Answer: To find A and B such that g(x) = f(x) is bijective, we need to ensure that g(x) satisfies the conditions for a bijective function, namely, that it is both injective and surjective.
To show that g(x) is injective, we need to show that for any distinct x1, x2 in A, g(x1) ≠ g(x2). We can do this by assuming that g(x1) = g(x2) and then showing that it leads to a contradiction.
So, let's assume that g(x1) = g(x2). Then, we have:
f(x1) = f(x2)
x1(x1-1) = x2(x2-1)
x1^2 - x1 = x2^2 - x2
x1^2 - x2^2 - x1 + x2 = 0
(x1 - x2)(x1 + x2 - 1) = 0
Since x1 and x2 are distinct, we must have x1 + x2 = 1.
But this is impossible, since x1 and x2 are both real numbers, and the sum of two real numbers cannot equal 1 unless one of them is complex. Therefore, our assumption that g(x1) = g(x2) must be false, and g(x) is injective.
To show that g(x) is surjective, we need to show that for any y in B, there exists at least one x in A such that g(x) = y. In other words, we need to find an expression for x in terms of y.
So, let's solve the equation f(x) = y for x:
x(x-1) = y
x^2 - x - y = 0
Using the quadratic formula, we get:
x = (1 ± √(1 + 4y))/2
Since we want to define g(x) on R, we need to ensure that the expression under the square root is non-negative. This means that 1 + 4y ≥ 0, or y ≥ -1/4.
Therefore, we can define A = [-1/4, ∞) and B = [0, ∞), and g(x) = f(x) is a bijective function from A to B.