x>3,y>2√5
y^2=x^2+2x+5=(x+1)^2+4
x+1=±√(y^2-4)
x=-1±√(y^2-4)
根據(jù)題意得
x=-1+√(y^2-4)
交換變量次序得
y=-1+√(x^2-4) (x>2√5)
求y=√(x^2+2x+5)的反函數(shù)(x>3)
求y=√(x^2+2x+5)的反函數(shù)(x>3)
數(shù)學(xué)人氣:910 ℃時(shí)間:2019-11-15 04:09:46
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