Composition of Functions

Consider the functions \( f(x) = 3x + 2 \) and \( g(x) = 5x + 4 \). Find the function \( h(x) = (f \circ g)(x) \).

Solution:

To find the composite, replace each \( x \) in function \( f(x) \) with the the function \( g(x) \).

\[ \begin{align} h(x) &= (f \circ g)(x) \\ &= 3g(x) + 2 \\ &= 3(5x + 4) + 2 . \end{align}\]

Then it's just a case of simplifying by expanding the brackets and collecting like terms.

\[ \begin{align} h(x) &= 3(5x + 4) + 2 \\ &=15x + 12 + 2 \\ &= 15x + 14 . \end{align} \]

Given the functions

\[ f(x) = \frac{1}{x-2}, \quad \{ x \in \mathbb{R} | x \neq 2 \} \]

and

\[ g(x) = x + 2, \quad \{ x \in \mathbb{R} \}, \]

what is the domain of composite function \( f(g(x)) \)?

Solution:

The outputs of the function \( g(x) \) must fall within the domain of \( f(x) \), therefore they must be a subsection of the set of all real numbers.

\[ \{ x \in \mathbb{R}\}\]

You also know that the output of function \( g(x) \) must not equal \( 2 \).

\[ g(x) \neq 2 \]

\[ \frac{1}{x-2} \neq 2 \]

\[ x \neq 2.5\]

Therefore the domain of the function \( f(g(x)) \) is everything except \(2.5\), or on other words

\[ \{ x \in \mathbb{R} | x \neq 2.5 \}. \]

Take the functions \( g(x) = \sqrt{x}\) and

\[ f(x) = \frac{1}{x-2}.\]

Find \( (f \circ g)(x) \) and the domain of the composite function.

Solution:

Notice that the function \(g(x)\) is a square root, so its domain is \([0, \infty)\). So already you know that the domain of the composition function isn't all real numbers. If you do the composition,

\[ (f \circ g) (x) = \frac{1}{\sqrt{x}-2}.\]

You know that you can't have a zero in the denominator, so the domain of \(f \circ g\) needs to have

\[ \sqrt{x} - 2 \not= 0,\]

which means

\[ \sqrt{x} \not= 2,\]

so

\[ x \not= 4.\]

Combining the restrictions on the domain of \(g(x)\) with the restriction that \(x \not= 4\) gives you that the composition \( (f \circ g)(x)\) has a domain of \( [0, 4) \cup (4, \infty)\).

First, let's look at two functions with different variables. Consider the functions

\[ f(x) = \sqrt{x + 2} \]

and

\[ g(y) = 10y. \]

Let's find the composition of these functions \( f(g(y)) \). Notice that the composite function is only a function of \( y \), and not of \( x \).

Solution:

As before, you substitute the value of \( g(y) \) for the \( x \) in \( f(x) \) to find the resultant composite function.

\[ f(g(y)) = \sqrt{10y + 2} \]

The domain of this composite function will also be in terms of \( y \). As you cannot take the square root of a negative (assuming you neglect imaginary numbers) the domain is as follows:

\[ \begin{align} 10y + 2 &> 0\\ 10y &> -2 \\ y &>-0.2 . \end{align}\]

So the whole composite function, including the function domain is

\[f(g(y)) = \sqrt{10y + 2}, \quad\{y \in \mathbb{R}, y > -0.2 \}. \]

Given the functions \( f(x) = 5x \) and \( g(x) = x + 4 \), evaluate the function \( h(x) = (f \circ g)(x)) \) at \( x = 3 \).

Solution:

First do the composition:

\[ \begin{align} h(x) &= (f \circ g)(x)\\ &= 5(g(x))\\ & = 5(x + 4) \\ &= 5x + 20. \end{align} \]

Then plug in \(x = 3\) to get

\[ \begin{align} h(3) & = 5\cdot 3 + 20 \\ &=15 + 20\\ & = 35. \end{align}\]

Given the function

\[ f(x) = \frac{x}{x+1},\]

show that \( f^{-1}(f(x)) = x \).

Solution:

First you find \( f^{-1}(x) \) by swapping \( x\) and \( y \):

\[ f(x) = y = \frac{x}{x+1}, \]

so

\[ x = \frac{y}{y + 1}. \]

Then you rearrange for \( y \) in terms of \( x \) to get:

\[ \begin{align} \frac{(y+1)}{y} &= \frac{1}{x} \\ 1 + \frac{1}{y} &= \frac{1}{x} \\ \frac{1}{y} &= \frac{1}{x} - 1 \\ \frac{1}{y} &= \frac{1-x}{x} \\ y &= \frac{x}{1-x} . \end{align}\]

Now, let's check and see if you have found the inverse correctly by evaluating \( h(x) = f\left(f^{-1}(x)\right) \):

\[ \begin{align} h(x) &= \frac{x}{x+1}\\ & = \frac{\frac{x}{1-x}}{\frac{x}{1-x}+1} \\ &= \frac{x}{x+1-x} \\ & = x. \end{align} \]

Since \(f\left(f^{-1}(x)\right) = x\) you know that you have found the inverse correctly.

