# Finding the inverse of this function

• Sep 21st 2009, 07:32 PM
letzdiscuss
Finding the inverse of this function
f(x) = 3 + x^2 + tan(pi(x/2)) , -1 < x < 1

I found the inverse and it was:

3 + y^2 + tan(pi(y/2)), -1 < y < 1
Correct me if I was wrong.

Then the question asks me to find f(f^-1(5))
And here's what I did.

f^-1 (5) = 5 = 3 + y^2 + tan(pi(y/2))
2 = y^2 + tan(pi(y/2))
f^-1 (5) = 1

f (f^-1 (5)) = f(1) = 3 + (1)^2 + tan(pi(1/2))
= 3 + 1 + 1
= 5

But the domain clearly states that x has to be between -1 and 1. What am I doing wrong?
• Sep 21st 2009, 09:08 PM
mr fantastic
Quote:

Originally Posted by letzdiscuss
f(x) = 3 + x^2 + tan(pi(x/2)) , -1 < x < 1

I found the inverse and it was:

3 + y^2 + tan(pi(y/2)), -1 < y < 1 Mr F says: The inverse is the function y such that x = 3 + y^2 + tan(pi(y/2)) where -1 < y < 1.

Correct me if I was wrong.

Then the question asks me to find f(f^-1(5))
And here's what I did.

f^-1 (5) = 5 = 3 + y^2 + tan(pi(y/2)) Mr F says: f^-1 (5) is the value of y such that 5 = 3 + y^2 + tan(pi(y/2)).

2 = y^2 + tan(pi(y/2))
f^-1 (5) = 1 Mr F says: This is clearly wrong. x = 5 does not give y = 1. In other words, y = 1 is NOT a solution to 2 = y^2 + tan(pi(y/2)). In fact, y = 0.64216.

f (f^-1 (5)) = f(1) = 3 + (1)^2 + tan(pi(1/2))
= 3 + 1 + 1
= 5

But the domain clearly states that x has to be between -1 and 1. What am I doing wrong?

The logic of your setting out leaves something to be desired. Be that as it may, none of the above is necessary since by definition f(f^-1(a)) = a.
• Sep 22nd 2009, 12:56 AM
letzdiscuss
Thank you ! I know what I did wrong (Cool)

but how to I calculate what y is from 2 = y^2 + tan(pi(y/2))?
• Sep 22nd 2009, 01:17 AM
mr fantastic
Quote:

Originally Posted by letzdiscuss
Thank you ! I know what I did wrong (Cool)

but how to I calculate what y is from 2 = y^2 + tan(pi(y/2))?

An exact answer using algebra is not possible. You have to settle for a numerical approximation using technnology. But as I said earlier, the solution to this equation is not required in this case.
• Sep 22nd 2009, 01:27 AM
letzdiscuss
Quote:

Originally Posted by mr fantastic
An exact answer using algebra is not possible. You have to settle for a numerical approximation using technnology. But as I said earlier, the solution to this equation is not required in this case.

Once again, Thanks a lot !!!