# Inverse Function!!

• January 9th 2008, 04:00 PM
aquaglass88
Inverse Function!!
Hi, I am in the middle of solving this problem:
Let f(x) = 3 +x^2 + tan (πx/2), where -1 < x < 1.
a.) Find the(inverse) f^–1 (3).
b.) Find f(f^-1(5)).

I know the steps on how to find the inverse equation, but I do not know how to solve for x, in order to get it alone, because of the tan[(pi)(x/2)]. I do not remember how to get x alone when it is in sin (x), tan (x), etc. Do you multiply everything by tan x in order to cancel it out?

I'm pretty sure the second part requires a cancellation equation, but then again I'm stuck on the tan issue. Trigonemtric functions confuse me :( .

Thank you so much for your help!! :)
• January 9th 2008, 04:21 PM
Quote:

I do not know how to solve for x, in order to get it alone, because of the tan[(pi)(x/2)].
This cannot be done. In this case there is no way of expressing $f^-1(x)$ in terms of x. However, you do not need to solve for x to get an answer.

It sounds like you have already got to $x = 3 +y^2+\tan (\frac { \pi y}{2})$
Now substitute x = 3 to get

$3 = 3 +y^2+\tan (\frac {\pi y}{2})$
$y^2+\tan (\frac {\pi y}{2}) = 0$

It is easy to see that y = 0 is a solution to this, just by recognising that tan (0) = 0.

Quote:

b.) Find f(f^-1(5)).
You are expected to know that $f(f^-1(x)) = x$