Answer :
Let's analyze if each given table of values represents an even function, an odd function, or neither.
1. Analysis of [tex]\( f(x) \)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -4 & -2 & 0 & 2 & 4 \\ \hline f(x) & 8 & 1 & 0 & -1 & -8 \\ \hline \end{array} \][/tex]
To determine if [tex]\( f(x) \)[/tex] is even, odd, or neither, we need to check the properties:
- Even Function: [tex]\( f(x) = f(-x) \)[/tex]
- Odd Function: [tex]\( f(x) = -f(-x) \)[/tex]
Let’s check:
- [tex]\( f(4) = -8 \)[/tex] and [tex]\( f(-4) = 8 \)[/tex] → [tex]\( f(4) = -f(-4) \)[/tex]
- [tex]\( f(2) = -1 \)[/tex] and [tex]\( f(-2) = 1 \)[/tex] → [tex]\( f(2) = -f(-2) \)[/tex]
- [tex]\( f(0) = 0 \)[/tex]
Since [tex]\( f(x) = -f(-x) \)[/tex] for all values, [tex]\( f(x) \)[/tex] is an odd function.
2. Analysis of [tex]\( g(x) \)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -4 & -2 & 0 & 2 & 4 \\ \hline g(x) & -4 & 2 & 4 & 2 & -4 \\ \hline \end{array} \][/tex]
Let’s check:
- [tex]\( g(4) = -4 \)[/tex] and [tex]\( g(-4) = -4 \)[/tex] → [tex]\( g(4) = g(-4) \)[/tex]
- [tex]\( g(2) = 2 \)[/tex] and [tex]\( g(-2) = 2 \)[/tex] → [tex]\( g(2) = g(-2) \)[/tex]
- [tex]\( g(0) = 4 \)[/tex]
Since [tex]\( g(x) = g(-x) \)[/tex] for all values, [tex]\( g(x) \)[/tex] is an even function.
3. Analysis of [tex]\( j(x) \)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline j(x) & 2 & -4 & -6 & -4 & 2 \\ \hline \end{array} \][/tex]
Let’s check:
- [tex]\( j(2) = 2 \)[/tex] and [tex]\( j(-2) = 2 \)[/tex] → [tex]\( j(2) = j(-2) \)[/tex]
- [tex]\( j(1) = -4 \)[/tex] and [tex]\( j(-1) = -4 \)[/tex] → [tex]\( j(1) = j(-1) \)[/tex]
- [tex]\( j(0) = -6 \)[/tex]
Since [tex]\( j(x) = j(-x) \)[/tex] for all values, [tex]\( j(x) \)[/tex] is an even function.
4. Analysis of [tex]\( k(x) \)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -4 & -2 & 0 & 2 & 4 \\ \hline k(x) & 9 & 4 & 1 & 0 & 1 \\ \hline \end{array} \][/tex]
Let’s check:
- [tex]\( k(4) = 1 \)[/tex] and [tex]\( k(-4) = 9 \)[/tex] → [tex]\( k(4) \neq k(-4) \)[/tex] and [tex]\( k(4) \neq -k(-4) \)[/tex]
- [tex]\( k(2) = 0 \)[/tex] and [tex]\( k(-2) = 4 \)[/tex] → [tex]\( k(2) \neq k(-2) \)[/tex] and [tex]\( k(2) \neq -k(-2) \)[/tex]
- [tex]\( k(0) = 1 \)[/tex]
Since neither [tex]\( k(x) = k(-x) \)[/tex] nor [tex]\( k(x) = -k(-x) \)[/tex] is true for all values, [tex]\( k(x) \)[/tex] is neither an even function nor an odd function.
Summary:
- [tex]\( f(x) \)[/tex] is odd.
- [tex]\( g(x) \)[/tex] is even.
- [tex]\( j(x) \)[/tex] is even.
- [tex]\( k(x) \)[/tex] is neither.
