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?