I need to find $f^{-1}(2)$ where $f(x) = 2 + x^2 + tan(πx/2)$
I know can substitute $f(x)$ with $y$ and swap $x$ and $y$:
$$x = 2 + y^2 + tan(πy/2)$$
But I'm having trouble eliminating the tangent:
$$y^2 + tan(πy/2) = x - 2$$
Could someone please tell me how I should continue? Is there any way to factor y out of $tan(πy/2)$?
In this case- it's actually easy. Note that If $0 \leq x \leq 1$ then $f(x) \geq 2$ with equality only when $x=0$. So that gives you one solution $f^{-1}(2)=0$. Question is: are there other solutions? When $1<x<2$ the tan term is negative, and we might have a solution, if $x^2=tan\dfrac{\pi x}{2}$ with $1<x<2$. Is there such a solution? Can you proceed?