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a) f (2x+a) = √2 (2x + a) +a = 4x + 2a +a = √4x+3a
b) { f(x+h) - f(x) } / h =
primero hallar f (x+h)
f (x+h)= √2(x + h) +a = √2x + 2h + a
ahora f(x) es √(2x + a)
entonces queda √2x + 2h +a - (√2x+a )
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h
c. f (3a/2)= √2(3a/2) +a = √6a/2 +a = √3a+a = √4a = 2√a