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$f (x) = 4 x - 1 i f x > 4 = x^{2} - 2 i f - 2 \leq x \leq 3 = 3 x + 4 i f x < - 2 t h e n f (5) + f (2) + f (- 3)$

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answer is C.

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Detailed Solution

$f (5) + f (2) + f (- 3) = [4 (5) - 1] + [2^{2} - 2] + [3 (- 3) + 4] = 19 + 2 - 5 = 16$

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