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Math Help - One-to-One / Value of the Inverse

  1. #1
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    One-to-One / Value of the Inverse

    I don't know where to start with this one:

    f(x) = the integral from 1 to x of square root of (1+t^3) for x greater than or equal to -1

    I need to show that it's one to one and then find f^-1 of 0

    Any thoughts?

    (Sorry for the formatting, I don't know how to get the formulae and whatnot)

    -B
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  2. #2
    Member Abu-Khalil's Avatar
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    It's easy to show that f'(x)>0 when x\geq -1 so it must be one to one.
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  3. #3
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    Not necessarily so easy to show. Don't I need to take the antiderivative and then differentiate because x isn't the variable we're integrating, or do I just look at the function inside the integrl as being the derivative and replace t with x?
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  4. #4
    Member Abu-Khalil's Avatar
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    f'(x)=\frac{1}{\sqrt{1+x^3}}>0, \forall x \geq -1
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