guys could you help me with this inverse http://assets.openstudy.com/updates/attachments/54178fd2e4b0e37da3691c7b-st4lk3r-1410830636775-img_1447.jpg
looks good to me!
oh really?
what might be missing is a \(\pm\) before the radical, but not if you have restricted the domain
yep, all good
I dont know the thing is the graphing software shows this
the 5th line should just be x+1 not (x+1)^2 but otherwise you've got the inverse
ohh i see
but how about how the inverse is being displayed?
oh nice
guys and how many inverses can i consider from that equation?
you have to restrict the range of \(y=\sqrt{8-2x-x^2}\) to make it one to one
also what would be the domain of the inverse function? do i solve for 9-x^2 > 0 right ?.
two issues here \[f(x)=\sqrt{8-2x-x^2}\] is not a one to one function it is increasing, then decreasing, so the inverse is not a function that is why you get the \[y=-1\pm\sqrt{9-x^2}\]
you can restrict the domain of \[f(x)=\sqrt{8-2x-x^2}\]to make it one to one, that will determine whether you get \[f^{-1}(x)=-1+\sqrt{9-x^2}\] or \[f^{-1}(x)=-1-\sqrt{9-x^2}\]
as for the domain of either, you are right, you need to solve \[9-x^2\geq 0\]
or else determine the range of \[\sqrt{8-2x-x^2}\] it is the same problem
oh right but I didn't get quite well how would i restrict the main function do i have to take the positive side or the negative side?
@satellite73 hey friend how can i find the range of the first function, what i did was finding the vertex and then looking at the graph but is it possible to do i algebraically??
yes, find the vertex of \(8-2x-x^2\)
the second coordinate of the vertex is the largest value of \[8-2x-x^2\]so the largest value is the square root of that number
which is 3!!
oh so that's it
@satellite73 Thank you.
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