AOS1 Topic 7: Combinations of Functions

The combination of functions is the act of combining multiple functions into a single function. This usually entails using basic mathematical operators such as addition or multiplication.

Arithmetic Combinations of Functions

Arithmetic combination of functions is the creation of new functions by combining existing functions using basic arithmetic operations such as: addition, subtraction, multiplication, and division.

How to Combine Functions by Addition

Combining functions by addition involves creating a new function that is the sum of two existing functions. If we have two functions, \( f(x) \) and \( g(x) \), their sum is represented as:

\((f + g)(x) = f(x) + g(x)\)

Here are the steps to combine functions by addition:

  1. Identify the functions: Determine the two functions that you want to add. For example, let \( f(x) = 2x + 3 \) and \( g(x) = x^2 - 1 \).
  2. Write the sum of the functions: Combine the functions by adding their expressions. Using the example functions:

    \((f + g)(x) = (2x + 3) + (x^2 - 1)\)

  3. Simplify the expression: Combine like terms to simplify the new function. For the example:

    \((f + g)(x) = x^2 + 2x + 2\)

  4. Determine the domain: The domain of the new function is the intersection of the domains of the original functions. For most polynomial functions, the domain is all real numbers \( \mathbb{R} \). In this case, both \( f(x) \) and \( g(x) \) are polynomials, so the domain of \((f + g)(x)\) is all real numbers \( \mathbb{R} \).

Combining functions by addition is a straightforward process that allows you to create new functions that encapsulate the behaviors of the original functions.

How to Combine Functions by Subtraction

Combining functions by subtraction involves creating a new function that is the difference of two existing functions. If we have two functions, \( f(x) \) and \( g(x) \), their difference is represented as:

\((f - g)(x) = f(x) - g(x)\)

Here are the steps to combine functions by subtraction:

  1. Identify the functions: Determine the two functions that you want to subtract. For example, let \( f(x) = 2x + 3 \) and \( g(x) = x^2 - 1 \).
  2. Write the difference of the functions: Combine the functions by subtracting their expressions. Using the example functions:

    \((f - g)(x) = (2x + 3) - (x^2 - 1)\)

  3. Simplify the expression: Combine like terms to simplify the new function. For the example:

    \((f - g)(x) = 2x + 3 - x^2 + 1 = -x^2 + 2x + 4\)

  4. Determine the domain: The domain of the new function is the intersection of the domains of the original functions. For most polynomial functions, the domain is all real numbers \( \mathbb{R} \). In this case, both \( f(x) \) and \( g(x) \) are polynomials, so the domain of \((f - g)(x)\) is all real numbers \( \mathbb{R} \).

Combining functions by subtraction allows you to create new functions that highlight the differences between the original functions.

How to Combine Functions by Multiplication

Combining functions by multiplication involves creating a new function that is the product of two existing functions. If we have two functions, \( f(x) \) and \( g(x) \), their product is represented as:

\((f \cdot g)(x) = f(x) \cdot g(x)\)

Here are the steps to combine functions by multiplication:

  1. Identify the functions: Determine the two functions that you want to multiply. For example, let \( f(x) = 2x + 3 \) and \( g(x) = x^2 - 1 \).
  2. Write the product of the functions: Combine the functions by multiplying their expressions. Using the example functions:

    \((f \cdot g)(x) = (2x + 3) \cdot (x^2 - 1)\)

  3. Expand the expression: Use the distributive property to expand the product. For the example:

    \((f \cdot g)(x) = 2x \cdot x^2 + 2x \cdot (-1) + 3 \cdot x^2 + 3 \cdot (-1) = 2x^3 - 2x + 3x^2 - 3\)

  4. Combine like terms: Simplify the new function by combining like terms, if necessary. In this example, the expression is already simplified:

    \((f \cdot g)(x) = 2x^3 + 3x^2 - 2x - 3\)

  5. Determine the domain: The domain of the new function is the intersection of the domains of the original functions. For most polynomial functions, the domain is all real numbers \( \mathbb{R} \). In this case, both \( f(x) \) and \( g(x) \) are polynomials, so the domain of \((f \cdot g)(x)\) is all real numbers \( \mathbb{R} \).

Combining functions by multiplication allows you to create new functions that incorporate the behavior of both original functions through their product.

