# Composite and Inverse Functions

## Composite and Inverse Functions Revision

**Composite and Inverse Functions**

A mapping takes an input in one set of values to an output in another set. A **function** is a type of mapping. You need to be able to understand the language of functions and how to work with **composite functions** and **inverse functions**.

**Language of Functions**

There are some **key terms** that you need to understand before we look at **composite **and **inverse functions**:

- A
**function**is an operation that maps each number to only one number, e.g. x^3 is f(x) = x^3 or f : x \rightarrow x^3 - The
**domain**is the set of input (starting) values. - The
**range**is the set of possible output values. - The domain and/or range is usually the set of
**real numbers**, denoted \mathbb{R} (i.e. any number, decimal, fraction, surd etc.). If x is any real number then we write x \in \mathbb{R} - A function that maps
**one**element in the domain to**one**element in the range is called a**one-to-one function**. - A function that maps
**more than one**element in the domain to**one**element in the range is called a**many-to-one function**.

**Example:** State the domain and range of the following functions seen in the graphs:

Both functions have domain x \in \mathbb{R}.

f(x) has a range f(x) \in \mathbb{R} and it is a one-to-one function.

g(x) has a range g(x) \geq 0 and it is a many-to-one function.

**Composite Functions**

A **composite function** is a combination of two or more functions, say f and g, that makes a new function.

We write **composite functions** as fg(x) which means that we do g first, and then do f – we can rewrite this as f(g(x)) to make things clearer. The order is vital – in most cases fg(x) \neq gf(x) (they are different functions).

You may see squared functions, e.g. f^2 (x) – this is the same as ff(x) (you do f twice)

**Example:** For the functions \textcolor{red}{f} : x \rightarrow 5x^2 \,\,\, \{ x \in \mathbb{R} \} and \textcolor{limegreen}{g} : x \rightarrow x-1 \,\,\, \{ x\in \mathbb{R} \}, find

\textcolor{red}{f} \textcolor{limegreen}{g}(3), \textcolor{limegreen}{g} \textcolor{red}{f}(3), \textcolor{red}{f}^2(x) and \textcolor{red}{f} \textcolor{limegreen}{g}(x)

\textcolor{red}{f} \textcolor{limegreen}{g}(3) = \textcolor{red}{f} (\textcolor{limegreen}{g}(3)) = \textcolor{red}{f}(3-1) = \textcolor{red}{f}(2) = 5(2)^2 = 20

\textcolor{limegreen}{g} \textcolor{red}{f}(3) = \textcolor{limegreen}{g}( \textcolor{red}{f}(3)) = \textcolor{limegreen}{g}(5(3)^2) = \textcolor{limegreen}{g}(45) = 45 - 1 = 44

\textcolor{red}{f} ^2(x) = \textcolor{red}{f} ( \textcolor{red}{f}(x)) = 5(5x^2)^2 = 125x^4

\textcolor{red}{f} \textcolor{limegreen}{g} (x) = \textcolor{red}{f} (\textcolor{limegreen}{g}(x)) = \textcolor{red}{f}(x-1) = 5(x-1)^2

**Solving Composite Functions Equations**

You may be asked to **solve equations** involving **composite functions**, e.g. fg(x) = 2. Just find fg(x) first and then rearrange and solve to find x.

**Example:** For the functions \textcolor{red}{f} : x \rightarrow x^2 \,\,\, \{ x \in \mathbb{R} \} and \textcolor{limegreen}{g} : x \rightarrow x+2 \,\,\, \{ x\in \mathbb{R} \}, solve

\textcolor{red}{f} \textcolor{limegreen}{g} (x) = 4

Also, state the range of fg(x).

\textcolor{red}{f} \textcolor{limegreen}{g} (x) = \textcolor{red}{f} ( \textcolor{limegreen}{g} (x)) = \textcolor{red}{f} (x+2) = (x+2)^2

So,

(x+2)^2 = 4

Rearrange this and solve to find the values of x:

\begin{aligned} (x+2)^2 &= 4 \\ x+2 &= \pm 2 \end{aligned}

x = -4 or x = 0

To find the range, it may be helpful to draw the graph of \textcolor{red}{f} \textcolor{limegreen}{g} (x):

Hence, the range is \textcolor{red}{f} \textcolor{limegreen}{g} (x) \geq 0.

**Inverse Functions**

An **inverse** of a function does the opposite of that function. The inverse of a function f(x), is written as f^{-1} (x).

An **inverse function** maps an element in the range to an element in the domain (the opposite of a function). Hence, only **one-to-one functions** have inverses.

The **domain** of the **inverse** is the same as the **range** of the function. The **range** of the **inverse** is the same as the **domain** of the function.

A **composite function** of a **function** and its **inverse**, and vice versa, gives x – i.e. f^{-1}f(x) = ff^{-1} (x) = x

To work out the **inverse** of a function, you need to rearrange the function and change the subject.

**Example:** Find the inverse of f(x) = 2x^2 + 5, with domain x \textcolor{red}{\geq 0}. State the domain and range of f^{-1} (x).

