f(x) = 3 + x^2 + tan(pi(x/2)), where -1 < x < 1.
Find f^-1 (3).
I got up to:
f^-1 (x) = 3 + y^2 + tan(pi(y/2)), where -1 < y < 1
then subing x= 3
I get 0 = y^2 + tan(pi(y/2))
And I'm stuck again!
What should I do with the y within the tangent
Any advice and hint is greatly appreciated ! Thanks !