Hi, does anybody know how to find out the "inverse" function of this function: $\displaystyle f(x) = 3 + \arcsin (2x + 1) $ I have absolutely no idea how to start. Any help is welcome. Thanks.
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Originally Posted by petike Hi, does anybody know how to find out the "inverse" function of this function: $\displaystyle f(x) = 3 + \arcsin (2x + 1) $ I have absolutely no idea how to start. Any help is welcome. Thanks. no surprises here. as always, to find the inverse function, switch x and y and solve for y
Originally Posted by Jhevon no surprises here. as always, to find the inverse function, switch x and y and solve for y Ok, so I know how to start. But what then. I know I should make $\displaystyle y$ alone. But how? I would like to see the whole process of that. Thanks anyway.
Originally Posted by petike Ok, so I know how to start. But what then. I know I should make $\displaystyle y$ alone. But how? I would like to see the whole process of that. Thanks anyway. note that if $\displaystyle \theta = \arcsin \varphi$, then $\displaystyle \sin \theta = \varphi$ now can you try?
Originally Posted by Jhevon note that if $\displaystyle \theta = \arcsin \varphi$, then $\displaystyle \sin \theta = \varphi$ now can you try? Ah, thanks, now it's clearer. You've got my thanks.
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