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Fie functia bijectiva f:(1,∞) [2,∞) f(x)= x^2-2x+3 determinati f^-1

Răspuns :

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f^-1:[2,∞)->(1,∞)
x^2-2x+3=y
x^2-2x+3-y=0
x = √(y - 2) + 1
f^-1=√(x-2)+1

Verificare:f(f^-1(x))=(√(x-2)+1)^2-2(√(x-2)+1)+3=x-2+2√(x-2)+1-2√(x-2)-2+3=x

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