Final answer:
A one-to-one function that maps the open interval (1,3) to (2,5) and is still one-to-one when extended to [1,3) → [2,5) is f(x) = ¾(x-1) + 2. This linear function has a unique output for each input and smoothly transitions between the defined intervals.
Step-by-step explanation:
To construct a one-to-one function f: (1,3) → (2,5) that remains one-to-one when extended to f: [1,3) → [2,5), we can use a linear function that maps the interval (1,3) onto the interval (2,5). Since one-to-one functions require each x-value to map to a unique y-value, we need a function with a constant rate of change that doesn't repeat y-values.
An example of such a function is f(x) = ¾(x-1) + 2. This function maps 1 to 2 and approaches mapping 3 to 5 as x approaches 3. To verify that this function is indeed one-to-one, we can check that for any two different x-values in the domain, the function produces two different y-values.
Function Verification
Let's take two x-values, x1 and x2, such that 1 < x1 < x2 < 3. Applying the function, we get:
f(x1) = ¾(x1-1) + 2
f(x2) = ¾(x2-1) + 2
Since x1 < x2, it is clear that f(x1) < f(x2), which confirms that f(x) is one-to-one on this interval. This function is a linear transformation from the interval (1,3) to (2,5), retaining the one-to-one property even when restricted to [1,3) mapped to [2,5).