If f : R rightarrow R is defined by f(x) = be | vedclass

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(D) Given $f(x) = \begin{cases} x + 4, & x < -4 \ 3x + 2, & -4 \leq x < 4 \ x - 4, & x \geq 4 \end{cases}$ $(A) f(-5) + f(-4) = (-5 + 4) + (3(-4) +

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(D) Given $f(x) = \begin{cases} x + 4, & x $(A) f(-5) + f(-4) = (-5 + 4) + (3(-4) + 2) = -1 + (-12 + 2) = -1 - 10 = -11$. Thus, $(A) \rightarrow (iii)$. $(B) f(|f(-8)|) = f(|-8 + 4|) = f(|-4|) = f(4) = 4 - 4 = 0$. Thus, $(B) \rightarrow (vi)$. $(C) f(f(-7) + f(3)) = f((-7 + 4) + (3(3) + 2)) = f(-3 + 11) = f(8) = 8 - 4 = 4$. Thus, $(C) \rightarrow (ii)$. $(D) f(f(f(f(0)))) + 1$. First, $f(0) = 3(0) + 2 = 2$. Then $f(f(0)) = f(2) = 3(2) + 2 = 8$. Then $f(f(f(0))) = f(8) = 8 - 4 = 4$. Then $f(f(f(f(0)))) = f(4) = 4 - 4 = 0$. Finally, $0 + 1 = 1$. Thus, $(D) \rightarrow (v)$. The correct matching is $(A)-(iii), (B)-(vi), (C)-(ii), (D)-(v)$.

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