求反函数y=lg(x+√x²-1)

1个回答

  • ∵y=lg[x+√(x²-1)] ==>x+√(x²-1)=10^y.(1)

    ==>[x-√(x²-1)]/{[x+√(x²-1)][x-√(x²-1)]}=10^y

    ==>1/[x-√(x²-1)]=10^y

    ==>x-√(x²-1)=10^(-y).(2)

    ∴由(1)式+(2)得 2x=10^y+10^(-y) ==>x=[10^y+10^(-y)]/2

    故所求反函数是 x=[10^y+10^(-y)]/2.

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