Let's take a look at the function \( h(x) = 2x^2 + 4 \). Now, you can decompose this function by factorising.

\[ h(x) = 2(x^2 + 2) \]

Now, if the function was in the form \( f(g(x)) \), what would its parts be? Well, you can say that inside the brackets is one function, \( g(x) \), and out of the brackets is \( f(x) \).

Let's try

\[ f(x) = 2x \]

and

\[ g(x) = x^2 + 2. \]

Evaluating \( f(g(x)) \) you get

\[ \begin{align} f(g(x)) &= 2(x^2 + 2) \\ &= 2x^2 + 4 \\ &= h(x). \end{align}\]

Therefore you have found the correct decomposition.

Consider a rocket moving through space with a uniform acceleration, \( a \) and mass, \( m \), from an initial velocity, \( u \). Given that its velocity at time \( t \) is described by \( v = u + at \), and its kinetic energy is described by

\[ E_K = \frac{1}{2} mv^2 ,\]

find an expression for the rocket's kinetic energy at time \( t \).

Solution:

You can rewrite each of these equations to function notation, recognising that \( u \), \( a \), and \( m \) are all constants, \( t \) and \( v \) are your two independent variables, and \( v \) and \( E_K \) are your dependent variables. So in functional notation:

\[ \begin{align} &v(t) = u + at \\ & E_K(v) = \frac{1}{2} mv^2 .\end{align}\]

Now, you want to find \( E_K(v) \) terms of \( t \), so you create a composite function \( E_K(v(t)) \) to achieve this. So let's substitute the value of the function \( v(t) \) wherever there is a \( v \) in \( E_K(v) \). This gives you

\[ E_K(v(t)) = \frac{1}{2} m(u + at) .\]

Given the functions

\[ f(x) = \frac{1}{x^2 + 6x + 9}, \quad \{ x \in \mathbb{N} | x \neq 3 \} \]

and

\[ g(x) = \sqrt{x}, \quad \{ x \in \mathbb{R}, \quad x\ge 0 \} ,\] what is the domain of composite function \( f(g(x)) \)?

Solution:

The output of \( g(x) \) must be within the domain of \( f(x) \). Therefore the outputs of \( g(x) \) must belong to the set \( \mathbb{N}\setminus \{3 \} \).

That means the output of \( g(x) \) cannot be \( 3 \), since the output of \(g(x)\) must be within the domain of \(f(x)\) and the domain of \(f(x)\) does not include \(3\). Therefore

\[ g(x) \neq 3, \]

so

\[ 3 \neq \sqrt{x} ,\]

which means

\[ x \neq 9. \]

The output of \( g(x) \) must be a natural number, so the output must also be greater than zero. Therefore

\[ \begin{align} & g(x) > 0 \\ & \sqrt{x} > 0\\ & x > 0. \end{align} \]

That actually doesn't give you much new information since you already knew the output needed to be a natural number!

Putting it all together, yur final domain for \( f(g(x) \) is therefore

\[ \{ x \in \mathbb{N} | x \neq 9 \}. \]

Let's look at two functions with multiple shared variables. Consider the functions

\[ f(x,y) = 2x + 9y \]

and

\[ g(x,y) = \frac{15x}{y - 10} .\]

Let's find the composite function \( f(g(x,y),y) \).

Solution:

In this case, the \(y\) in the second spot of \( f(g(x,y),y) \) tells you that two things need to occur:

1. the value of \( g(x,y) \) is substituted into \( f(x,y) \) where any \( x \) is present; and
2. \(y\) remains where any \( y \) is present.

Performing those two steps gives you

\[ f(g(x,y),y) = 2 \left( \frac{15x}{y-10} \right) + 9y. \]

Now, simplify by expanding brackets and collecting like terms:

\[ \begin{align} f(g(x,y), y) &= \frac{30x}{y-10} + 9y \\ &= \frac{30x}{y-10} + \frac{9y(y-10)}{y-10}\\ & = \frac{30x + 9y^2 -90y}{y-10} . \end{align}\]

Decompose the function

\[ f(g(x)) = 4 \left( \frac{3x}{4x + 5} \right)^2 \]

into constituent functions.

Solution:

You can see by observation that the inner function is

\[ g(x) = \frac{3x}{4x + 5}, \]

And the outer function is therefore

\[ f(x) = 4x^2. \]