Answer :
To determine which table correctly represents the function values for the piecewise function [tex]\( f(x) \)[/tex], we need to evaluate [tex]\( f(x) \)[/tex] at the given [tex]\( x \)[/tex] values from each table and compare the results.
The piecewise function is provided as:
[tex]\[ f(x)=\left\{\begin{array}{ll} -\frac{5}{4} x+90, & 0 \leq x<40 \\ -\frac{3}{8} x+75, & 40 \leq x \leq 200 \end{array}\right. \][/tex]
Let's verify each table step-by-step:
### Table 1
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 90 \\ \hline 16 & 85 \\ \hline 40 & 75 \\ \hline \end{array} \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -\frac{5}{4}(0) + 90 = 90 \quad (\text{matches}) \][/tex]
- For [tex]\( x = 16 \)[/tex]:
[tex]\[ f(16) = -\frac{5}{4}(16) + 90 = -20 + 90 = 70 \quad (\text{does not match}) \][/tex]
- For [tex]\( x = 40 \)[/tex]:
[tex]\[ f(40) = -\frac{3}{8}(40) + 75 = -15 + 75 = 60 \quad (\text{does not match}) \][/tex]
Since [tex]\( f(x) \)[/tex] does not match for all [tex]\( x \)[/tex] values in the table, Table 1 is incorrect.
### Table 2
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 90 \\ \hline 40 & 40 \\ \hline 200 & 0 \\ \hline \end{array} \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -\frac{5}{4}(0) + 90 = 90 \quad (\text{matches}) \][/tex]
- For [tex]\( x = 40 \)[/tex]:
[tex]\[ f(40) = -\frac{3}{8}(40) + 75 = -15 + 75 = 60 \quad (\text{does not match}) \][/tex]
- For [tex]\( x = 200 \)[/tex]:
[tex]\[ f(200) = -\frac{3}{8}(200) + 75 = -75 + 75 = 0 \quad (\text{matches}) \][/tex]
Since [tex]\( f(x) \)[/tex] does not match for all [tex]\( x \)[/tex] values in the table, Table 2 is incorrect.
### Table 3
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 40 & 75 \\ \hline 120 & 30 \\ \hline 200 & 0 \\ \hline \end{array} \][/tex]
- For [tex]\( x = 40 \)[/tex]:
[tex]\[ f(40) = -\frac{3}{8}(40) + 75 = -15 + 75 = 60 \quad (\text{does not match}) \][/tex]
- For [tex]\( x = 120 \)[/tex]:
[tex]\[ f(120) = -\frac{3}{8}(120) + 75 = -45 + 75 = 30 \quad (\text{matches}) \][/tex]
- For [tex]\( x = 200 \)[/tex]:
[tex]\[ f(200) = -\frac{3}{8}(200) + 75 = -75 + 75 = 0 \quad (\text{matches}) \][/tex]
Since [tex]\( f(x) \)[/tex] does not match for all [tex]\( x \)[/tex] values in the table, Table 3 is incorrect.
### Table 4
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 40 & 60 \\ \hline 160 & 15 \\ \hline 200 & 0 \\ \hline \end{array} \][/tex]
- For [tex]\( x = 40 \)[/tex]:
[tex]\[ f(40) = -\frac{3}{8}(40) + 75 = -15 + 75 = 60 \quad (\text{matches}) \][/tex]
- For [tex]\( x = 160 \)[/tex]:
[tex]\[ f(160) = -\frac{3}{8}(160) + 75 = -60 + 75 = 15 \quad (\text{matches}) \][/tex]
- For [tex]\( x = 200 \)[/tex]:
[tex]\[ f(200) = -\frac{3}{8}(200) + 75 = -75 + 75 = 0 \quad (\text{matches}) \][/tex]
Since [tex]\( f(x) \)[/tex] matches for all [tex]\( x \)[/tex] values in the table, Table 4 is correct.
