Let f(x) the function defined below, where -1 < x < 1.
a) Find f -1(7).
(b) Find f(f -1(8)).
I'm having a tough time finding the inverse.
Find x such that :
$\displaystyle 7=7+x^2+\tan\left(\frac{\pi x}{2}\right)$
One advice : $\displaystyle x^2 \ge 0$ and since $\displaystyle -1<x<1$, $\displaystyle -\frac \pi 2 < \underbrace{\frac{\pi x}{2}}_{t} < \frac \pi 2$
One knows that from $\displaystyle -\frac \pi 2$ and $\displaystyle \frac \pi 2$, tan(t) is strictly increasing (from - infinity to + infinity). So it has only one solution if you want to solve tan(t)=...
Try for x=0.
Huh ? Isn't $\displaystyle f(f^{-1}(x))=x$ ? (if f is a bijection ^-^)(b) Find f(f -1(8)).
And it will be very hard to find itI'm having a tough time finding the inverse.