To compare the functions \( f(x) \) and \( g(x) \) based on the provided values in the table:
| \( x \) | \( f(x) \) | \( g(x) \) | |-----|--------|--------| | -2 | -1 | -1 | | -1 | 34 | 0 | | 0 | 1 | 1 | | 1 | 114 | 2 | | 2 | 3 | 3 |
Now, let's compare \( f(x) \) and \( g(x) \):
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For \( x = -2 \):
- \( f(-2) = -1 \)
- \( g(-2) = -1 \)
- \( f(-2) \) is not greater than \( g(-2) \).
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For \( x = -1 \):
- \( f(-1) = 34 \)
- \( g(-1) = 0 \)
- \( f(-1) \) is greater than \( g(-1) \).
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For \( x = 0 \):
- \( f(0) = 1 \)
- \( g(0) = 1 \)
- \( f(0) \) is not greater than \( g(0) \).
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For \( x = 1 \):
- \( f(1) = 114 \)
- \( g(1) = 2 \)
- \( f(1) \) is greater than \( g(1) \).
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For \( x = 2 \):
- \( f(2) = 3 \)
- \( g(2) = 3 \)
- \( f(2) \) is not greater than \( g(2) \).
Based on the comparisons:
- \( f(x) > g(x) \) when \( x = -1 \) and \( x = 1 \).
Thus, the answer is: f(x) is greater than g(x) when \( -1 < x < 2 \) (taking into account that it does not hold exactly at 2).