2 - 6}{-0.5} = \frac{-3.5}{-0.5} = 7.0 \][/tex]
3. For \( x = 0.9 \):
[tex]\[ \frac{2 \cdot 0.9^2 + 4 \cdot 0.9 - 6}{0.9 - 1} = \frac{2 \cdot 0.81 + 3.6 - 6}{-0.1} = \frac{1.62 + 3.6 - 6}{-0.1} = \frac{-0.78}{-0.1} \approx 7.8 \][/tex]
(\(7.8\) is rounded from \(7.799999999999995\))
4. For \( x = 0.99 \):
[tex]\[ \frac{2 \cdot 0.99^2 + 4 \cdot 0.99 - 6}{0.99 - 1} = \frac{2 \cdot 0.9801 + 3.96 - 6}{-0.01} = \frac{1.9602 + 3.96 - 6}{-0.01} = \frac{-0.0798}{-0.01} \approx 7.98 \][/tex]
(\(7.98\) is approximately \(7.980000000000047\))
5. For \( x = 0.999 \):
[tex]\[ \frac{2 \cdot 0.999^2 + 4 \cdot 0.999 - 6}{0.999 - 1} = \frac{2 \cdot 0.998001 + 3.996 - 6}{-0.001} = \frac{1.996002 + 3.996 - 6}{-0.001} = \frac{-0.007998}{-0.001} \approx 8.0 \][/tex]
(\(8.0\) is approximately \(7.99799999999972\))
6. For \( x = 1.001 \):
[tex]\[ \frac{2 \cdot 1.001^2 + 4 \cdot 1.001 - 6}{1.001 - 1} = \frac{2 \cdot 1.002001 + 4.004 - 6}{0.001} = \frac{2.004002 + 4.004 - 6}{0.001} = \frac{0.008002}{0.001} \approx 8.0 \][/tex]
(\(8.0\) is approximately \(8.00200000000028\))
7. For \( x = 1.01 \):
[tex]\[ \frac{2 \cdot 1.01^2 + 4 \cdot 1.01 - 6}{1.01 - 1} = \frac{2 \cdot 1.0201 + 4.04 - 6}{0.01} = \frac{2.0402 + 4.04 - 6}{0.01} = \frac{0.0802}{0.01} \approx 8.0 \][/tex]
(\(8.0\) is approximately \(8.019999999999953\))
8. For \( x = 1.1 \):
[tex]\[ \frac{2 \cdot 1.1^2 + 4 \cdot 1.1 - 6}{1.1 - 1} = \frac{2 \cdot 1.21 + 4.4 - 6}{0.1} = \frac{2.42 + 4.4 - 6}{0.1} = \frac{0.82}{0.1} \approx 8.2 \][/tex]
(\(8.2\) is rounded from \(8.199999999999996\))
9. For \( x = 1.5 \):
[tex]\[ \frac{2 \cdot 1.5^2 + 4 \cdot 1.5 - 6}{1.5 - 1} = \frac{2 \cdot 2.25 + 6 - 6}{0.5} = \frac{4.5 + 6 - 6}{0.5} = \frac{4.5}{0.5} = 9.0 \][/tex]
10. For \( x = 2 \):
[tex]\[ \frac{2 \cdot 2^2 + 4 \cdot 2 - 6}{2 - 1} = \frac{2 \cdot 4 + 8 - 6}{1} = \frac{8 + 8 - 6}{1} = \frac{10}{1} = 10.0 \][/tex]
Now, we can fill in the table with these evaluations:
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$\frac{2 x^2+4 x-6}{x-1}$[/tex] \\
\hline
0 & 6.0 \\
0.5 & 7.0 \\
0.9 & 7.8 \\
0.99 & 7.98 \\
0.999 & 8.0 \\
1.001 & 8.0 \\
1.01 & 8.0 \\
1.1 & 8.2 \\
1.5 & 9.0 \\
2 & 10.0 \\
\hline
\end{tabular}
And that's the completed solution for each [tex]\( x \)[/tex] in the given table.
