Composite and Inverse Functions Revision Notes for VCE SSCE Mathematical Methods

Composite and Inverse Functions

Composition of functions

When we compose two functions, we apply one function to the output of another function. This creates a new function called a composite function.

The composition of $g$ with $f$ is written as $g \circ f$ (read as "composition of $f$ followed by $g$ "). The rule for the composite function is:

g \circ f(x) = g(f(x))

This means we first apply function $f$ to $x$ , then apply function $g$ to the result.

infoNote

Think of composition as a "function pipeline": the output of one function becomes the input to the next. Reading $g \circ f$ from right to left helps remember the order: " $f$ first, then $g$ ".

Domain requirements for composition

For the composition $g \circ f$ to be defined, the range of $f$ must be contained within the domain of $g$ . In mathematical notation:

If $\text{ran } f \subseteq \text{dom } g$ , then the composition $g \circ f$ is defined and $\text{dom}(g \circ f) = \text{dom } f$

This makes sense because every output from $f$ needs to be a valid input for $g$ .

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Worked Example: Expressing Functions as Compositions

Express each of the following as the composition of two functions:

a) $h(x) = e^{x^2}$

Choose $f(x) = x^2$ and $g(x) = e^x$

Then $h(x) = g \circ f(x) = g(f(x)) = g(x^2) = e^{x^2}$

We first square the input ( $f$ ), then take the exponential of the result ( $g$ ).

b) $h(x) = \sin(x^2)$

Choose $f(x) = x^2$ and $g(x) = \sin x$

Then $h(x) = g \circ f(x) = g(f(x)) = g(x^2) = \sin(x^2)$

We first square the input ( $f$ ), then take the sine of the result ( $g$ ).

c) $h(x) = (x^2 - 2)^n$ , where $n \in \mathbb{N}$

Choose $f(x) = x^2 - 2$ and $g(x) = x^n$

Then $h(x) = g \circ f(x) = g(f(x)) = g(x^2 - 2) = (x^2 - 2)^n$

We first compute $x^2 - 2$ ( $f$ ), then raise the result to the power $n$ ( $g$ ).

Note: These are not the only possible answers, but they represent the most natural choices for breaking down each function.

Inverse functions

An inverse function reverses the action of the original function. If a function $f$ takes an input $a$ and produces an output $b$ , then the inverse function $f^{-1}$ takes $b$ back to $a$ .

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For a function to have an inverse, it must be one-to-one. This means different inputs must produce different outputs. In mathematical terms, a function $f$ is one-to-one if $a \neq b$ implies $f(a) \neq f(b)$ , for all $a, b \in \text{dom } f$ .

Definition of inverse function

If $f$ is a one-to-one function, then the inverse function $f^{-1}$ is defined by:

f^{-1}(x) = y \text{ if } f(y) = x, \text{ for } x \in \text{ran } f \text{ and } y \in \text{dom } f

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The domain of $f^{-1}$ is the range of $f$ , and the range of $f^{-1}$ is the domain of $f$ . Think of the inverse as "swapping" the domain and range of the original function.

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Worked Example: Finding Inverses and Compositions

Let $f(x) = e^{2x}$ and let $g(x) = \frac{1}{\sqrt{x}}$ for $x \in \mathbb{R}^+$ . Find:

a) $f^{-1}$

To find the inverse, we set $x = e^{2y}$ and solve for $y$ :

Taking natural logarithm of both sides:

\log_e x = 2y

y = \frac{1}{2}\log_e x

Therefore, $f^{-1}(x) = \frac{1}{2}\log_e x$ , where $x \in \mathbb{R}^+$

b) $g^{-1}$

Set $x = \frac{1}{\sqrt{y}}$ and solve for $y$ :

\sqrt{y} = \frac{1}{x}

Squaring both sides:

y = \frac{1}{x^2}

Therefore, $g^{-1}(x) = \frac{1}{x^2}$ , where $x \in \mathbb{R}^+$

c) $f \circ g$

(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{\sqrt{x}}\right) = e^{2 \cdot \frac{1}{\sqrt{x}}} = e^{\frac{2}{\sqrt{x}}}

Domain: $x \in \mathbb{R}^+$

d) $g \circ f$

(g \circ f)(x) = g(f(x)) = g(e^{2x}) = \frac{1}{\sqrt{e^{2x}}} = \frac{1}{e^x}

Domain: $x \in \mathbb{R}$

e) $(f \circ g)^{-1}$

To find this inverse, let $x = e^{\frac{2}{\sqrt{y}}}$

Taking natural logarithm:

\log_e x = \frac{2}{\sqrt{y}}

\sqrt{y} = \frac{2}{\log_e x}

Squaring both sides:

y = \left(\frac{2}{\log_e x}\right)^2

Therefore, $(f \circ g)^{-1}(x) = \left(\frac{2}{\log_e x}\right)^2$ , where $x \in (1, \infty)$

f) $(g \circ f)^{-1}$

Let $x = \frac{1}{e^y}$

Then:

e^y = \frac{1}{x}

Taking natural logarithm:

y = \log_e\left(\frac{1}{x}\right) = -\log_e x

Therefore, $(g \circ f)^{-1}(x) = -\log_e x$ , where $x \in \mathbb{R}^+$

Strictly increasing and strictly decreasing functions

Understanding whether a function is increasing or decreasing helps us determine if it has an inverse and understand the behaviour of that inverse.

Definitions

A function $f$ is strictly increasing if whenever $a > b$ , we have $f(a) > f(b)$ , for all $a, b \in \text{dom } f$ .

