To determine which of the given function rules could not satisfy the condition that \( f(2) = 7 \), we will evaluate each function at \( x = 2 \).
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Function 1: \( f(x) = |2x - 5| + 6 \) \[ f(2) = |2(2) - 5| + 6 = |4 - 5| + 6 = | -1 | + 6 = 1 + 6 = 7 \]
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Function 2: \( f(x) = [2x + 3] \) (assuming [ ] denotes floor function) \[ f(2) = [2(2) + 3] = [4 + 3] = [7] = 7 \]
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Function 3: \( f(x) = 2[x] + 3 \) (assuming [ ] denotes floor function) \[ f(2) = 2[2] + 3 = 2(2) + 3 = 4 + 3 = 7 \]
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Function 4: \[ f(x) = \begin{cases} 3x + 1 & \text{if } x < 2 \ x^2 + 1 & \text{if } x \geq 2 \end{cases} \] Since \( 2 \geq 2 \), we use the second part of the piecewise function: \[ f(2) = 2^2 + 1 = 4 + 1 = 5 \]
Based on the evaluations:
- Functions 1, 2, and 3 yield \( f(2) = 7 \).
- Function 4 yields \( f(2) = 5 \).
Thus, the function rule that could not be a function rule for \( f(x) \) if \( f(2) = 7 \) is:
\( f(x) = { 3x + 1 \text{ if } x < 2 \text{ over } x^2 + 1 \text{ if } x \geq 2 } \).