DOMAIN, CODOMAIN, RANGE

Contents

• Introduction
• Domain
• Codomain and Range
• Domain and Range on a Graph
• Domain and Range Calculator
• Summary
• Practice Questions
• Recommended Videos
• Test Questions

Learning Objectives

By the end of this section, you should be able to:

• Explain Domain of functions.
• Explain Codomain and Range of Functions.

Introduction

In the previous blog post, we discussed functions, how to identify functions; we saw that a function is any rule that maps each element in one set, to one corresponding element in another set. 

At a certain point, we saw that not all input variables can enter into a function, these are input variables that make a function to be undefined, hence, we give a name to the set of all input variables which can go into a particular function (to produce valid output), this is called the Domain of the function. 

A set of all possible output variables which can come out of a function is called the Codomain of the function, whereas the set of the actual output variables which come out of a function is called the Range of the function.

Domain

You will find a lot of questions where you will be asked to find the domain of a function. What does this mean?

All that the question requires of you is to find values of $x$ that can be plugged into the function, to produce defined outputs.

This is useful to know, because as stated earlier, what is permissible as input into some functions may be limited.

For instance, consider the function:

$f(x)=\frac{1}{x}$

We know that it is impossible to divide by 0, therefore, our domain here, can not include the value $x=0$ (because this would lead to undefined, you will get a math error on a calculator).

However, all other values of $x$ would be OK. We can plug in any other number into this function and we would get a valid output.

If we ever put a number into a function and we can’t get an output, then we know that there is some sort of domain issue.

The most common occurrences of domain issues are:

 • Dividing by 0 (e.g $\frac{x+2}{0}$, $\frac{x^2}{0}$) ❌

 • Square root of negative numbers (e.g $\sqrt{–2}$, $\sqrt{–0.001}$) ❌

 • Logarithm of 0 or numbers less than 0 (e.g $\log 0$, $\log –2$, $\log –0.001$) ❌

How do we write the domain of a function?

There are two main ways to write the domain of a function:

 • Interval Notation

 • Set Notation

Interval Notation:

Whenever interval Notation is used to show the domain of a function, parentheses (or brackets) are used. 

For instance, considering the function $f(x)=\frac{1}{x}$ shown above, which permits any input variable but 0, using interval Notation, we write its domain as

$\text{dom}(f)=(–\infty ,0)\cup (0,\infty )$

What this notation implies is that: 

$(–\infty ,0)$:

all values from $–\infty$ to $0$ can enter into the function $f(x)=\frac{1}{x}$, but 0 is not included,

$\cup$:

union, it means and,

$(0,\infty )$:

all values from 0 (0 is not included) to positive infinity can enter into the function as input variables.

PRO TIP: if you see an interval Notation that is shown as $[0,\infty )$, this type of bracket, $[$, shows that all numbers from $0$ to infinity can be plugged into the function, with $0$ included; but when we have $(0,\infty )$, the bracket, $($, means that $0$ is not included.

Set Notation:

Set notation uses sets to say explicitly where a function is defined or not. 

For instance, in $f(x)=\frac{1}{x}$, any value but 0 is permissible into the function, hence, we write the Domain using set Notation as

$\text{dom}(f)=\{x\in \mathbb{R}:x\ne 0\}$

This is read as domain of the function $f$ is any value, $x$, which is an element of the set of all real numbers, ($\mathbb{R}$), such that $x$ is not equal to $0$.

What this notation implies is that:

Our domain includes any value for $x$ except $0$.

PRO TIP:

• Every input value can enter a linear function, hence, domain of all linear functions is the set of all real numbers {$\mathbb{R}$}.

• Every input value can enter a quadratic, polynomial function, hence, domain of all quadratic or polynomial functions is the set of all real numbers {$\mathbb{R}$}.

• A Function which is a fraction (or Quotient) has limited domain, its denominator must not be equal to 0, hence, domain of all rational functions is the set of all real numbers excluding any value that will make the denominator of the function, equal to 0.

• A Function which has a square root, or an even root, cannot produce an output for a negative radicand (=number or expression inside a root), hence, domain of any even root functions is the set of all values that will not make its radicand equal to negative.

• A Logarithmic Function, or a function which involves Logarithm, cannot produce an output for the $\log$ of any expression that yields 0 or less than 0, hence, domain of all logarithmic functions is the set of all values that will render the expression (bounded by the $\log$), as greater than 0.

Examples

Using interval and set Notation, show the domain of:

A. $f(x)=x+7$

B. $f(x)=x^2+7x+7$

C. $\text{g}(x)=x^9–18x^5+6x^4–8$

D. $f(x)=\frac{x+7}{x–2}$

E. $\text{g}(x)=\frac{x}{x^2–3x+2}$

F. $h(x)=\sqrt{x–5}$

G. $h(x)=\frac{7x–\sqrt{x^2–4}}{x}$

H. $h(x)=\log (x+2)+\frac{x}{x+2}$

I. $f(x)=e^x$

J. $f(x)=\frac{1}{2}(3^x+3^{–x})$

Solution

A.

