7.1 Inverse Functions and Domain Restrictions

7.1 Inverse Functions and Domain Restrictions

Quick Answer: The inverse function $f^{-1}(x)$ reverses the operations of a function $f(x)$ by exchanging the roles of inputs and outputs: $(x, y) \in f \iff (y, x) \in f^{-1}$. Graphically, $f$ and $f^{-1}$ are symmetric reflections across the diagonal identity line $y = x$. A function possesses an inverse function if and only if it is one-to-one (passes both the Vertical and Horizontal Line Tests). To find an inverse algebraically, replace $f(x)$ with $y$, swap $x$ and $y$, and solve for $y$. Two functions are confirmed as inverses if and only if $f(g(x)) = x$ and $g(f(x)) = x$. Non-one-to-one functions (such as parabolas) require domain restrictions (e.g., $x \ge 0$ on $f(x) = x^2 + 1$) so each output corresponds to a unique input.

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Formal Definition of an Inverse Function (AII-F.BF.4a)

[!NOTE] Standard note. The assessed component is AII-F.BF.4a, and NYSED’s instructional note states that inverses of one-to-one functions will be found algebraically and graphically. The June 2026 Part I asked directly for $f^{-1}(x)$ when $f(x) = 3^x$.

In mathematics, a relation maps elements from an input set (the domain) to an output set (the range). The inverse relation reverses this process entirely: it maps every output back to its originating input. If an ordered pair $(a, b)$ belongs to the original function $f$, then the ordered pair $(b, a)$ belongs to the inverse relation $f^{-1}$:

$f(a) = b \iff f^{-1}(b) = a$

Because all coordinates swap places, the domain and range of the function and its inverse are mutually interchanged:

$\text{Domain}(f^{-1}) = \text{Range}(f) \quad \text{and} \quad \text{Range}(f^{-1}) = \text{Domain}(f)$

[!WARNING] Inverse Notation vs. Exponents: The notation $f^{-1}(x)$ denotes the functional inverse, NOT a negative exponent. It does not mean the reciprocal: $f^{-1}(x) \neq \frac{1}{f(x)} = [f(x)]^{-1}$. The superscript $-1$ signifies that the mapping operation is being reversed.

Geometric Reflection Across the Identity Line $y = x$

When a function and its inverse are graphed on the Cartesian plane, they exhibit bilateral symmetry across the line $y = x$:

• The line segment connecting any point $(a, b)$ on $f(x)$ to its corresponding point $(b, a)$ on $f^{-1}(x)$ has a slope of $\frac{a - b}{b - a} = -1$.
• Because the line $y = x$ has a slope of $1$, this connecting segment is perpendicular to $y = x$.
• The midpoint of the segment connecting $(a, b)$ and $(b, a)$ is $\left(\frac{a + b}{2}, \frac{a + b}{2}\right)$, which lies precisely on the line $y = x$.
• Therefore, every point on $f^{-1}(x)$ is the exact geometric reflection of a point on $f(x)$ across the diagonal line of symmetry $y = x$.

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One-to-One Functions & The Horizontal Line Test

Not every function possesses an inverse that is also a function. For an inverse relation to qualify as a valid function, each input in its domain must map to exactly one output. Reversing this statement back to the original function yields the definition of a one-to-one (injective) function.

Definition of a One-to-One Function

A function $f$ is one-to-one if no two distinct inputs produce the same output:

$f(x_1) = f(x_2) \implies x_1 = x_2$

The Horizontal Line Test (HLT)

While the Vertical Line Test (VLT) verifies whether a relation is a function, the Horizontal Line Test (HLT) determines whether a function's inverse is also a function:

• Horizontal Line Test Theorem: A function $f(x)$ has an inverse function $f^{-1}(x)$ if and only if no horizontal line intersects the graph of $f(x)$ at more than one point.
• Mathematical Justification: A horizontal line represents the equation $y = k$. If a horizontal line intersects the curve at two points $(x_1, k)$ and $(x_2, k)$ where $x_1 \neq x_2$, the inverse relation would contain $(k, x_1)$ and $(k, x_2)$. When inverted, the single input $k$ would produce two different outputs, failing the Vertical Line Test.

