Let's complete the table of values for the functions \( f(x) = 3x^2 + 5 \) and \( g(x) = 2x + 8 \).
First, we'll calculate the values for each function at \( x = 0, 1, 2, \) and \( 3 \).
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For \( x = 0 \):
- \( f(0) = 3(0)^2 + 5 = 5 \)
- \( g(0) = 2(0) + 8 = 8 \)
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For \( x = 1 \):
- \( f(1) = 3(1)^2 + 5 = 3 + 5 = 8 \)
- \( g(1) = 2(1) + 8 = 2 + 8 = 10 \)
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For \( x = 2 \):
- \( f(2) = 3(2)^2 + 5 = 3(4) + 5 = 12 + 5 = 17 \)
- \( g(2) = 2(2) + 8 = 4 + 8 = 12 \)
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For \( x = 3 \):
- \( f(3) = 3(3)^2 + 5 = 3(9) + 5 = 27 + 5 = 32 \)
- \( g(3) = 2(3) + 8 = 6 + 8 = 14 \)
Now we can fill out the completed table:
| x | f(x) | g(x) | |---|------|------| | 0 | 5 | 8 | | 1 | 8 | 10 | | 2 | 17 | 12 | | 3 | 32 | 14 |
Now, let's analyze the growth of both functions:
- At \( x = 0 \), \( g(x) \) is greater (8 > 5).
- At \( x = 1 \), \( g(x) \) is again greater (10 > 8).
- At \( x = 2 \), \( f(x) \) exceeds \( g(x) \) (17 > 12).
- At \( x = 3 \), \( f(x) \) is much greater than \( g(x) \) (32 > 14).
From this analysis, we can see that \( g(x) \) starts out greater for small values of \( x \), but as \( x \) increases, \( f(x) \) eventually exceeds \( g(x) \) and continues to grow at a faster rate due to its quadratic nature. Thus, \( f(x) \) will eventually exceed \( g(x) \) for sufficiently large \( x \).
In summary, the function \( f(x) \) eventually exceeds the function \( g(x) \).
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