$f(x)=x+7$

This is a linear function. 

All linear functions can accept any input value for $x$,

Domain for the above function is:

Interval Notation: $\text{dom}(f)=(–\mathrm{\infty },\mathrm{\infty })$

Set Notation: $\text{dom}(f)=\{x:x\in \mathbb{R}\}$

B.

$f(x)=x^2+7x+7$

This is a quadratic function. 

All quadratic functions can accept any input value for x,

Domain for the above function is:

Interval Notation: $\text{dom}(f)=(–\mathrm{\infty },\mathrm{\infty })$

Set Notation: $\text{dom}(f)=\{x:x\in \mathbb{R}\}$

C.

$\text{g}(x)=x^9–18x^5+6x^4–8$

This is a polynomial function. 

All polynomial functions can accept any input value for x,

Domain for the above function is:

Interval Notation: $\text{dom}\left(\text{g}\right)=(–\mathrm{\infty },\mathrm{\infty })$

Set Notation: $\text{dom}(\text{g})=\{x:x\in \mathbb{R}\}$

D.

$f(x)=\frac{x+7}{x–2}$

This is a rational (or Quotient) function. 

A rational function can only accept input values for x that will not make its denominator to equal zero.

Thus, 

$x–2\ne 0$

We can simplify this simple equation, we have;

$x\ne 2$

Domain for the above function is then shown as:

Interval Notation: $\text{dom}\left(f\right)=(–\mathrm{\infty },2)\cup (2,\mathrm{\infty })$

Set Notation: $\text{dom}\left(f\right)=\{x\in \mathbb{R}:x\ne 2\}$

You can see that we were not concerned with the numerator, we are concerned only with the denominator.

E.

$\text{g}(x)=\frac{x}{x^2–3x+2}$

This is another Quotient. This means that for the function to be defined, the denominator must not be equal to zero.

Thus,

$x^2–3x+2\ne 0$

We can factorise this quadratic equation,

$x^2–2x–x+2\ne 0$

$(x^2–2x)(–x+2)\ne 0$

$x(x–2)–1(x–2)\ne 0$

$(x–1)(x–2)\ne 0$

$x\ne 1,x\ne 2$

We can then write the domain for the function as:

Interval Notation: $\text{dom}\left(\text{g}\right)=(–\mathrm{\infty },1)\cup (1,2)\cup (2,\mathrm{\infty })$

Set Notation: $\text{dom}\left(\text{g}\right)=\{x\in \mathbb{R}:x\ne 1,x\ne 2\}$

F.

$h(x)=\sqrt{x–5}$

This is a square root function. 

A square root can only accept values starting from zero, upwards, i.e, no negative value

Thus,

$x–5\ge 0$

$x\ge 5$

The domain for the function is:

Interval Notation: $\text{dom}(h)=[5,\mathrm{\infty })$

Set Notation: $\text{dom}(h)=\{x\in \mathbb{R}:x\ge 5\}$

G.

$h(x)=\frac{7x–\sqrt{x^2–4}}{x}$

This is a quotient, and it contains a square root expression; a quotient must not have its denominator equal to zero, and a square root must not have its radicand as negative, we will analyse these two separately;

For quotient, $x\ne 0$

For square root, $x^2–4\ge 0$

$x^2–4\ge 0$

$x^2\ge 4$

$x\le –2,x\ge 2$

This means that all values between –2 and +2 cannot enter into the function as input.

Thus, domain is written as:

Interval Notation: $\text{dom}(h)=(–\mathrm{\infty },–2]\cup [2,\mathrm{\infty })$

Set Notation: $\text{dom}(h)=\{x\in \mathbb{R}:–2\ge x,2\le x\}$

H.

$h(x)=\log (x+2)+\frac{x}{x+2}$

In this function, we have a Logarithmic expression, and a fraction:

In $\log (x+2)$,

$x+2>0$

Therefore, $x>–2$

In Fraction $\frac{x}{x+2}$, $x+2\ne 0$

$x\ne –2$

Domain is therefore:

Interval Notation: $\text{dom}(h)=(–2,\mathrm{\infty })$

Set Notation: $\text{dom}(h)=\{x\in \mathbb{R}:x>–2\}$

I.

$f(x)=e^x$

All exponential functions can accept any input value.

Domain is:

Interval Notation: $\text{dom}\left(f\right)=(–\mathrm{\infty },\mathrm{\infty })$

Set Notation: $\text{dom}\left(f\right)=\{x:x\in \mathbb{R}\}$

J.

