Analyzing the Machine's Behavior
Imagine you are an engineer designing a machine that takes an input x and transforms it into an output defined by the function:
You are told that b is a constant, a fixed parameter such that 0<b<1. To determine if this machine is a perfect mirror—that is, if it is invertible—we must analyze its mathematical properties.
Phase 1
The Monotonicity Check
To determine if the function is one-to-one, we examine its derivative. We apply the quotient rule, where u=b−x and v=1−bx. The derivative formula is:
Substituting our values, we obtain:
f′(x)=(1−bx)2(1−bx)(−1)−(b−x)(−b)
Expanding the numerator, we get −1+bx+b2−bx. The bx terms cancel out, leaving us with the elegant expression:
Because 0<b<1, we know b2<1, which implies that b2−1 is strictly negative. Since the denominator is a square, it is always positive. Thus, f′(x)<0 for all x in the domain, confirming that the function is strictly decreasing and therefore one-to-one.
Phase 2
The Range Trap
A function is invertible only if it is both one-to-one and onto. We must now determine if the range of f(x) covers the entire codomain, R.
Given that the function is strictly decreasing, we examine the behavior at the boundaries of the domain (0,1):
The range of the function is the open interval (−1,b). Since $(-1, b)
eq \mathbb{R}$, the function is not onto. Consequently, the machine fails the test for global invertibility.
Phase 3
The Inverse Paradox
Although the function is not invertible on R, we can derive its inverse relation by solving y=1−bxb−x for x:
The algebraic rule for the inverse is identical to the original function. However, we must note that the domain of f is (0,1), while the domain of f−1 is (−1,b).
Because these domains differ, f and f−1 are not the same function. You have successfully navigated the traps of this problem by respecting the strict definitions of functions, domains, and ranges.