To analyze the given function \( f(x) = 2x + 1 \) and the values provided in the table, we first need to compute the differences in the \( f(x) \) values over the specified intervals.
The \( x \) values and their corresponding \( f(x) \) values from the table are:
- For \( x = 2 \), \( f(2) = 5 \)
- For \( x = 7 \), \( f(7) = 15 \)
- For \( x = 12 \), \( f(12) = 25 \)
- For \( x = 17 \), \( f(17) = 35 \)
Now we will calculate the differences in \( f(x) \) for the intervals of 2 and 5.
For intervals of 5:
- From \( x = 2 \) to \( x = 7 \):
- \( f(7) - f(2) = 15 - 5 = 10 \)
- From \( x = 7 \) to \( x = 12 \):
- \( f(12) - f(7) = 25 - 15 = 10 \)
- From \( x = 12 \) to \( x = 17 \):
- \( f(17) - f(12) = 35 - 25 = 10 \)
The difference in values of \( f(x) \) for \( x \)-values over equal intervals of 5 is consistently 10.
For intervals of 2:
- From \( x = 2 \) to \( x = 4 \) (not in the table; assumed for calculation):
- Let's calculate \( f(4) = 2(4) + 1 = 9 \)
- \( f(4) - f(2) = 9 - 5 = 4 \) (this is hypothetical)
- From \( x = 2 \) to \( x = 7 \):
- \( f(7) - f(2) = 15 - 5 = 10 \)
- From \( x = 7 \) to \( x = 12 \):
- \( f(12) - f(7) = 25 - 15 = 10 \)
- From \( x = 12 \) to \( x = 17 \):
- \( f(17) - f(12) = 35 - 25 = 10 \)
While we have calculated differences over different hypothetical intervals, the highlighted consistency shows values increase by 10 for all interval calculations of 5. Therefore, we can confirm:
The correct statement is: The table can be used to show that for x-values over equal intervals of 5, the difference in the values of f(x) for the function y=2x+1 is 10.
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