Review of Functions: Learn It 2

Lumen Learning

Review of Functions: Learn It 2

The Domain and Range of a Function

The domain of a function is the complete set of values that can be input into the function. It answers the question, “What values am I allowed to put into this function?” To identify the domain, we look for values that will make the function work without causing mathematical errors such as division by zero or taking the square root of a negative number.

The range of a function, conversely, is the complete set of all output values that the function can produce. When we ask, “What values can come out of this function?” we are essentially inquiring about the range. The range is determined by taking all the possible values of the domain and observing what outputs are produced by the function.

domain and range of a function

The set of inputs is called the domain of the function.

The set of outputs is called the range of the function.

How to: Determine the Domain and Range of a Function

Determining the Domain:

Examine the function for any mathematical restrictions (like division by zero or even roots of negative numbers).
Exclude these restricted values from the domain.
Express the domain using set notation or interval notation based on what the question is asking for, considering whether the endpoints are included (closed interval) or not (open interval)

Determining the Range:

Use the domain to calculate possible outputs.
Consider the nature of the function: Is it linear, quadratic, exponential? What behavior do these types of functions typically exhibit?
Determine if there are any maximum or minimum values the outputs cannot exceed.
Express the range using set notation or interval notation based on what the question is asking for, including endpoints where appropriate.

When describing domains and ranges, both “set notation” and “interval notation” are commonly used.

For the functions [latex]f(x)=x^2[/latex] and [latex]f(x)=\sqrt{x}[/latex], the domains are sets with an infinite number of elements. Clearly we cannot list all these elements. When describing a set with an infinite number of elements, it is often helpful to use set-builder or interval notation. When using set-builder notation to describe a subset of all real numbers, denoted [latex]ℝ[/latex], we write

[latex]\{x|x \, \text{has some property}\}[/latex]

We read this as the set of real numbers [latex]x[/latex] such that [latex]x[/latex] has some property. For example, if we were interested in the set of real numbers that are greater than one but less than five, we could denote this set using set-builder notation by writing

[latex]\{x|1 < x < 5\}[/latex]

A set such as this, which contains all numbers greater than [latex]a[/latex] and less than [latex]b[/latex], can also be denoted using the interval notation [latex](a,b)[/latex]. Therefore,

[latex](1,5)=\{x|1 < x < 5\}[/latex]

The numbers 1 and 5 are called the endpoints of this set. If we want to consider the set that includes the endpoints, we would denote this set by writing

[latex][1,5]=\{x|1\le x\le 5\}[/latex]

We can use similar notation if we want to include one of the endpoints, but not the other. To denote the set of nonnegative real numbers, we would use the set-builder notation

[latex]\{x|0\le x\}[/latex]

The smallest number in this set is zero, but this set does not have a largest number. Using interval notation, we would use the symbol [latex]\infty[/latex], which refers to positive infinity, and we would write the set as

[latex][0,\infty)=\{x|0\le x\}[/latex]

It is important to note that [latex]\infty[/latex] is not a real number. It is used symbolically here to indicate that this set includes all real numbers greater than or equal to zero. Similarly, if we wanted to describe the set of all nonpositive numbers, we could write

[latex](−\infty ,0]=\{x|x\le 0\}[/latex]

Here, the notation [latex]−\infty[/latex] refers to negative infinity, and it indicates that we are including all numbers less than or equal to zero, no matter how small. The set

[latex](−\infty ,\infty)=\{x|x \, \text{is any real number}\}[/latex]

refers to the set of all real numbers.

Using the union symbol allows us to describe the domain and range of functions that aren’t continuous across all numbers and have breaks or gaps in between.

Consider a function [latex]h(x)[/latex] that represents the reciprocal squared, defined as [latex]h(x) = \frac{1}{(x-1)^2}[/latex].

For the function [latex]h(x)[/latex], the domain excludes [latex]x=1[/latex] because the denominator becomes zero at this point, which is undefined in real number arithmetic.

Domain of [latex]h(x)[/latex]: [latex](−\infty,1)\cup(1,\infty)[/latex]

The domain notation here, using the union symbol [latex]\cup[/latex] communicates that [latex]h(x)[/latex] is defined for all real numbers except [latex]x=1[/latex].

For each of the following functions, determine the domain and range.