How to Combine Functions by Division

Combining functions by division involves creating a new function that is the quotient of two existing functions. If we have two functions, \( f(x) \) and \( g(x) \), their quotient is represented as:

\(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}\)

Here are the steps to combine functions by division:

  1. Identify the functions: Determine the two functions that you want to divide. For example, let \( f(x) = 2x + 3 \) and \( g(x) = x^2 - 1 \).
  2. Write the quotient of the functions: Combine the functions by dividing their expressions. Using the example functions:

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

  3. Simplify the expression: If possible, simplify the new function by factoring and reducing. In this example, the expression is already in its simplest form:

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

  4. Determine the domain: The domain of the new function is the intersection of the domains of the original functions, excluding any values that make the denominator zero. For the example functions, \( g(x) = x^2 - 1 \) is zero when \( x = 1 \) or \( x = -1 \):

    \( x^2 - 1 = (x - 1)(x + 1) \)

    So, the domain of \(\left(\frac{f}{g}\right)(x)\) is all real numbers except \( x \neq 1 \) and \( x \neq -1 \). In interval notation, the domain is \( (-\infty, -1) \cup (-1, 1) \cup (1, \infty) \).

Combining functions by division allows you to create new functions that incorporate the behavior of both original functions through their quotient, but it's important to carefully consider the domain to avoid division by zero.

Domain for the combining functions:

When we arithmetically combine functions, the domain of the new function will contain the x-values that are common between the original functions. In other words, both functions must be defined at a point for their combination to be defined.

Additionally, when dividing functions, the domain is further restricted so that the denominator isn't equal to zero.

Basically, what this means is when we evaluate combined functions, we can either:

  • Combine the functions and then evaluate, or
  • We can evaluate the functions and then combine them.

Domain for Addition and Subtraction of Functions:

If the domain of a function, \(f\), is the set, \(A\), and the domain of the function, \(g\), is the set, \(B\), then the domain of \(f + g\) is the intersection \(A \cap B\) (note that the symbol, \(\cap\), just means "intersection") because both \(f(x)\) and \(g(x)\) have to be defined.

Domain for Multiplication and Division of Functions:

As with the addition and subtraction of functions, the domain for the multiplication and division of functions is the intersection.

However, when we divide functions, we need to further restrict the domain of the combined function since we can't divide by zero. So, for the division of functions, the domain is.

This is read as "the set of all values of \(x\) such that \(x\) is an element of the intersection of \(A\) and \(B\), as long as \(g(x)\) does not equal zero."

Composite Functions

A composite function is formed when one function is applied to the result of another function. In mathematical terms, if you have two functions f and g, the composite function f ∘ g is defined as:

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

Here, \( g(x) \) is evaluated first and then \( f \) is applied to the result of \( g(x) \). For the composite function \( f \circ g \) to be defined, the output of \( g(x) \) must lie within the domain of \( f \).

For example, if \( f(x) = 2x + 3 \) and \( g(x) = x^2 \), the composite function \( f \circ g \) would be:

\( (f \circ g)(x) = f(g(x)) = f(x^2) = 2(x^2) + 3 = 2x^2 + 3 \)

In this example, you first apply \( g \) to \( x \) to get \( x^2 \), and then apply \( f \) to the result, which is \( x^2 \), yielding \( 2x^2 + 3 \).

In general, for functions f and g such that

\(\text{ran } f \subseteq \text{dom } g\),

we define the composite function of g with f by

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

The domain of the composite function \( g \circ f \) is given by:

\(\text{dom}(g \circ f) = \text{dom } f\)

\( x \)

\(\text{domain of } f\)

\(\text{range of } f\)

\(\text{domain of } g\)

\( g(f(x)) \)

\( \longrightarrow \)

\( \longrightarrow \)

\( \longrightarrow \)

\( \longrightarrow \)

\( \longrightarrow \)

\( f \)

\( g \)

\( g(f(x)) \)

Example 1

Function Operations

If \( f(x) = \sqrt{x - 2} \) for all \( x \geq 2 \) and \( g(x) = \sqrt{4 - x} \) for all \( x \leq 4 \), find:

  1. \( f + g \)
  2. \( (f + g)(3) \)
  3. \( f \cdot g \)
  4. \( (f \cdot g)(3) \)

Solution:

Note that \( \text{dom } f \cap \text{dom } g = [2, 4] \).

\( (f + g)(x) = f(x) + g(x) = \sqrt{x - 2} + \sqrt{4 - x} \)

\( \text{dom}(f + g) = [2, 4] \)

a. \( (f + g)(3) = \sqrt{3 - 2} + \sqrt{4 - 3} = 1 + 1 = 2 \)

b. \( (f \cdot g)(x) = f(x) \cdot g(x) = \sqrt{(x - 2)(4 - x)} \)

\( \text{dom}(f \cdot g) = [2, 4] \)

c. \( (f \cdot g)(3) = \sqrt{(3 - 2)(4 - 3)} = \sqrt{1 \cdot 1} = 1 \)

Example 2

Composite Functions

Find both \( f \circ g \) and \( g \circ f \), stating the domain and range of each, where:

\( f : \mathbb{R} \to \mathbb{R}, f(x) = 2x - 1 \) and \( g: \mathbb{R} \to \mathbb{R}, g(x) = 3x^2 \)

Solution

To determine the existence of a composite function, it is useful to form a table of domains and ranges.

