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URL: https://nakafa.com/en/subjects/mathematics/function-composition-inverse-function/properties-of-inverse-function
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Learn inverse function properties, including composition identity, double reversal, and domain-range relationships.

---

## 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 $$I(x) = x$$.

Visible text: 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 .

1.  **Composition of $$f$$ with $$f^{-1}$$:**

    
    
    ```math
    (f \circ f^{-1})(x) = f(f^{-1}(x)) = x
    ```

    This holds for all $$x$$ in the domain of $$f^{-1}$$ (which is the range of $$f$$).

2.  **Composition of $$f^{-1}$$ with $$f$$:**

    
    
    ```math
    (f^{-1} \circ f)(x) = f^{-1}(f(x)) = x
    ```

    This holds for all $$x$$ in the domain of $$f$$.

Visible text: 1. **Composition of with :**

 
 

 This holds for all in the domain of (which is the range of ).

2. **Composition of with :**

 
 

 This holds for all in the domain of .

**Example:**

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

Visible text: We know that if , its inverse is . Let's verify the composition property:

- $$f(f^{-1}(x)) = f\left(\frac{x - 3}{2}\right) = 2\left(\frac{x - 3}{2}\right) + 3 = (x - 3) + 3 = x$$
- $$f^{-1}(f(x)) = f^{-1}(2x + 3) = \frac{(2x + 3) - 3}{2} = \frac{2x}{2} = x$$

Visible text: - 
-

Both compositions result in $$x$$, as expected.

Visible text: Both compositions result in , as expected.

## Property of the Inverse of an Inverse

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

```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 $$f$$ and $$g$$ be two functions with inverses $$f^{-1}$$ and $$g^{-1}$$. Then the inverse of the composition $$f \circ g$$ is:

Visible text: Let and be two functions with inverses and . Then the inverse of the composition is:

```math
(f \circ g)^{-1}(x) = (g^{-1} \circ f^{-1})(x)
```

Note the reversed order: $$g^{-1}$$ is applied first, then $$f^{-1}$$.

Visible text: Note the reversed order: is applied first, then .

Imagine putting on socks ($$g$$) and then shoes ($$f$$). To undo this (the inverse), you must take off the shoes ($$f^{-1}$$) first, then take off the socks ($$g^{-1}$$). The order is reversed.

Visible text: Imagine putting on socks () and then shoes (). To undo this (the inverse), you must take off the shoes () first, then take off the socks (). The order is reversed.

**Example:**

Let $$f(x) = x + 1$$ (its inverse is $$f^{-1}(x) = x - 1$$) and $$g(x) = 3x$$ (its inverse is $$g^{-1}(x) = \frac{x}{3}$$).

Visible text: Let (its inverse is ) and (its inverse is ).

1.  **Find $$(f \circ g)(x)$$:**

    
    
    ```math
    (f \circ g)(x) = f(g(x)) = f(3x) = 3x + 1
    ```

2.  **Find the inverse of $$(f \circ g)(x)$$:**

    Let $$y = 3x + 1$$. Swap $$x$$ and $$y$$: $$x = 3y + 1$$.

    Solve for $$y$$: $$x - 1 = 3y \implies y = \frac{x - 1}{3}$$.

    So, $$(f \circ g)^{-1}(x) = \frac{x - 1}{3}$$.

3.  **Find $$(g^{-1} \circ f^{-1})(x)$$:**

    <MathContainer>
      
    
    ```math
    (g^{-1} \circ f^{-1})(x) = g^{-1}(f^{-1}(x)) = g^{-1}(x - 1)
    ```

      
    
    ```math
    g^{-1}(x - 1) = \frac{x - 1}{3}
    ```

    </MathContainer>

Visible text: 1. **Find :**

 
 

2. **Find the inverse of :**

 Let . Swap and : .

 Solve for : .

 So, .

3. **Find :**

 <MathContainer>
 
 

 
 

 </MathContainer>

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

Visible text: Since the results from steps and are the same, it is proven that .

## Domain and Range Relationship

The domain of the original function $$f$$ becomes the range of its inverse function $$f^{-1}$$, and the range of the original function $$f$$ becomes the domain of its inverse function $$f^{-1}$$.

Visible text: The domain of the original function becomes the range of its inverse function , and the range of the original function becomes the domain of its inverse function .

Component: MathContainer
Children:

```math
\text{Domain}(f) = \text{Range}(f^{-1})
```

```math
\text{Range}(f) = \text{Domain}(f^{-1})
```