Expert: Sherman D. - 9/10/2005

Question
find f^-1 and verify that (f o f^-1)(x)= (f^-1 o f)(x)=x

1) f(x)=x^2, x < or equal to 0.
2) f(x)= -(x-2)^2, x < or equal to 2.

Answer
if you are saying (f(f(x)'))) = (f(x)'(f(x))) also (f(f(x)'))) = x

f(x)' means the inverse of f(x)

1.)
f(x) = x^2
x <= 0

f(x) = x^2
y = x^2
x = y^2
y = sqrt(x)

ANS : f(x)' = sqrt(x)

Proofs
f(f(x)')
f(sqrt(x)) = x^2
f(sqrt(x)) = (sqrt(x))^2

the sqrt and ^2 cancels out, leaving you with "x" as the answer.

f(f(x))'
f(x^2) = sqrt(x)
f(x^2) = sqrt(x^2)
the sqrt and ^2 cancel out leaving you with "x" as the answer.

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2.)
f(x) = -(x - 2)^2
x <= 2

y = -(x - 2)^2
x = -(y - 2)^2
-x = (y - 2)^2
sqrt(-x) = y - 2
y = 2 + isqrt(x)

ANS : f(x)' = 2 + isqrt(x)

Proofs
f(f(x)')
f(2 + isqrt(x)) = -((2 + isqrt(x)) - 2)^2
f(2 + isqrt(x)) = -(2 + isqrt(x) - 2)^2
f(2 + isqrt(x)) = -(isqrt(x))^2
f(2 + isqrt(x)) = -(sqrt(-x))^2
the sqrt and ^2 cancel out
f(2 + isqrt(x)) = -(-x)
f(2 + isqrt(x)) = x

f(f(x))' = 2 + isqrt(2 + isqrt(x))
f(f(x))' = 2 + isqrt(-(x - 2)^2)
f(f(x))' = 2 + (i^2)sqrt((x - 2)^2)
f(f(x))' = 2 + (-1)sqrt((x - 2)^2)
f(f(x))' = 2 - sqrt((x - 2)^2)
the sqrt and ^2 cancel out and since the only way for this to work is if you took the positive side of the sqrt since sqrt's have a negative and a positive side.
f(f(x))' = 2 + (x - 2)
f(f(x))' = 2 + x - 2
f(f(x))' = x