For the plecewise linear function, find \( (a) f(-4),(b) f(-2) \), (c) \( f(0) \), (d) \( f(2) \), and \( (e) f(5) \) \( f(x)=\left\{\begin{array}{ll}2 x & \text { if } x \leq-2 \\ x-2 & \text { if } x>-2\end{array}\right. \) (a) \( f(-4)=\square \) (b) \( f(-2)=\square \) (c) \( f(0)=\square \) (d) \( f(2)=\square \)

To solve for \( f(x) \) at the given values, we need to determine which piece of the piecewise function to use for each input. For \( (a) f(-4) \): Since \(-4 \leq -2\), we use the first piece: \[ f(-4) = 2(-4) = -8 \] So, \( f(-4) = -8 \). For \( (b) f(-2) \): Here, we still use the first piece since \(-2\) is equal to \(-2\): \[ f(-2) = 2(-2) = -4 \] Thus, \( f(-2) = -4 \). For \( (c) f(0) \): Since \(0 > -2\), we switch to the second piece: \[ f(0) = 0 - 2 = -2 \] Therefore, \( f(0) = -2 \). For \( (d) f(2) \): Again, since \(2 > -2\), we use the second piece: \[ f(2) = 2 - 2 = 0 \] So, \( f(2) = 0 \). Thus, we have summarized the results as: (a) \( f(-4)=-8 \) (b) \( f(-2)=-4 \) (c) \( f(0)=-2 \) (d) \( f(2)=0 \)