Answers: 1
with its inverse 
you'll get the identity function:![\mathsf{(f\circ f^{-1})(x)=x}\\\\ \mathsf{f\big[f^{-1}(x)\big]=x}\\\\](/tpl/images/0269/7351/63987.png)
![\mathsf{f\big[f^{-1}(x)\big]=10-[f^{-1}(x)]^2}\\\\ \mathsf{x=10-[f^{-1}(x)]^2}\\\\ \mathsf{[f^{-1}(x)]^2=10-x}\\\\ \mathsf{f^{-1}(x)=\pm \sqrt{10-x}}\\\\ \mathsf{f^{-1}(x)=-\sqrt{10-x}~~~or~~~f^{-1}(x)=\sqrt{10-x}}](/tpl/images/0269/7351/8aa78.png)
itself, and where it's invertible.Another question on Mathematics
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What is the inverse of f (x)=10-×^2...
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