Thread: 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?

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.

Thank you ! I know what I did wrong

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

Originally Posted by letzdiscuss

Thank you ! I know what I did wrong

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.

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 !!!