$f(x)=\frac{1}{2}(3^x+3^{–x})$

This is also an exponential function.

Domain is:

Interval Notation: $\text{dom}\left(f\right)=(–\mathrm{\infty },\mathrm{\infty })$

Set Notation: $\text{dom}\left(f\right)=\{x:x\in \mathbb{R}\}$

Codomain and Range

In simple terms, Codomain is usually a larger set which contains the Range of a function, and many other times, Codomain may be equal to Range.

Codomain is the set of all possible outputs that a function can produce. It includes values that will be produced by a function, and values which may not produced by the function.

Range on the other hand is a set of all the actual output that a function produces. It includes only values that are produced by a function. It is also called the image of a function.

PRO TIP: where you see preimage of a function, we are talking about the domain of the function, while the image of a function is the range of the function.

Mapping of domain X to Codomain Y

Mapping of domain X to range Z

It is important for you to note that a Codomain can change a relation that is not a function, to one that is a function. 

For instance, we learnt from the previous post that $f(x)=\sqrt{x}$ is not a function, because we can have positive and negative outputs for every valid input; however, when we limit the Codomain for $f(x)=\sqrt{x}$ as {positive real numbers}, this means that every input will have an output of only positive numbers, we then see that $f(x)=\sqrt{x}$ will have only one output for every input, therefore actively depicting a function relation.

A function maps its domain (inputs) to its Range (outputs), we can normally show it as $f:\text{input}\to \text{output}$

In other words, if we have a function $f(x)=x^2+7$, all input (domain) are $x$, while all output (range) are $f(x)$ which is equal to $x^2+7$, we then show the relation as:

$f:x\to x^2+7$

This is still equal to $f(x)=x^2+7$, same thing, but different depiction.

More acceptably, We map the domain of a function to a Codomain. For instance, say the domain of a function is the set of all real numbers, we may not know the range of the function, but we can assume that the Codomain of the function is also set of all real numbers, since the range will always be a subset of the codomain. We then depict this as:

$f:\text{domain}\to \text{codomain}$

$f:\mathbb{R}\to \mathbb{R}$

This is generally and considerably more accepted than the earlier depiction of mapping a domain to its range, because many attimes, the range of a function may be complicated.

The Codomain of a function is not usually found or calculated, because it can usually be assumed, however range which is specific, needs to be gotten.

How do we get the range of a function?

Normally, we get the range of a function by taking a look at the graph of the function.

But in our case here, we cannot always plot the the graph of functions, especially when you have just a minute to attempt such a question in your test or examination, we employ the use of inverse of functions.

Recall from the previous post, we said that while a function maps an input to an output, an inverse function does the reverse, by mapping an output to an input. In this reverse sense, we can see output values which will be valid, hence, getting the range of functions.

For instance,

Considering the function $f(x)=\frac{1}{x}$, we know the domain of this function is the set of all $\mathbb{R}$ except 0; we can get the range by first solving for the inverse of the function;

Recall how to get the inverse of a function? We make the input $x$ as the subject of the formula:

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

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

Taking a look at this simple inverse function, all values except 0 can enter as input for $y$; this means that from the original function $f(x)=\frac{1}{x}$, we could get other output values but we will never get zero, we can then interpret that the range of the above function is all $\mathbb{R}$ except 0.

It's quite simple to do, to get the range of a function:

1. Find the inverse of the function.

2. Check for all valid input that can enter the inverse function. Remember that we follow the rules stated while discussing domain; quotients must not have a denominator equal to zero, radicand of all square roots (or even roots) must be greater than or equal to zero, logarithmic expression must be greater than 0.

Other methods are by graph checking, or checking for maximum and minimum values using differential calculus.

Examples

Using interval and set Notation, show the range of:

A. $f(x)=x+7$

B. $f(x)=x^2+7x+7$

C. $\text{g}(x)=x^9–18x^5+6x^4–8$

D. $f(x)=\frac{x+7}{x–2}$

E. $\text{g}(x)=\frac{x}{x^2–3x+2}$

F. $h(x)=\sqrt{x–5}$

G. $h(x)=\frac{7x–\sqrt{x^2–4}}{x}$

H. $h(x)=\log (x+2)+\frac{x}{x+2}$

I. $f(x)=e^x$

J. Prove that $f(x)\ge 1$ in $f(x)=\frac{1}{2}(3^x+3^{–x})$

Solution

A.

$f(x)=x+7$

To get $\text{rng}(f)$, we'll need to get the inverse of this function;

$y=x+7$

$x=y–7$

This is a linear inverse function, and every linear function can accept all real numbers, hence,

Range is:

Interval Notation: $\text{rng}\left(f\right)=(–\mathrm{\infty },\mathrm{\infty })$

Set Notation: $\text{rng}\left(f\right)=\{y:y\in \mathbb{R}\}$

B.

