Thread: Tangent problem

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 !

Hint: The easiest way for is for both terms to be zero in the first place.

Originally Posted by letzdiscuss

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 !

By inspection, y = 0 is a solution. You should prove that this is the only solution that satisfies -1 < x < 1.

I'm sorry but how would I prove y = 0. Could you please give me some more hints?

Originally Posted by letzdiscuss

I'm sorry but how would I prove y = 0. Could you please give me some more hints?

Try substituting y = 0 into 0 = y^2 + tan(pi(y/2)). Does it work?

Yes it worked ! So my answer would be f^-1 (3) = 0.
Thanks a lot !!!