Let's first calculate the values of \( f(x) = 3x + 1 \) and \( g(x) = x^2 + 5 \) for the given values of \( x \).
Calculating Values
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For \( x = 0 \):
- \( f(0) = 3(0) + 1 = 1 \)
- \( g(0) = 0^2 + 5 = 5 \)
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For \( x = 1 \):
- \( f(1) = 3(1) + 1 = 4 \)
- \( g(1) = 1^2 + 5 = 6 \)
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For \( x = 2 \):
- \( f(2) = 3(2) + 1 = 7 \)
- \( g(2) = 2^2 + 5 = 9 \)
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For \( x = 3 \):
- \( f(3) = 3(3) + 1 = 10 \)
- \( g(3) = 3^2 + 5 = 14 \)
Filling the Table
Now we can complete the table:
\[ \begin{array}{|c|c|c|} \hline x & f(x) & g(x) \ \hline 0 & 1 & 5 \ 1 & 4 & 6 \ 2 & 7 & 9 \ 3 & 10 & 14 \ \hline \end{array} \]
Analyzing Growth of Functions
Next, we need to determine which function eventually exceeds the other as \( x \) becomes larger.
- \( f(x) \) is a linear function, growing at a constant rate of 3.
- \( g(x) \) is a quadratic function, which grows faster as \( x \) increases because quadratic functions increase faster than linear functions for sufficiently large \( x \).
To find out when \( g(x) \) will exceed \( f(x) \), we can set them equal to each other and solve for \( x \):
\[ 3x + 1 = x^2 + 5 \]
Rearranging gives us:
\[ x^2 - 3x + 4 = 0 \]
Calculating the discriminant:
\[ D = b^2 - 4ac = (-3)^2 - 4(1)(4) = 9 - 16 = -7 \]
Since the discriminant is negative, this means the equation has no real solutions which indicates that \( g(x) \) starts exceeding \( f(x) \) at all \( x \) greater than zero.
Conclusion
Thus, while both functions grow, \( g(x) \) will eventually exceed \( f(x) \) as \( x \) increases, particularly noticeable after \( x = 3 \).
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