Question 919113
If {{{x >=0}}}, then 


{{{f(x) = x^2 - 16}}}


{{{y = x^2 - 16}}}


{{{x = y^2 - 16}}} Swap x and y. Then solve for y.


{{{x+16 = y^2}}}


{{{y^2=x+16}}}


{{{y=sqrt(x+16)}}} Since {{{x >= 0}}} this means {{{y >= 0}}} (remember x and y swapped).


The inverse function is *[Tex \LARGE f^{-1}(x) = \sqrt{x+16}]


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Thanks,


Jim