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Algebraic Determination of Inverse Functions

To find the formula for an inverse function algebraically, follow a structured four-stage algorithm:

1. Replace Function Notation: Substitute the dependent variable $y$ for $f(x)$.
2. Interchange Variables: Swap every occurrence of $x$ and $y$. This step executes the fundamental inverse operation $(x, y) \leftrightarrow (y, x)$.
3. Solve for $y$: Use algebraic manipulations to isolate the new dependent variable $y$. For rational expressions, this involves clearing the denominator, expanding, grouping all terms with $y$ on one side, factoring out $y$, and dividing.
4. Re-establish Notation: Replace $y$ with $f^{-1}(x)$, and explicitly state any domain restrictions inherited from the range of the original function.

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Verifying Inverses via Mutual Function Composition

A central Regents requirement is proving or verifying that two candidate functions are inverses. Stating that their formulas look reversed is insufficient.

The Inverse Composition Theorem

Two functions $f$ and $g$ are inverse functions if and only if both composite expressions simplify to the identity function $I(x) = x$:

$(f \circ g)(x) = f(g(x)) = x \quad \text{for all } x \text{ in the domain of } g$

$\text{AND}$

$(g \circ f)(x) = g(f(x)) = x \quad \text{for all } x \text{ in the domain of } f$

[!IMPORTANT] Both Compositions Are Mandatory: You must evaluate both $f(g(x))$ and $g(f(x))$. Verifying only one composition does not constitute a complete mathematical proof, as certain functions may be one-sided inverses on restricted subsets.

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Restricting Domains of Non-One-to-One Functions

Parabolas of the form $f(x) = a(x - h)^2 + k$ fail the Horizontal Line Test because they are symmetric across their axis of symmetry $x = h$. Every output value above the vertex (for $a > 0$) is produced by two distinct input values: one to the left of the vertex and one to the right.

To construct an invertible function from a quadratic parent, we restrict the domain to a subset where the graph is strictly monotonic (strictly increasing or strictly decreasing):

The Vertex Partitioning Strategy

1. Identify the vertex $(h, k)$ and axis of symmetry $x = h$.
2. Restrict the domain to one half of the parabola: typically the right branch $[h, \infty)$, or alternatively the left branch $(-\infty, h]$.
3. Determine the resulting range: for $f(x) = (x - h)^2 + k$ on $[h, \infty)$, the range is $[k, \infty)$.
4. Swap $x$ and $y$: $x = (y - h)^2 + k \implies x - k = (y - h)^2$.
5. Take the square root of both sides: $y - h = \pm\sqrt{x - k}$.
6. Because the original domain was restricted to $x \ge h$, the inverse must satisfy $y \ge h$. Therefore, select the positive principal square root: $f^{-1}(x) = \sqrt{x - k} + h \quad \text{with Domain } [k, \infty) \text{ and Range } [h, \infty)$

If the domain had been restricted to the left branch $(-\infty, h]$, we would select the negative root: $f^{-1}(x) = -\sqrt{x - k} + h$.

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Summary of Inverse Function Properties

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Worked Examples

Worked Problem 1: Finding the Inverse of a Rational Function

Problem: Algebraically find the inverse function $f^{-1}(x)$ for $f(x) = \frac{3x - 5}{2x + 1}$, where $x \neq -\frac{1}{2}$. State its domain.

