# Nakafa Framework: LLM

URL: /en/subject/high-school/11/mathematics/function-composition-inverse-function/properties-of-inverse-function
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Output docs content for large language models.

---

export const metadata = {
  title: "Properties of Inverse Function",
  description: "Discover inverse function properties: composition identity, double reversal, and domain-range relationships. Understand f∘f⁻¹=x with examples.",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "04/27/2025",
  subject: "Function Composition and Inverse Function",
};

## Property of Composition with Inverse

This property is the core of the inverse function definition: the inverse function "undoes" the effect of the original function, and vice versa. If we compose a function with its inverse (in any order), we get the identity function <InlineMath math="I(x) = x" />.

1.  **Composition of <InlineMath math="f" /> with <InlineMath math="f^{-1}" />:**

    <BlockMath math="(f \circ f^{-1})(x) = f(f^{-1}(x)) = x" />

    This holds for all <InlineMath math="x" /> in the domain of <InlineMath math="f^{-1}" /> (which is the range of <InlineMath math="f" />).

2.  **Composition of <InlineMath math="f^{-1}" /> with <InlineMath math="f" />:**

    <BlockMath math="(f^{-1} \circ f)(x) = f^{-1}(f(x)) = x" />

    This holds for all <InlineMath math="x" /> in the domain of <InlineMath math="f" />.

**Example:**

We know that if <InlineMath math="f(x) = 2x + 3" />, its inverse is <InlineMath math="f^{-1}(x) = \frac{x - 3}{2}" />. Let's verify the composition property:

- <InlineMath math="f(f^{-1}(x)) = f\left(\frac{x - 3}{2}\right) = 2\left(\frac{x - 3}{2}\right) + 3 = (x - 3) + 3 = x" />
- <InlineMath math="f^{-1}(f(x)) = f^{-1}(2x + 3) = \frac{(2x + 3) - 3}{2} = \frac{2x}{2} = x" />

Both compositions result in <InlineMath math="x" />, as expected.

## Property of the Inverse of an Inverse

If we find the inverse of an inverse function, we get back the original function.

<BlockMath math="(f^{-1})^{-1}(x) = f(x)" />

This makes sense because the process of finding an inverse is a "reversal". If we reverse something twice, we return to the original state.

## Property of the Inverse of a Composition

If we have a composition of two functions, both of which have inverses, the inverse of the composition is the composition of their inverses, but in **reverse order**.

Let <InlineMath math="f" /> and <InlineMath math="g" /> be two functions with inverses <InlineMath math="f^{-1}" /> and <InlineMath math="g^{-1}" />. Then the inverse of the composition <InlineMath math="f \circ g" /> is:

<BlockMath math="(f \circ g)^{-1}(x) = (g^{-1} \circ f^{-1})(x)" />

Note the reversed order: <InlineMath math="g^{-1}" /> is applied first, then <InlineMath math="f^{-1}" />.

**Analogy:** Imagine putting on socks (<InlineMath math="g" />) and then shoes (<InlineMath math="f" />). To undo this (the inverse), you must take off the shoes (<InlineMath math="f^{-1}" />) first, then take off the socks (<InlineMath math="g^{-1}" />). The order is reversed.

**Example:**

Let <InlineMath math="f(x) = x + 1" /> (its inverse is <InlineMath math="f^{-1}(x) = x - 1" />) and <InlineMath math="g(x) = 3x" /> (its inverse is <InlineMath math="g^{-1}(x) = \frac{x}{3}" />).

1.  **Find <InlineMath math="(f \circ g)(x)" />:**

    <BlockMath math="(f \circ g)(x) = f(g(x)) = f(3x) = 3x + 1" />

2.  **Find the inverse of <InlineMath math="(f \circ g)(x)" />:**

    Let <InlineMath math="y = 3x + 1" />. Swap <InlineMath math="x" /> and <InlineMath math="y" />: <InlineMath math="x = 3y + 1" />.

    Solve for <InlineMath math="y" />: <InlineMath math="x - 1 = 3y \implies y = \frac{x - 1}{3}" />.

    So, <InlineMath math="(f \circ g)^{-1}(x) = \frac{x - 1}{3}" />.

3.  **Find <InlineMath math="(g^{-1} \circ f^{-1})(x)" />:**

    <div className="flex flex-col gap-4">
      <BlockMath math="(g^{-1} \circ f^{-1})(x) = g^{-1}(f^{-1}(x)) = g^{-1}(x - 1)" />
      <BlockMath math="g^{-1}(x - 1) = \frac{x - 1}{3}" />
    </div>

Since the results from steps 2 and 3 are the same, it is proven that <InlineMath math="(f \circ g)^{-1}(x) = (g^{-1} \circ f^{-1})(x)" />.

## Domain and Range Relationship

The domain of the original function <InlineMath math="f" /> becomes the range of its inverse function <InlineMath math="f^{-1}" />, and the range of the original function <InlineMath math="f" /> becomes the domain of its inverse function <InlineMath math="f^{-1}" />.

<div className="flex flex-col gap-4">
  <BlockMath math="\text{Domain}(f) = \text{Range}(f^{-1})" />
  <BlockMath math="\text{Range}(f) = \text{Domain}(f^{-1})" />
</div>