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Mathematics 21 Online
OpenStudy (anonymous):

guys could you help me with this inverse http://assets.openstudy.com/updates/attachments/54178fd2e4b0e37da3691c7b-st4lk3r-1410830636775-img_1447.jpg

OpenStudy (anonymous):

looks good to me!

OpenStudy (anonymous):

oh really?

OpenStudy (anonymous):

what might be missing is a \(\pm\) before the radical, but not if you have restricted the domain

OpenStudy (alekos):

yep, all good

OpenStudy (anonymous):

I dont know the thing is the graphing software shows this

OpenStudy (anonymous):

OpenStudy (alekos):

the 5th line should just be x+1 not (x+1)^2 but otherwise you've got the inverse

OpenStudy (anonymous):

ohh i see

OpenStudy (anonymous):

but how about how the inverse is being displayed?

OpenStudy (anonymous):

oh nice

OpenStudy (anonymous):

guys and how many inverses can i consider from that equation?

OpenStudy (anonymous):

you have to restrict the range of \(y=\sqrt{8-2x-x^2}\) to make it one to one

OpenStudy (anonymous):

also what would be the domain of the inverse function? do i solve for 9-x^2 > 0 right ?.

OpenStudy (anonymous):

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 \[y=-1\pm\sqrt{9-x^2}\]

OpenStudy (anonymous):

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 \[f^{-1}(x)=-1-\sqrt{9-x^2}\]

OpenStudy (anonymous):

as for the domain of either, you are right, you need to solve \[9-x^2\geq 0\]

OpenStudy (anonymous):

or else determine the range of \[\sqrt{8-2x-x^2}\] it is the same problem

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

@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??

OpenStudy (anonymous):

yes, find the vertex of \(8-2x-x^2\)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

which is 3!!

OpenStudy (anonymous):

oh so that's it

OpenStudy (anonymous):

@satellite73 Thank you.

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