Function Domain Range
\( g \) \( \mathbb{R} \) \( \mathbb{R}^+ \cup \{0\} \)
\( f \) \( \mathbb{R} \) \( \mathbb{R} \)

We see that \( f \circ g \) is defined since \( \text{ran } g \subseteq \text{dom } f \), and that \( g \circ f \) is defined since \( \text{ran } f \subseteq \text{dom } g \).

Calculations

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

\( = f(3x^2) \)

\( = 2(3x^2) - 1 \)

\( = 6x^2 - 1 \)

\( \text{dom}(f \circ g) = \text{dom } g = \mathbb{R} \)

\( \text{ran}(f \circ g) = [-1, \infty) \)

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

\( = g(2x - 1) \)

\( = 3(2x - 1)^2 \)

\( = 12x^2 - 12x + 3 \)

\( \text{dom}(g \circ f) = \text{dom } f = \mathbb{R} \)

\( \text{ran}(g \circ f) = [0, \infty) \)

Note:It can be seen from this example that in general \( f \circ g \neq g \circ f \).

Example 3

For the functions \( g(x) = 2x - 1 \), \( x \in \mathbb{R} \), and \( f(x) = \sqrt{x} \), \( x \geq 0 \):

a. State which of \( f \circ g \) and \( g \circ f \) is defined.

b.For the composite function that is defined, state the domain and rule.

Step 2

Solution

a. State which of \( f \circ g \) and \( g \circ f \) is defined.

Range of \( f \subseteq \) domain of \( g \)
Range of \( g \not\subseteq \) domain of \( f \)

Thus \( g \circ f \) is defined, but \( f \circ g \) is not defined.

Function Domain Range
\( g \) \( \mathbb{R} \) \( \mathbb{R} \)
\( f \) \( \mathbb{R}^+ \cup \{0\} \) \( \mathbb{R}^+ \cup \{0\} \)

Step 3

b.

For the composite function that is defined, state the domain and rule.

\( g \circ f(x) = g(f(x)) \)
\( = g(\sqrt{x}) \)
\( = 2\sqrt{x} - 1 \)

\( \text{dom}(g \circ f) = \text{dom } f = \mathbb{R}^+ \cup \{0\} \)


Solution

a. State which of \( f \circ g \) and \( g \circ f \) is defined.

Range of \( f \subseteq \) domain of \( g \)
Range of \( g \not\subseteq \) domain of \( f \)

Thus \( g \circ f \) is defined, but \( f \circ g \) is not defined.

Function Domain Range
\( g \) \( \mathbb{R} \) \( \mathbb{R} \)
\( f \) \( \mathbb{R}^+ \cup \{0\} \) \( \mathbb{R}^+ \cup \{0\} \)

b.

For the composite function that is defined, state the domain and rule.

\( g \circ f(x) = g(f(x)) \)
\( = g(\sqrt{x}) \)
\( = 2\sqrt{x} - 1 \)

\( \text{dom}(g \circ f) = \text{dom } f = \mathbb{R}^+ \cup \{0\} \)

Example 4

Solution

a. State which of \( f \circ g \) and \( g \circ f \) is defined.

Range of \( f \subseteq \) domain of \( g \)
Range of \( g \not\subseteq \) domain of \( f \)

Thus \( g \circ f \) is defined, but \( f \circ g \) is not defined.

Function Domain Range
\( g \) \( \mathbb{R} \) \( \mathbb{R} \)
\( f \) \( \mathbb{R}^+ \cup \{0\} \) \( \mathbb{R}^+ \cup \{0\} \)

b.

For the composite function that is defined, state the domain and rule.

\( g \circ f(x) = g(f(x)) \)
\( = g(\sqrt{x}) \)
\( = 2\sqrt{x} - 1 \)

\( \text{dom}(g \circ f) = \text{dom } f = \mathbb{R}^+ \cup \{0\} \)

Exercise &&1&& (&&1&& Question)

What is the domain of the function \((f + g)(x)\) if \(f(x) = \sqrt{x-1}\) and \(g(x) = \ln(x)\)?

1
Submit

Exercise &&2&& (&&1&& Question)

If \(f(x) = 2x\) and \(g(x) = x^2\), what is \((f \circ g)(x)\)?

2
Submit

Exercise &&3&& (&&1&& Question)

If \(f(x) = x^2\) and \(g(x) = 3x + 1\), what is \((f \circ g)(2)\)?

3
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Exercise &&4&& (&&1&& Question)

For the functions \(f(x) = x + 3\) and \(g(x) = x^2\), what is \((f+g)(x)\)?

4
Submit