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?

Quote:

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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?

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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 (Cool)

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

Quote:

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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))?

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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.

Quote:

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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.

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Once again, Thanks a lot !!!