Given the continuous functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex], let's determine the [tex]\( x \)[/tex]-intercepts of each function and compare them.
### For Function [tex]\( f \)[/tex]:
The provided table of values shows the [tex]\( x \)[/tex] and [tex]\( f(x) \)[/tex]:
[tex]\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
2 & \\
\hline
-1 & -4 \\
\hline
0 & -2 \\
\hline
3 & 0 \\
\hline
8 & 2 \\
\hline
15 & 4 \\
\hline
24 & 6 \\
\hline
\end{array}
\][/tex]
To find the [tex]\( x \)[/tex]-intercept of [tex]\( f \)[/tex], we look for where [tex]\( f(x) = 0 \)[/tex]. From the table, we see that [tex]\( f(3) = 0 \)[/tex]. Therefore, the [tex]\( x \)[/tex]-intercept of [tex]\( f \)[/tex] is [tex]\( 3 \)[/tex].
### For Function [tex]\( g \)[/tex]:
The function [tex]\( g \)[/tex] is defined as:
[tex]\[ g(x) = 2 + (3x + 1)^{1/3} \][/tex]
To find the [tex]\( x \)[/tex]-intercept of [tex]\( g \)[/tex], we solve for [tex]\( x \)[/tex] such that [tex]\( g(x) = 0 \)[/tex]:
[tex]\[
0 = 2 + (3x + 1)^{1/3}
\][/tex]
Rearranging the equation:
[tex]\[
(3x + 1)^{1/3} = -2
\][/tex]
Cubing both sides gives:
[tex]\[
3x + 1 = (-2)^3
\][/tex]
[tex]\[
3x + 1 = -8
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
3x = -8 - 1
\][/tex]
[tex]\[
3x = -9
\][/tex]
[tex]\[
x = -3
\][/tex]
Therefore, the [tex]\( x \)[/tex]-intercept of [tex]\( g \)[/tex] is [tex]\(-3\)[/tex].
### Comparison:
We now compare the [tex]\( x \)[/tex]-intercept of [tex]\( f \)[/tex] (which is [tex]\( 3 \)[/tex]) with the [tex]\( x \)[/tex]-intercept of [tex]\( g \)[/tex] (which is [tex]\(-3\)[/tex]).
Clearly, [tex]\( 3 \)[/tex] is greater than [tex]\(-3\)[/tex].
So the completed statement should read:
The [tex]\( x \)[/tex]-intercept of function [tex]\( f \)[/tex] is greater than the [tex]\( x \)[/tex]-intercept of function [tex]\( g \)[/tex].
