Restricted Domain Math Example 4

Follow the full solution, then compare it with the other examples linked below.

hard

Restrict the domain of

f(x) = (x - 1)^2 + 3

to the largest interval containing

x = 4

on which

f

is one-to-one, then find

f^{-1}(x)

1
The vertex of this parabola is at $x = 1$ . Since $x = 4 > 1$ , restrict to $[1, \infty)$ where $f$ is increasing.
2
Solve for the inverse: $y = (x-1)^2 + 3 \Rightarrow (x-1)^2 = y - 3 \Rightarrow x - 1 = \sqrt{y-3}$ (positive root since $x \ge 1$ ). So $f^{-1}(x) = 1 + \sqrt{x-3}$ with domain $[3, \infty)$ .

Answer

f^{-1}(x) = 1 + \sqrt{x-3}, \quad \text{domain: } [3, \infty)

To make a parabola one-to-one, restrict to one side of the vertex. Since we need the interval containing

x = 4

, we use the right side

[1, \infty)

. The inverse's domain equals the original function's range on the restricted domain, which starts at the vertex value

f(1) = 3

About Restricted Domain

Restricting a domain limits allowable inputs so a function has desired properties, often invertibility.

Find the natural (implied) domain of [formula].

Restrict the domain of [formula] so that the function has an inverse. Find the inverse on this restr

Find the domain of [formula].