In other words, larger inputs always produce larger outputs. The function consistently goes up as we move from left to right.

A function $f$ is strictly decreasing if whenever $a > b$ , we have $f(a) < f(b)$ , for all $a, b \in \text{dom } f$ .

In other words, larger inputs always produce smaller outputs. The function consistently goes down as we move from left to right.

Relationship to one-to-one functions

Both strictly increasing and strictly decreasing functions are always one-to-one. This means they always have inverse functions.

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Key Results:

If $f$ is strictly increasing, then it is a one-to-one function.

If $f$ is strictly decreasing, then it is a one-to-one function.

Proof that strictly increasing functions are one-to-one

Assume $f$ is strictly increasing and let $a, b \in \text{dom } f$ with $a \neq b$ .

Then either $a > b$ or $b > a$ .

If $a > b$ , then $f(a) > f(b)$ since $f$ is strictly increasing. Therefore $f(a) \neq f(b)$ .

If $b > a$ , then $f(b) > f(a)$ since $f$ is strictly increasing. Therefore $f(a) \neq f(b)$ .

In both cases, we have f(a) ≠ f(b).

Hence $f$ is a one-to-one function.

infoNote

The proof for strictly decreasing functions follows similar logic. In that case, if $a > b$ , then $f(a) < f(b)$ , which still gives us $f(a) \neq f(b)$ , confirming the function is one-to-one.

Behaviour of inverse functions

An important property connects the monotonicity of a function to the monotonicity of its inverse:

chatImportant

Key Results:

If $f$ is strictly increasing, then $f^{-1}$ is also strictly increasing.

If $f$ is strictly decreasing, then $f^{-1}$ is also strictly decreasing.

Proof that inverse of strictly increasing is strictly increasing

Assume $f$ is strictly increasing and let $a, b \in \text{dom } f^{-1}$ with $a > b$ .

We need to show that $f^{-1}(a) > f^{-1}(b)$ .

If $f^{-1}(a) = f^{-1}(b)$ , then applying $f$ to both sides gives:

f(f^{-1}(a)) = f(f^{-1}(b))

This means $a = b$ , which contradicts our assumption that $a > b$ .

If $f^{-1}(a) < f^{-1}(b)$ , then since $f$ is strictly increasing:

f(f^{-1}(a)) < f(f^{-1}(b))

This means $a < b$ , which contradicts our assumption that $a > b$ .

Therefore, we must have f⁻¹(a) > f⁻¹(b), which proves that $f^{-1}$ is strictly increasing.

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Worked Example: Inverse Functions and Monotonicity

For each function $f$ , find the inverse function $f^{-1}$ , and state whether $f$ and $f^{-1}$ are strictly increasing, strictly decreasing or neither.

a) $f : \mathbb{R}^+ \cup \{0\} \to \mathbb{R}$ , $f(x) = x^{\frac{2}{3}}$

To find the inverse, write $x = y^{\frac{2}{3}}$ and solve for $y$ :

x = (y^{\frac{1}{3}})^2

y^{\frac{1}{3}} = \pm\sqrt{x}

y = (\pm\sqrt{x})^3 = \pm x^{\frac{3}{2}}

Since the domain of $f$ is $\mathbb{R}^+ \cup \{0\}$ (non-negative numbers), we need the positive solution.

The range of $f$ is $\mathbb{R}^+ \cup \{0\}$ .

Therefore, $f^{-1} : \mathbb{R}^+ \cup \{0\} \to \mathbb{R}$ , $f^{-1}(x) = x^{\frac{3}{2}}$

Both $f$ and $f^{-1}$ are strictly increasing.

b) $f : \mathbb{R}^- \cup \{0\} \to \mathbb{R}$ , $f(x) = x^{\frac{2}{3}}$

Write $x = y^{\frac{2}{3}}$ . Then $y = \pm x^{\frac{3}{2}}$ .

Since the domain of $f$ is $\mathbb{R}^- \cup \{0\}$ (non-positive numbers), we need the negative solution.

The range of $f$ is $\mathbb{R}^+ \cup \{0\}$ .

Therefore, $f^{-1} : \mathbb{R}^+ \cup \{0\} \to \mathbb{R}$ , $f^{-1}(x) = -x^{\frac{3}{2}}$

Both $f$ and $f^{-1}$ are strictly decreasing.

The graphs above illustrate these relationships. The top graph shows the strictly increasing function $f(x) = x^{\frac{2}{3}}$ (defined on non-negative numbers) and its strictly increasing inverse $f^{-1}(x) = x^{\frac{3}{2}}$ . The bottom graph shows the strictly decreasing function $f(x) = x^{\frac{2}{3}}$ (defined on non-positive numbers) and its strictly decreasing inverse $f^{-1}(x) = -x^{\frac{3}{2}}$ . Notice how the functions and their inverses are reflections of each other across the line $y = x$ (shown as a dashed line).

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Key Points to Remember:

The composition $g \circ f(x) = g(f(x))$ means "apply $f$ first, then apply $g$ to the result"
For composition to be defined, the range of $f$ must be contained in the domain of $g$
A function has an inverse if and only if it is one-to-one (different inputs give different outputs)
Strictly increasing and strictly decreasing functions are always one-to-one, so they always have inverses
If $f$ is strictly increasing, then $f^{-1}$ is also strictly increasing; if $f$ is strictly decreasing, then $f^{-1}$ is also strictly decreasing
The domain of $f^{-1}$ equals the range of $f$ , and the range of $f^{-1}$ equals the domain of $f$

Composite and Inverse Functions (VCE SSCE Mathematical Methods): Revision Notes