# Proving the area function has an inverse

I am able to differentiate A at x using the FTC, but then I was wondering how one could show that A was one to one and prove that it has an inverse. If anybody could please help.

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You have $A'(x)=\frac {1-2x}{2\sqrt{1-x^2}}+\sqrt {1-x^2}=\frac {1-2x-2-2x^2}{\sqrt{1-x^2}}=\frac {-(1+x)^2-x^2}{\sqrt{1-x^2}}\lt 0$ Since $A$ is continuous and monotonic, you can't have $A(x)=A(y)$ with $x \ne y$. Given that it is one-to-one, it has an inverse over its range.
@Dick: because the derivative doesn't change sign, the function is monotonic, in this case decreasing. So for $x \lt y,\ \ A(x) \gt A(y)$ and it is one-to-one. –  Ross Millikan Feb 21 at 1:11