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Pre Calculus

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Inverse functions

what is the inverse function of
f(x)=1/square root of x +1?


and f(x)=1-x^2, x<0?

  • Pre Calculus - ,

    f(x) = 1/sqrt(x+1)
    we can rewrite this as
    y = 1/sqrt(x+1)
    to get the inverse function, we replace x by y, and y by x, and solve for y:
    x = 1/sqrt(y+1)
    sqrt(y+1) = 1/x
    y + 1 = 1 / x^2
    y = (1/(x^2)) - 1
    f'(x) = (1/(x^2)) - 1

    for #2, we do the same:
    f(x) = 1 - x^2
    y = 1 - x^2
    x = 1 - y^2
    y^2 = 1 - x
    y = sqrt(1-x)
    f'(x) = sqrt(1-x)

    hope this helps~ :)

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