Jayhla please come back u are the nicest sweetest

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mark177

09.07.2021

Possible answer: $f(x) = \begin{cases} -2x + 5 & \text{ for }x\le 0 \\ 5 & \text{ for }0\ \textless \ x \ \textless \ 5 \\ 2x-5 & \text{ for } x \ge 5 \end{cases}$ =====definition of absolute value in terms of a piecewise expression $|\text{stuff}| = \begin{cases} \text{stuff} & \text{ if } \text{stuff} \ge 0 \\ -\text{stuff} & \text{ if } \text{stuff} \ \textless \ 0 \end{cases}$ there are two different absolute value parts in the equation for function f.since x - 5 ≥ 0 has solution set x ≥ 5 and x - 5 < 0 has solution set x < 0,for x ≥ 5, |x − 5| = x - 5 because the inside will be positive.for x < 5, |x - 5| = -(x-5) because the inside will be negative; that negative coefficient over the parentheses will turn the negatives into positive.for x ≥ 0, |x| = xfor x < 0, |x| = -xtherefore, we can create a piecewise definition broken apart into three intervals: x ≤ 0, where |x| = -x and |x-5| = -(x-5) 0 < x < 5, where |x| = x and |x-5| = -(x-5) x ≥ 5, where |x| = x and |x-5| = x-5a possible function rule is $f(x) = \begin{cases} (x) + (x-5) & \text{ for } x \ge 5 \\ (x) -(x-5) & \text{ for }0 \ \textless \ x \ \textless \ 5 \\ -(x) -(x-5) & \text{ for }x \le 0 \end{cases}$ which can be simplified if we distribute negatives and combine like terms $f(x) = \begin{cases} 2x-5 & \text{ for } x \ge 5 \\ 5 & \text{ for }0\ \textless \ x \ \textless \ 5 \\ -2x + 5 & \text{ for }x\le 0 \end{cases}$

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Jayhla please come back u are the nicest sweetest person I have ever know so pwase come back! Don't answer this but 15+6

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