• Step 1: Replace $f(x)$ with $y$. $y = \frac{3x - 5}{2x + 1}$

• Step 2: Swap $x$ and $y$. $x = \frac{3y - 5}{2y + 1}$

• Step 3: Clear the denominator. Multiply both sides by $(2y + 1)$: $x(2y + 1) = 3y - 5 \implies 2xy + x = 3y - 5$

• Step 4: Group terms containing $y$ on one side. Subtract $3y$ and subtract $x$: $2xy - 3y = -x - 5$

• Step 5: Factor out $y$ and isolate. $y(2x - 3) = -(x + 5) \implies y = \frac{-x - 5}{2x - 3} = \frac{x + 5}{3 - 2x}$

• Step 6: Write in inverse notation with domain restriction. $f^{-1}(x) = \frac{x + 5}{3 - 2x} \quad \text{where } x \neq \frac{3}{2}$

Worked Problem 2: Verifying an Inverse Relationship via Composition

Problem: Algebraically verify whether $f(x) = \frac{x + 7}{4}$ and $g(x) = 4x - 7$ are inverse functions.

• Step 1: Evaluate the composition $f(g(x))$. Substitute $g(x) = 4x - 7$ into the input slot of $f$: $f(g(x)) = f(4x - 7) = \frac{(4x - 7) + 7}{4}$ Simplify the numerator: $f(g(x)) = \frac{4x}{4} = x$

• Step 2: Evaluate the composition $g(f(x))$. Substitute $f(x) = \frac{x + 7}{4}$ into the input slot of $g$: $g(f(x)) = g\left(\frac{x + 7}{4}\right) = 4\left(\frac{x + 7}{4}\right) - 7$ Cancel the factor of 4: $g(f(x)) = (x + 7) - 7 = x$

• Step 3: State the conclusion. Since $f(g(x)) = x$ and $g(f(x)) = x$ for all real numbers, $f(x)$ and $g(x)$ are confirmed inverse functions.

Worked Problem 3: Restricting a Quadratic Domain and Deriving its Inverse

Problem: Consider the quadratic function $f(x) = (x - 3)^2 + 4$ restricted to the domain $[3, \infty)$. Determine the equation for $f^{-1}(x)$ and state its domain and range.

• Step 1: Identify vertex, restricted domain, and range. The vertex is $(3, 4)$. On $[3, \infty)$, the function values begin at $f(3) = 4$ and increase without bound. Thus, the range of $f$ is $[4, \infty)$.

• Step 2: Swap variables. $x = (y - 3)^2 + 4$

• Step 3: Isolate the squared binomial. $x - 4 = (y - 3)^2$

• Step 4: Take the square root and select the correct branch. $\pm\sqrt{x - 4} = y - 3$ Because the original domain was restricted to $x \ge 3$, the inverse output must satisfy $y \ge 3$. Therefore, select the positive square root: $y - 3 = \sqrt{x - 4} \implies y = \sqrt{x - 4} + 3$

• Step 5: State the final inverse with domain and range. $f^{-1}(x) = \sqrt{x - 4} + 3$ $\text{Domain: } [4, \infty), \quad \text{Range: } [3, \infty)$

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Common Regents Pitfalls & Exam Strategies

• Pitfall 1: Confusing functional inverse notation with reciprocals. Students frequently write $f^{-1}(x) = \frac{1}{f(x)}$. The reciprocal flips the fraction, but the inverse function swaps input and output variables. For example, if $f(x) = 2x + 6$, the reciprocal is $\frac{1}{2x + 6}$, but the inverse is $\frac{x - 6}{2}$.
• Pitfall 2: Forgetting to choose a sign when inverting restricted quadratics. Writing $f^{-1}(x) = 3 \pm \sqrt{x - 4}$ fails the Vertical Line Test and is not a function. You must inspect the restricted domain ($x \ge h$ vs. $x \le h$) to select the positive or negative root.
• Pitfall 3: Evaluating only one composition during verification. On Part II and III constructed-response items, proving that $f(g(x)) = x$ without also proving that $g(f(x)) = x$ results in a 1-credit deduction. Both compositions must be fully worked out.