Restricting a Domain So an Inverse Exists

You have f(x)=x²-4x+3. You need its inverse. The function fails the horizontal line test, so no inverse exists over its full domain. Restricting a domain so an inverse exists is the step that fixes this: you cut the domain to a single branch where the function is one-to-one, then write that restriction as an interval. The standard branch for this parabola is the right side of its vertex.

Complete the square: f(x)=(x-2)²-1. The vertex is at x=2. On (-∞,2] the function decreases; on [2,∞) it increases. Either branch is one-to-one, but the conventional choice is the increasing branch. Write the restriction as [2,∞). That interval tells you exactly which x-values the inverse will accept.

The inverse function calculator uses that same branch choice as its default. When you enter f(x)=x²-4x+3, it checks for one-to-one behaviour, computes the inverse on [2,∞), and outputs f⁻¹(y)=2+√(y+1). The interval [2,∞) is the key that makes the calculation valid.

Take f(x)=sin(x) over all real numbers. The function is periodic and fails the horizontal line test everywhere. To define arcsin(x), you restrict the domain of sin(x) to the interval where it is one-to-one and covers the full range of the inverse.

The chosen branch for sin(x) is [-π/2, π/2]. That is a closed interval: both endpoints are included. Brackets are correct because sin(-π/2)=-1 and sin(π/2)=1, and the inverse arcsin has to output those exact values. If you used parentheses, (-π/2, π/2), the inverse would miss -1 and 1, making it incomplete.

For cos(x), the principal branch is [0, π]. For tan(x), it is (-π/2, π/2), an open interval because tan(x) is undefined at those endpoints. The pattern is consistent: the interval brackets match the strictness of the inequality at the endpoint where the function is defined.

Follow the calculation for f(x)=x²-4x+3 step by step. This is the same parabola used earlier, worked out with the numbers.

Step 1: Find the Vertex

Write f(x)=x²-4x+3 as (x-2)²-1. The vertex is at x=2. The function is symmetric around this point.

Step 2: Identify the One-to-One Branches

The function decreases on (-∞,2] and increases on [2,∞). On either interval, every y-value appears exactly once. Both are valid restrictions.

Step 3: Choose the Standard Branch

Use [2,∞). This is the increasing branch, and it matches the convention for quadratic functions with a positive leading coefficient. Interval notation: left endpoint 2, bracket, right endpoint ∞, parenthesis.

Step 4: Write the Inverse

Solve y=(x-2)²-1 for x. Add 1: y+1=(x-2)². Take the positive square root because x ≥ 2: √(y+1)=x-2. Then x=2+√(y+1). The inverse is f⁻¹(y)=2+√(y+1), defined for y ≥ -1. That matches the range of the original restricted function.

The restriction must be a single interval. A non-one-to-one function like f(x)=x² has two branches; you pick one and write it in interval notation. The rules for brackets and parentheses are the same as for any domain: use a bracket when the endpoint is included, a parenthesis when it is not.

For f(x)=1/x, the function is already one-to-one on its domain (-∞,0) ∪ (0,∞). But there is no single interval that covers both pieces. The inverse exists on the same union, and interval notation handles this with the union symbol ∪. The one case where a single interval is impossible is when restricting a function to a subinterval that is too small, the inverse still exists but you lose data. For example, restrict f(x)=x²-4x+3 to [3,4]. The function is one-to-one there, and the inverse exists, but it misses all y-values below the range on that interval. The standard rule is to use the maximal interval where the function is one-to-one, preserving as much of the original function as possible.

Identify the vertex, local extremum, or point where the function changes direction. For a quadratic, complete the square to find the vertex. For a trigonometric function, use the principal branch defined by convention.

• Quadratic f(x)=ax²+bx+c: Vertex at x=-b/(2a). Choose x ≥ vertex if a>0, x ≤ vertex if a<0. Write as [vertex, ∞) or (-∞, vertex].
• Sine and cosine: Use the closed intervals [-π/2, π/2] for sine, [0, π] for cosine.
• Tangent: Use the open interval (-π/2, π/2).
• Rational like 1/x: The function is one-to-one on each piece, but no single interval works. Use the union (-∞,0) ∪ (0,∞).

Writers choose the correct branch but use the wrong bracket for the vertex. For f(x)=x²-4x+3, the vertex at x=2 is included in the increasing branch because the function is defined there and the inverse needs that value. Write [2,∞), not (2,∞). The bracket costs nothing and preserves the endpoint. A parenthesis at the vertex drops the point where the function turns, and the inverse loses that value forever.

The same mistake happens with sin(x): the interval is [-π/2, π/2], not (-π/2, π/2). Check the endpoint bracket before you move on. If the function is defined at the endpoint, use a bracket.