[latex]f(x)=(x-4)^2+5[/latex]
[latex]f(x)=\sqrt{3x+2}-1[/latex]
[latex]f(x)=\dfrac{3}{x-2}[/latex]

Consider [latex]f(x)=(x-4)^2+5[/latex].
1. Since [latex]f(x)=(x-4)^2+5[/latex] is a real number for any real number [latex]x[/latex], the domain of [latex]f[/latex] is the interval [latex](−\infty ,\infty)[/latex].
2. Since [latex](x-4)^2\ge 0[/latex], we know [latex]f(x)=(x-4)^2+5\ge 5[/latex]. Therefore, the range must be a subset of [latex]\{y|y\ge 5\}[/latex]. To show that every element in this set is in the range, we need to show that for a given [latex]y[/latex] in that set, there is a real number [latex]x[/latex] such that [latex]f(x)=(x-4)^2+5=y[/latex]. Solving this equation for [latex]x[/latex], we see that we need [latex]x[/latex] such that
  [latex](x-4)^2=y-5[/latex].
  
  This equation is satisfied as long as there exists a real number [latex]x[/latex] such that
  
  [latex]x-4=±\sqrt{y-5}[/latex].
  
  Since [latex]y\ge 5[/latex], the square root is well-defined. We conclude that for [latex]x=4±\sqrt{y-5}[/latex], [latex]f(x)=y[/latex], and therefore the range is [latex]\{y|y\ge 5\}[/latex] in set notation or [latex][5,\infty)[/latex] in interval notation.
Consider [latex]f(x)=\sqrt{3x+2}-1[/latex].
1. To find the domain of [latex]f[/latex], we need the expression [latex]3x+2\ge 0[/latex]. Solving this inequality, we conclude that the domain is [latex]\{x|x\ge -\frac{2}{3}\}[/latex] in set notation or [latex][−\frac{2}{3},\infty)[/latex] in interval notation.
2. To find the range of [latex]f[/latex], we note that since [latex]\sqrt{3x+2}\ge 0[/latex], [latex]f(x)=\sqrt{3x+2}-1\ge -1[/latex]. Therefore, the range of [latex]f[/latex] must be a subset of the set [latex]\{y|y\ge -1\}[/latex]. To show that every element in this set is in the range of [latex]f[/latex], we need to show that for all [latex]y[/latex] in this set, there exists a real number [latex]x[/latex] in the domain such that [latex]f(x)=y[/latex]. Let [latex]y\ge -1[/latex]. Then, [latex]f(x)=y[/latex] if and only if
  [latex]\sqrt{3x+2}-1=y[/latex].
  
  Solving this equation for [latex]x[/latex], we see that [latex]x[/latex] must solve the equation
  
  [latex]\sqrt{3x+2}=y+1[/latex].
  
  Since [latex]y\ge -1[/latex], such an [latex]x[/latex] could exist. Squaring both sides of this equation, we have [latex]3x+2=(y+1)^2[/latex].
  Therefore, we need
  
  [latex]3x=(y+1)^2-2[/latex],
  
  which implies
  
  [latex]x=\frac{1}{3}(y+1)^2-\frac{2}{3}[/latex].
  
  We just need to verify that [latex]x[/latex] is in the domain of [latex]f[/latex]. Since the domain of [latex]f[/latex] consists of all real numbers greater than or equal to [latex]-\frac{2}{3}[/latex], and
  
  [latex]\frac{1}{3}(y+1)^2-\frac{2}{3}\ge -\frac{2}{3}[/latex],
  
  there does exist an [latex]x[/latex] in the domain of [latex]f[/latex]. We conclude that the range of [latex]f[/latex] is [latex]\{y|y\ge -1\}[/latex] in set notation or [latex][−1,\infty)[/latex] in interval notation.
Consider [latex]f(x)=\dfrac{3}{x-2}[/latex].
1. Since [latex]\frac{3}{x-2}[/latex] is defined when the denominator is nonzero, the domain is [latex]\{x|x\ne 2\}[/latex] in set notation or [latex](−\infty,2)\cup(2,\infty)[/latex] in interval notation.
2. To find the range of [latex]f[/latex], we need to find the values of [latex]y[/latex] such that there exists a real number [latex]x[/latex] in the domain with the property that
  [latex]\dfrac{3}{x-2}=y[/latex].
  
  Solving this equation for [latex]x[/latex], we find that
  
  [latex]x=\dfrac{3}{y}+2[/latex].
  
  Therefore, as long as [latex]y\ne 0[/latex], there exists a real number [latex]x[/latex] in the domain such that [latex]f(x)=y[/latex]. Thus, the range is [latex]\{y|y\ne 0\}[/latex] in set notation or [latex](−\infty,0)\cup(0,\infty)[/latex] in interval notation.