**Step 1:** Replace f(x) with y in the equation:

y = 2x^2 + 5

**Step 2:** Rearrange to make x the subject:

\begin{aligned} y &= 2x^2 + 5 \\[1.1em] y-5 &= 2x^2 \\[1.1em] \dfrac{y-5}{2} &= x^2 \\[1.1em] x &= \sqrt{\dfrac{y-5}{2}} \end{aligned}

x \geq 0, so we don’t need the negative square root.

**Step 3:** Replace x with f^{-1} (x) and y with x:

\textcolor{limegreen}{f^{-1} (x) = \sqrt{\dfrac{x-5}{2}}}

**Step 4:** Swap the domain and range:

The domain of f(x) is given as x \textcolor{red}{\geq 0} and its range is f(x) \textcolor{blue}{\geq 5}

Hence, the domain of f^{-1} (x) is x \textcolor{blue}{\geq 5} and its range is f^{-1} (x) \textcolor{red}{\geq 0}

**Note:** For simple functions, you can work out the inverse by looking at it – e.g. f(x) = x-2 has inverse f^{-1}(x) = x+2

**Drawing Inverse Functions**

For a **function** f(x) plotted on a graph, its **inverse** f^{-1} (x) is its reflection in the line y=x.

**Example:** Given that f(x) = x^2 - 1 with domain x \geq 0, sketch the graph of f^{-1}(x).

Firstly, draw \textcolor{red}{f(x)}. Then draw in the line y=x. Finally, reflect f(x) in y=x to get \textcolor{blue}{f^{-1}(x)}.

The inverse function is \textcolor{blue}{f^{-1} (x)} = \sqrt{x+1}

You can see from the graph that f(x) has domain x \geq 0 and range f(x) \geq -1, and f^{-1} (x) has domain x \geq -1 and range f^{-1} (x) \geq 0.

## Composite and Inverse Functions Example Questions

**Question 1:** For the functions f : x \rightarrow x+3, \{ x \in \mathbb{R} \} and g : x \rightarrow \dfrac{2}{2x+1}, \{ x \neq - \dfrac{1}{2} \}, find

**a)** gf(x)

**b)** fg(3)

**[4 marks]**

**a)**

\begin{aligned} gf(x) &= g(f(x)) \\[1.1em] &= g(x+3) \\[1.1em] &= \dfrac{2}{2(x+3)+1} \\[1.1em] &= \dfrac{2}{2x+7} \end{aligned}

**b)**

\begin{aligned} fg(3) &= f(g(3)) \\[1.1em] &= f \left( \dfrac{2}{2(3)+1} \right) \\[1.1em] &= f \left ( \dfrac{2}{7} \right) \\[1.1em] &= \dfrac{2}{7} + 3 \\[1.1em] &= \dfrac{23}{7} \end{aligned}

**Question 2:** For the function f : x \rightarrow 2x^2 - 3, \{ x \in \mathbb{R} \}, find

**a)** f^2(x)

**b)** f^2(2)

**[3 marks]**

a)

\begin{aligned} f^2(x) &= f(f(x)) \\ &= 2(2x^2 - 3)^2 - 3 \\ &= 2(4x^4 - 12x^2 + 9) - 3 \\ &= 8x^4 - 24x^2 + 18 - 3 \\ &= 8x^4 - 24x^2 + 15 \end{aligned}

b)

f^2(2) = 8(2)^4 - 24(2)^2 + 15 = 47

**Question 3:** For the functions f : x \rightarrow x^2 +3, \{ x \in \mathbb{R} \} and g : x \rightarrow \dfrac{6}{x-2}, \{ x \neq 2 \}, solve gf(x) = 2

**[3 marks]**

\begin{aligned} gf(x) &= g(f(x)) \\[1.1em] &= \dfrac{6}{(x^2 + 3) - 2} \\[1.1em] &= \dfrac{6}{x^2 + 1} \end{aligned}

So, we can solve the equation to find x

\begin{aligned} gf(x) &= 2 \\ \dfrac{6}{x^2 + 1} &= 2 \\ 6 &= 2(x^2 + 1) \\ 6 &= 2x^2 + 2 \\ 4 &= 2x ^2 \\ 2 &= x^2 \\ x &= \pm \sqrt{2} \end{aligned}

**Question 4: **

**a)** Find the inverse of f(x) = \dfrac{4}{x+3} with domain x > -3.

**b)** State the domain and range of the inverse.

**[4 marks]**

**a)** Replace f(x) with y:

y = \dfrac{4}{x+3}

Rearrange to make x the subject:

\begin{aligned} y(x+3) &= 4 \\ xy + 3y &= 4 \\ xy &= 4 - 3y \\ x &= \dfrac{4 - 3y}{y} \end{aligned}

Replace x with f^{-1} (x) and y with x:

f^{-1} (x) = \dfrac{4 - 3x}{x}

**b)** f(x) has domain x > -3 and range x > 0.

Hence, f^{-1} (x) has domain x>0 and range x > -3