# Thread: Inverse Function with multiple x terms.

1. ## Inverse Function with multiple x terms.

Hi I'm trying to find the inverse function of $\displaystyle f(x) = 8+x^2+tan{\frac{\pi*x}{2}}$ where $\displaystyle -1< x <1$ and $\displaystyle f^{-1}(8)$
The main place i am having trouble is I don't know how to rearrange for x when there is more than 1 x term. The best I've done is; by letting $\displaystyle f(x)= y$, get the equation $\displaystyle y-8=x^2+tan{\frac{\pi*x}{2}}$which doesn't seem to be a great deal better.

Any ideas would be greatly appreciated.

2. Originally Posted by mattty
Hi I'm trying to find the inverse function of $\displaystyle f(x) = 8+x^2+tan{\frac{\pi*x}{2}}$ where $\displaystyle -1< x <1$ and $\displaystyle f^{-1}(8)$
The main place i am having trouble is I don't know how to rearrange for x when there is more than 1 x term. The best I've done is; by letting $\displaystyle f(x)= y$, get the equation $\displaystyle y-8=x^2+tan{\frac{\pi*x}{2}}$which doesn't seem to be a great deal better.

Any ideas would be greatly appreciated.
please state the problem in its entirety. i suspect that what you were asked to find is something along the lines of $\displaystyle \frac d{dx}f^{-1}(8)$, in which case, finding the inverse function is not necessary. if all you are after is $\displaystyle f^{-1}(8)$, finding the inverse function is still not necessary

3. If , where , find .
That is the question verbatim. If finding the inverse function is not necessary then would you mind giving any hints as to the path i should be taking then? It wouldn't be as simple as letting $\displaystyle f(x)=8$ would it? and then since we know that x is between -1 and 1 sub 0 in and $\displaystyle f^{-1}=0$

If it's that simple I will quite happily kick myself

4. Originally Posted by mattty
If , where , find .
That is the question verbatim. If finding the inverse function is not necessary then would you mind giving any hints as to the path i should be taking then? It wouldn't be as simple as letting $\displaystyle f(x)=8$ would it? and then since we know that x is between -1 and 1 sub 0 in and $\displaystyle f^{-1}{\color{red}(8)}=0$

If it's that simple I will quite happily kick myself
start kicking

5. le sigh, thanks a bunch