So, the table should be filled as follows:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline x & -4 & -2 & 0 & 2 & 4 \\ \hline f(x) & 8 & 1 & 0 & -1 & -8 & \text{odd} \\ \hline \end{tabular} \][/tex]
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline x & -4 & -2 & 0 & 2 & 4 \\ \hline g(x) & -4 & 2 & 4 & 2 & -4 & \text{even} \\ \hline \end{tabular} \][/tex]
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline j(x) & 2 & -4 & -6 & -4 & 2 & \text{even} \\ \hline \end{tabular} \][/tex]
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline x & -4 & -2 & 0 & 2 & 4 \\ \hline k(x) & 9 & 4 & 1 & 0 & 1 & \text{neither} \\ \hline \end{tabular} \][/tex]
1. Analysis of [tex]\( f(x) \)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -4 & -2 & 0 & 2 & 4 \\ \hline f(x) & 8 & 1 & 0 & -1 & -8 \\ \hline \end{array} \][/tex]
To determine if [tex]\( f(x) \)[/tex] is even, odd, or neither, we need to check the properties:
- Even Function: [tex]\( f(x) = f(-x) \)[/tex]
- Odd Function: [tex]\( f(x) = -f(-x) \)[/tex]
Let’s check:
- [tex]\( f(4) = -8 \)[/tex] and [tex]\( f(-4) = 8 \)[/tex] → [tex]\( f(4) = -f(-4) \)[/tex]
- [tex]\( f(2) = -1 \)[/tex] and [tex]\( f(-2) = 1 \)[/tex] → [tex]\( f(2) = -f(-2) \)[/tex]
- [tex]\( f(0) = 0 \)[/tex]
Since [tex]\( f(x) = -f(-x) \)[/tex] for all values, [tex]\( f(x) \)[/tex] is an odd function.
2. Analysis of [tex]\( g(x) \)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -4 & -2 & 0 & 2 & 4 \\ \hline g(x) & -4 & 2 & 4 & 2 & -4 \\ \hline \end{array} \][/tex]
Let’s check:
- [tex]\( g(4) = -4 \)[/tex] and [tex]\( g(-4) = -4 \)[/tex] → [tex]\( g(4) = g(-4) \)[/tex]
- [tex]\( g(2) = 2 \)[/tex] and [tex]\( g(-2) = 2 \)[/tex] → [tex]\( g(2) = g(-2) \)[/tex]
- [tex]\( g(0) = 4 \)[/tex]
Since [tex]\( g(x) = g(-x) \)[/tex] for all values, [tex]\( g(x) \)[/tex] is an even function.
3. Analysis of [tex]\( j(x) \)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline j(x) & 2 & -4 & -6 & -4 & 2 \\ \hline \end{array} \][/tex]
Let’s check:
- [tex]\( j(2) = 2 \)[/tex] and [tex]\( j(-2) = 2 \)[/tex] → [tex]\( j(2) = j(-2) \)[/tex]
- [tex]\( j(1) = -4 \)[/tex] and [tex]\( j(-1) = -4 \)[/tex] → [tex]\( j(1) = j(-1) \)[/tex]
- [tex]\( j(0) = -6 \)[/tex]
Since [tex]\( j(x) = j(-x) \)[/tex] for all values, [tex]\( j(x) \)[/tex] is an even function.
4. Analysis of [tex]\( k(x) \)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -4 & -2 & 0 & 2 & 4 \\ \hline k(x) & 9 & 4 & 1 & 0 & 1 \\ \hline \end{array} \][/tex]
Let’s check:
- [tex]\( k(4) = 1 \)[/tex] and [tex]\( k(-4) = 9 \)[/tex] → [tex]\( k(4) \neq k(-4) \)[/tex] and [tex]\( k(4) \neq -k(-4) \)[/tex]
- [tex]\( k(2) = 0 \)[/tex] and [tex]\( k(-2) = 4 \)[/tex] → [tex]\( k(2) \neq k(-2) \)[/tex] and [tex]\( k(2) \neq -k(-2) \)[/tex]
- [tex]\( k(0) = 1 \)[/tex]
Since neither [tex]\( k(x) = k(-x) \)[/tex] nor [tex]\( k(x) = -k(-x) \)[/tex] is true for all values, [tex]\( k(x) \)[/tex] is neither an even function nor an odd function.
Summary:
- [tex]\( f(x) \)[/tex] is odd.
- [tex]\( g(x) \)[/tex] is even.
- [tex]\( j(x) \)[/tex] is even.
- [tex]\( k(x) \)[/tex] is neither.
So, the table should be filled as follows:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline x & -4 & -2 & 0 & 2 & 4 \\ \hline f(x) & 8 & 1 & 0 & -1 & -8 & \text{odd} \\ \hline \end{tabular} \][/tex]
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline x & -4 & -2 & 0 & 2 & 4 \\ \hline g(x) & -4 & 2 & 4 & 2 & -4 & \text{even} \\ \hline \end{tabular} \][/tex]
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline j(x) & 2 & -4 & -6 & -4 & 2 & \text{even} \\ \hline \end{tabular} \][/tex]
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline x & -4 & -2 & 0 & 2 & 4 \\ \hline k(x) & 9 & 4 & 1 & 0 & 1 & \text{neither} \\ \hline \end{tabular} \][/tex]
