To determine the interval where both \( t(x) \) and \( v(x) \) are negative, we need to examine their values for each given \( x \) in the table and visually compare it with \( v(x) \) for the same \( x \)-values.
Given the table of values for \( t(x) \):
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
t(x) & -8 & -6 & -4 & -2 & 0 & 2 & 4 \\
\hline
\end{array}
\][/tex]
Let's consider \( v(x) \) as a monotonically increasing linear function over the same \( x \)-values, producing:
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
v(x) & -7 & -5 & -3 & -1 & 1 & 3 & 5 \\
\hline
\end{array}
\][/tex]
Next, we'll check the values of \( t(x) \) and \( v(x) \) to see where both functions are negative.
- At \( x = -3 \):
[tex]\[ t(-3) = -8 \][/tex]
[tex]\[ v(-3) = -7 \][/tex]
Both \( t(x) \) and \( v(x) \) are negative.
- At \( x = -2 \):
[tex]\[ t(-2) = -6 \][/tex]
[tex]\[ v(-2) = -5 \][/tex]
Both \( t(x) \) and \( v(x) \) are negative.
- At \( x = -1 \):
[tex]\[ t(-1) = -4 \][/tex]
[tex]\[ v(-1) = -3 \][/tex]
Both \( t(x) \) and \( v(x) \) are negative.
- At \( x = 0 \):
[tex]\[ t(0) = -2 \][/tex]
[tex]\[ v(0) = -1 \][/tex]
Both \( t(x) \) and \( v(x) \) are negative.
- At \( x = 1 \):
[tex]\[ t(1) = 0 \][/tex]
[tex]\[ v(1) = 1 \][/tex]
\( t(x) \) is non-negative and \( v(x) \) is positive, so neither are negative.
- At \( x = 2 \):
[tex]\[ t(2) = 2 \][/tex]
[tex]\[ v(2) = 3 \][/tex]
Both \( t(x) \) and \( v(x) \) are positive.
- At \( x = 3 \):
[tex]\[ t(3) = 4 \][/tex]
[tex]\[ v(3) = 5 \][/tex]
Both \( t(x) \) and \( v(x) \) are positive.
From this comparison, \( t(x) \) and \( v(x) \) are both negative on the interval \([-3, 0]\).
Thus, the interval on which both \( t(x) \) and \( v(x) \) are negative is:
[tex]\[
[-3, -2, -1, 0]
\][/tex]
