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

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

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

2. It's easy to show that $f'(x)>0$ when $x\geq -1$ so it must be one to one.

3. 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?

4. $f'(x)=\frac{1}{\sqrt{1+x^3}}>0, \forall x \geq -1$
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