Complete the table of values for f(x)=3x+1 and g(x)=x2+5.

x f(x) g(x)
0
1
2
3
Both f(x) and g(x) grow as x gets larger and larger. Which function eventually exceeds the other?

1 answer

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

  1. For \( x = 0 \):

    • \( f(0) = 3(0) + 1 = 1 \)
    • \( g(0) = 0^2 + 5 = 5 \)
  2. For \( x = 1 \):

    • \( f(1) = 3(1) + 1 = 4 \)
    • \( g(1) = 1^2 + 5 = 6 \)
  3. For \( x = 2 \):

    • \( f(2) = 3(2) + 1 = 7 \)
    • \( g(2) = 2^2 + 5 = 9 \)
  4. 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 \( x = 3 \).