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f(x)=2-x^2, 0 is < and = to x, I need to find a formula for f^-1(x).
For our convenience, lets take,
y = 2-x^2
Rewrite the given equation as follows.
Add x^2 to both sides of the equation,
==> y + x^2 = 2-x^2 + x^2
==> x^2 + y = 2
Subtract y from both sides of the equation,
==> x^2 + y - y = 2 - y
==> x^2 = 2 - y
Take square root on both sides of the expression,
==> x = square root of ( 2 - y)
We can write this expressions(after replacing y by x ) as,
f-1(x) = sqrt (2-x)
Which is the required inverse function.
Here we can see that this inverse function is not defined when the value
of x is greater than 2 because then 2-x becomes a negative number and sqrt of
a negative number cannot be determined.
So we can write, inverse of the given function is sqrt of (2-x) and x is less
than or equal to 2.
Hope you Understood.