Thus, the correct table is:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 40 & 60 \\ \hline 160 & 15 \\ \hline 200 & 0 \\ \hline \end{array} \][/tex]
The piecewise function is provided as:
[tex]\[ f(x)=\left\{\begin{array}{ll} -\frac{5}{4} x+90, & 0 \leq x<40 \\ -\frac{3}{8} x+75, & 40 \leq x \leq 200 \end{array}\right. \][/tex]
Let's verify each table step-by-step:
### Table 1
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 90 \\ \hline 16 & 85 \\ \hline 40 & 75 \\ \hline \end{array} \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -\frac{5}{4}(0) + 90 = 90 \quad (\text{matches}) \][/tex]
- For [tex]\( x = 16 \)[/tex]:
[tex]\[ f(16) = -\frac{5}{4}(16) + 90 = -20 + 90 = 70 \quad (\text{does not match}) \][/tex]
- For [tex]\( x = 40 \)[/tex]:
[tex]\[ f(40) = -\frac{3}{8}(40) + 75 = -15 + 75 = 60 \quad (\text{does not match}) \][/tex]
Since [tex]\( f(x) \)[/tex] does not match for all [tex]\( x \)[/tex] values in the table, Table 1 is incorrect.
### Table 2
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 90 \\ \hline 40 & 40 \\ \hline 200 & 0 \\ \hline \end{array} \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -\frac{5}{4}(0) + 90 = 90 \quad (\text{matches}) \][/tex]
- For [tex]\( x = 40 \)[/tex]:
[tex]\[ f(40) = -\frac{3}{8}(40) + 75 = -15 + 75 = 60 \quad (\text{does not match}) \][/tex]
- For [tex]\( x = 200 \)[/tex]:
[tex]\[ f(200) = -\frac{3}{8}(200) + 75 = -75 + 75 = 0 \quad (\text{matches}) \][/tex]
Since [tex]\( f(x) \)[/tex] does not match for all [tex]\( x \)[/tex] values in the table, Table 2 is incorrect.
### Table 3
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 40 & 75 \\ \hline 120 & 30 \\ \hline 200 & 0 \\ \hline \end{array} \][/tex]
- For [tex]\( x = 40 \)[/tex]:
[tex]\[ f(40) = -\frac{3}{8}(40) + 75 = -15 + 75 = 60 \quad (\text{does not match}) \][/tex]
- For [tex]\( x = 120 \)[/tex]:
[tex]\[ f(120) = -\frac{3}{8}(120) + 75 = -45 + 75 = 30 \quad (\text{matches}) \][/tex]
- For [tex]\( x = 200 \)[/tex]:
[tex]\[ f(200) = -\frac{3}{8}(200) + 75 = -75 + 75 = 0 \quad (\text{matches}) \][/tex]
Since [tex]\( f(x) \)[/tex] does not match for all [tex]\( x \)[/tex] values in the table, Table 3 is incorrect.
### Table 4
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 40 & 60 \\ \hline 160 & 15 \\ \hline 200 & 0 \\ \hline \end{array} \][/tex]
- For [tex]\( x = 40 \)[/tex]:
[tex]\[ f(40) = -\frac{3}{8}(40) + 75 = -15 + 75 = 60 \quad (\text{matches}) \][/tex]
- For [tex]\( x = 160 \)[/tex]:
[tex]\[ f(160) = -\frac{3}{8}(160) + 75 = -60 + 75 = 15 \quad (\text{matches}) \][/tex]
- For [tex]\( x = 200 \)[/tex]:
[tex]\[ f(200) = -\frac{3}{8}(200) + 75 = -75 + 75 = 0 \quad (\text{matches}) \][/tex]
Since [tex]\( f(x) \)[/tex] matches for all [tex]\( x \)[/tex] values in the table, Table 4 is correct.
Thus, the correct table is:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 40 & 60 \\ \hline 160 & 15 \\ \hline 200 & 0 \\ \hline \end{array} \][/tex]
