For the function defined by f(x)=2-x^2, 0 less then or equal to x, find a formula for f^-1(x)?

Question

Thanks for the help.

a.f^-1(x)=x^2-2, x less then or equal to 2
b.f^-1(x)=1/(2-x^2), 0 less then or equal to X
c.f^-1(x)=-sqrt 2+sqrt x, 0 less / equal to X
d.f^-1(x)=sqrt(2-x), x less / equal to 0

Answer is either a,b,c,d

Answer 1

f(x) = 2 - x^2

x = 2 - (f^-1)^2
x - 2 = -(f^-1)^2
-x + 2 = (f^-1)^2
f^(-1) = sqrt(-x + 2) or sqrt(2 - x)

ANS : D

Answer 2

I don't quite agree with the question.
As other answerers have shown, you put
x = 2 - y^2, with y greater than or equal to 0 (because we've swapped x and y)
and then make y the subject, getting
y = √(2-x)

However the original function has x ≥ 0, and so f(x) ≤ 2, or y≤2.
When we swap x and y for the inverse, these become
y ≥ 0, and x ≤ 2, so I say the inverse function is

f^(-1)(x) = √(2-x), with x ≤ 2. D is the best choice, but it ain't quite right!!

Answer 3

2-x^2(x-1)

Answer 4

y = 2 - x^2
inverse:
x = 2 - y^2
y^2 = 2 - x
y = ±√(2 - x)