Question 941088
In this context, *[tex \large g^{-1} (x)] is not the reciprocal of g(x). It is the function such that *[tex \large g^{-1}(g(x)) = g(g^{-1}(x)) = x] for all x, if it exists.



To solve, let *[tex \large y = g(x)] and solve for x:


*[tex \large y = 1-2x]


*[tex \large 2x = 1-y]


*[tex \large x = \frac{1}{2}(1-y)]


Replace x with g^{-1} x and y with x:


*[tex \large g^{-1}(x) = \frac{1}{2}(1-x)]


As a check, *[tex \large g(g^{-1}(x)) = g(\frac{1}{2}(1-x)) = 1 - 2(\frac{1}{2}(1-x) = x] and *[tex \large g^{-1}(g(x)) = g^{-1}(1-2x) = \frac{1}{2}(1 - (1-2x)) = x].