Answer :
To determine where the function [tex]\( f(x) \)[/tex] crosses zero, we need to identify intervals where the value of [tex]\( f(x) \)[/tex] changes sign from negative to positive or from positive to negative. Let's examine the table of values:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -2 & -11.5 \\ \hline -1 & -0.5 \\ \hline 0 & 2.5 \\ \hline 1 & 3.5 \\ \hline 2 & 8.5 \\ \hline \end{tabular} \][/tex]
1. Examine the intervals:
- Between [tex]\( x = -2 \)[/tex] and [tex]\( x = -1 \)[/tex]:
- [tex]\( f(-2) = -11.5 \)[/tex] and [tex]\( f(-1) = -0.5 \)[/tex]
- Both are negative, so no zero crossing here.
- Between [tex]\( x = -1 \)[/tex] and [tex]\( x = 0 \)[/tex]:
- [tex]\( f(-1) = -0.5 \)[/tex] and [tex]\( f(0) = 2.5 \)[/tex]
- Here [tex]\( f(x) \)[/tex] changes from negative to positive, indicating a zero crossing.
- Between [tex]\( x = 0 \)[/tex] and [tex]\( x = 1 \)[/tex]:
- [tex]\( f(0) = 2.5 \)[/tex] and [tex]\( f(1) = 3.5 \)[/tex]
- Both are positive, so no zero crossing here.
- Between [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex]:
- [tex]\( f(1) = 3.5 \)[/tex] and [tex]\( f(2) = 8.5 \)[/tex]
- Both are positive, so no zero crossing here.
2. Identify interval with a zero crossing:
- The only interval where the sign of [tex]\( f(x) \)[/tex] changes is between [tex]\( x = -1 \)[/tex] and [tex]\( x = 0 \)[/tex].
3. Approximate the zero:
- Since within the interval [tex]\([-1, 0]\)[/tex], [tex]\( f(x) \)[/tex] changes from negative to positive, an approximate zero of the function is within this interval. Given the data available, we approximate the zero of the function as occurring at [tex]\( x = 0 \)[/tex].
So, based on this analysis, we can complete the statements as follows:
A zero can be found between input values of
[tex]\[ \boxed{-1 \text{ and } 0} \][/tex]
because
[tex]\[ f(x) \text{ changes sign from negative to positive.} \][/tex]
One zero of the function is approximately
[tex]\[ \boxed{0} \][/tex]
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -2 & -11.5 \\ \hline -1 & -0.5 \\ \hline 0 & 2.5 \\ \hline 1 & 3.5 \\ \hline 2 & 8.5 \\ \hline \end{tabular} \][/tex]
1. Examine the intervals:
- Between [tex]\( x = -2 \)[/tex] and [tex]\( x = -1 \)[/tex]:
- [tex]\( f(-2) = -11.5 \)[/tex] and [tex]\( f(-1) = -0.5 \)[/tex]
- Both are negative, so no zero crossing here.
- Between [tex]\( x = -1 \)[/tex] and [tex]\( x = 0 \)[/tex]:
- [tex]\( f(-1) = -0.5 \)[/tex] and [tex]\( f(0) = 2.5 \)[/tex]
- Here [tex]\( f(x) \)[/tex] changes from negative to positive, indicating a zero crossing.
- Between [tex]\( x = 0 \)[/tex] and [tex]\( x = 1 \)[/tex]:
- [tex]\( f(0) = 2.5 \)[/tex] and [tex]\( f(1) = 3.5 \)[/tex]
- Both are positive, so no zero crossing here.
- Between [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex]:
- [tex]\( f(1) = 3.5 \)[/tex] and [tex]\( f(2) = 8.5 \)[/tex]
- Both are positive, so no zero crossing here.
2. Identify interval with a zero crossing:
- The only interval where the sign of [tex]\( f(x) \)[/tex] changes is between [tex]\( x = -1 \)[/tex] and [tex]\( x = 0 \)[/tex].
3. Approximate the zero:
- Since within the interval [tex]\([-1, 0]\)[/tex], [tex]\( f(x) \)[/tex] changes from negative to positive, an approximate zero of the function is within this interval. Given the data available, we approximate the zero of the function as occurring at [tex]\( x = 0 \)[/tex].
So, based on this analysis, we can complete the statements as follows:
A zero can be found between input values of
[tex]\[ \boxed{-1 \text{ and } 0} \][/tex]
because
[tex]\[ f(x) \text{ changes sign from negative to positive.} \][/tex]
One zero of the function is approximately
[tex]\[ \boxed{0} \][/tex]