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2014-07-14T19:30:44-03:00
f(x)= \dfrac{2x+1}{x+9}

Determinação da função inversa de f(x):

y= \dfrac{2x+1}{x+9}

Trocando x com y: x= \dfrac{2y+1}{y+9}

x.(y+9)=2y+1

xy+9x=2y+1

xy-2y=1-9x

y.(x-2)=1-9x

a) y= f^{-1}(x)= \dfrac{1-9x}{x-2} → função inversa

b) D(f) = {x ∈ R | x ≠ - 9}

c) D(f⁻¹) = {x ∈ R | x ≠ 2}

d) Im (f) = {y ∈ R | y ≠ 2}

e) Im(f⁻¹) = {y ∈ R | y ≠ - 9}

f) (fof-¹)(x) = f(f⁻¹(x)) = \dfrac{2.(\frac{1-9x}{x-2})+1}{\frac{1-9x}{x-2}+9}=\dfrac{\frac{2-18x}{x-2}+1}{\frac{1-9x}{x-2}+9}=\dfrac{\frac{2-18x+x-2}{x-2}}{\frac{1-9x+9x-18}{x-2}}=\dfrac{\frac{-17x}{x-2}}{\frac{-17}{x-2}}=\dfrac{-17x}{x-2}. \dfrac{x-2}{-17} =x, para x≠2
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