Use the definition of the absolute value to expand and write the function \( f(x)=-2|4-2 x|+5 \) as a piecewise define function. Choose the most correct answer. (A) \( f(x)=\left\{\begin{array}{ll}4 x-3 & \text { if } x \geq 2 \\ 13-4 x & \text { if } x<2\end{array}\right. \) (B) \( f(x)=\left\{\begin{array}{ll}-4 x-3 & \text { if } x \leq 2 \\ 4 x+13 & \text { if } x>2\end{array}\right. \) (C) \( f(x)=\left\{\begin{array}{ll}4 x-3 & \text { if } x \leq 2 \\ 13-4 x & \text { if } x>2\end{array}\right. \) (D) \( f(x)=\left\{\begin{array}{ll}4 x-3 & \text { if } x \geq 0 \\ -4 x+13 & \text { if } x<0\end{array}\right. \) (D) None of these (E)

To express the function \( f(x) = -2|4 - 2x| + 5 \) as a piecewise function, we first evaluate the absolute value based on the critical point where \( 4 - 2x = 0 \) or \( x = 2 \). Thus, we can rewrite \( f(x) \) in two segments. For \( x \geq 2 \), \( |4 - 2x| = 2x - 4 \), leading to \( f(x) = -2(2x - 4) + 5 = -4x + 8 + 5 = -4x + 13 \). For \( x < 2 \), \( |4 - 2x| = 4 - 2x \), so \( f(x) = -2(4 - 2x) + 5 = -8 + 4x + 5 = 4x - 3 \). Thus, we have: \[ f(x) = \begin{cases} 4x - 3 & \text{if } x < 2 \\ -4x + 13 & \text{if } x \geq 2 \end{cases} \] Even though that analysis is crucial, here's a little historical nugget: did you know that the absolute value function has been a vital part of mathematics since the time of the ancient Greeks? They were investigating distances, and the idea of measuring "how far" something is without regard for direction laid the groundwork for our modern understanding! Additionally, in real-world applications, absolute values pop up everywhere! They're crucial in fields like physics when dealing with distances and magnitude irrespective of direction, or in economics when discussing profit or loss without concern for whether a figure is negative or positive. Understanding this can help you analyze data effectively!