To determine the [tex]\(x\)[/tex]-intercepts of the functions [tex]\(f\)[/tex] and [tex]\(g\)[/tex], let's go through the steps systematically:
### Step 1: Finding the [tex]\(x\)[/tex]-intercept of the function [tex]\(f\)[/tex]
Given the table of values for [tex]\(f\)[/tex]:
[tex]\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-1 & -4 \\
\hline
0 & -2 \\
\hline
3 & 0 \\
\hline
8 & 2 \\
\hline
15 & 4 \\
\hline
24 & 6 \\
\hline
\end{array}
\][/tex]
The [tex]\(x\)[/tex]-intercept of [tex]\(f\)[/tex] is the value of [tex]\(x\)[/tex] for which [tex]\(f(x) = 0\)[/tex]. From the table, we see that when [tex]\(x = 3\)[/tex], [tex]\(f(x) = 0\)[/tex]. Thus, the [tex]\(x\)[/tex]-intercept of [tex]\(f\)[/tex] is 3.
### Step 2: Finding the [tex]\(x\)[/tex]-intercept of the function [tex]\(g\)[/tex]
The function [tex]\(g\)[/tex] is given by:
[tex]\[ g(x) = 2 + \left(3x + 1\right)^{\frac{1}{3}} \][/tex]
To find the [tex]\(x\)[/tex]-intercept, set [tex]\(g(x) = 0\)[/tex]:
[tex]\[ 0 = 2 + \left(3x + 1\right)^{\frac{1}{3}} \][/tex]
Subtract 2 from both sides:
[tex]\[ -2 = \left(3x + 1\right)^{\frac{1}{3}} \][/tex]
Cube both sides to eliminate the cube root:
[tex]\[ (-2)^3 = 3x + 1 \][/tex]
Calculate [tex]\((-2)^3\)[/tex]:
[tex]\[ -8 = 3x + 1 \][/tex]
Subtract 1 from both sides:
[tex]\[ -9 = 3x \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{-9}{3} \][/tex]
[tex]\[ x = -3 \][/tex]
Thus, the [tex]\(x\)[/tex]-intercept of [tex]\(g\)[/tex] is -3.
### Conclusion
The [tex]\(x\)[/tex]-intercept of function [tex]\(f\)[/tex] is 3, and the [tex]\(x\)[/tex]-intercept of function [tex]\(g\)[/tex] is -3. Therefore, the correct statement is:
The [tex]\(x\)[/tex]-intercept of function [tex]\(f\)[/tex] is [tex]\(\boxed{3}\)[/tex], and the [tex]\(x\)[/tex]-intercept of function [tex]\(g\)[/tex] is [tex]\(\boxed{-3}\)[/tex].