$f(x)=x^2+7x+7$

To get $\text{rng}(f)$, we'll need to get the inverse of this function;

$y=x^2+7x+7$

Making $x$ the subject of the formulae:

$x^2+7x+7–y=0$

Using quadratic formula to simplify for $x$;

$x=\frac{–b\pm \sqrt{b^2–4ac}}{2a}$

$x=\frac{–7\pm \sqrt{7^2–4\times 1\times (7–y)}}{2\times 1}$

$x=\frac{–7\pm \sqrt{49–4(7–y)}}{2}$

$x=\frac{–7\pm \sqrt{49–28+4y}}{2}$

$x=\frac{–7\pm \sqrt{21+4y}}{2}$

$x=\frac{–7\pm \sqrt{21+4y}}{2}$

This inverse function has a square root; the expression inside the square root must not be less than 0;

$21+4y\ge 0$

$4y\ge –21$

$y\ge \frac{–21}{4}$

$y\ge –5.25$

This means that our output values cannot be less than –5.25

Range is therefore written as;

Interval Notation: $\text{rng}\left(f\right)=[–5.25,\mathrm{\infty })$

Set Notation: $\text{rng}\left(f\right)=\{y:y\in \mathbb{R},y\ge –5.25\}$

C.

$\text{g}(x)=x^9–18x^5+6x^4–8$

There's a simple trick for this; 

it is useful to know that when the highest exponent of a function is an odd number, the range of the function is the set of all real numbers.

When we try to plot a graph for the above function, or try to calculate for the inverse of the function, we will end up spending a whole lot of time, but we can quickly evaluate for the range of the polynomial function by looking at its highest power. 

In the polynomial function above, the highest power is 9, 9 is an odd number, then we can predict that the range of the function will be the set of all real numbers. The same does not occur if the highest exponent of a function is an even number.

Interval Notation: $\text{rng}\left(\text{g}\right)=(–\mathrm{\infty },\mathrm{\infty })$

Set Notation: $\text{rng}\left(\text{g}\right)=\{y:y\in \mathbb{R}\}$

D.

$f(x)=\frac{x+7}{x–2}$

We can quickly get the inverse of this function,

$y=\frac{x+7}{x–2}$

$xy–2y=x+7$

$xy–x=2y+7$

$x(y–1)=2y+7$

$x=\frac{2y+7}{y–1}$

From this inverse function, we can see that $y$ can be any value except value for which $y–1=0$; this means that $y\ne 1$

Range is therefore:

Interval Notation: $\text{rng}\left(f\right)=(–\mathrm{\infty },1)\cup (1,\mathrm{\infty })$

Set Notation: $\text{rng}\left(f\right)=\{y\in \mathbb{R}:y\ne 1\}$

E.

$\text{g}(x)=\frac{x}{x^2–3x+2}$

Calculating for inverse of the function;

$y=\frac{x}{x^2–3x+2}$

$x^{2}y–3xy+2y=x$

$x^{2}y–3xy–x+2y=0$

$x^{2}y–x(3y+1)+2y=0$

$x=\frac{–[–(3y+1)]\pm \sqrt{[–(3y+1){]}^2–4\times y\times 2y}}{2\times y}$

$x=\frac{3y+1\pm \sqrt{9y^2+6y+1–8y^2}}{2y}$

$x=\frac{3y+1\pm \sqrt{y^2+6y+1}}{2y}$

Radicand of the square root above must be non-negative;

$y^2+6y+1\ge 0$

Simplifying the inequality by using quadratic formula;

$y\ge \frac{–6\pm \sqrt{6^2–4}}{2}$

$y\ge \frac{–6\pm \sqrt{36–4}}{2}$

$y\ge \frac{–6\pm \sqrt{32}}{2}$

$y\ge \frac{–6\pm \sqrt{16\times 2}}{2}$

$y\ge \frac{–6\pm 4\sqrt{2}}{2}$

$y\ge –3\pm 2\sqrt{2}$

$y\ge –3+2\sqrt{2}$

$y\le –3–2\sqrt{2}$

Range is therefore;

Interval Notation: $\text{rng(g)}=(–\mathrm{\infty },–3–2\surd 2]\cup [–3+2\surd 2,\mathrm{\infty })$

Set Notation: $\text{rng(g)}=\{y\in \mathbb{R}:y\le –3–2\surd 2,y\ge –3+2\surd 2\}$

F.

$h(x)=\sqrt{x–5}$

There's something you need to note whenever you see a square root function:

Normally, a square root relation yields both positive and negative outputs for every valid (non-negative) input, which makes it (the square root relation) not to be a function.

But for a square root relation to become a function, we must have only one output for each valid input. 

This makes it that whenever we see a function that is a square root, or involves square root, we limit the outputs to be the set of positive real numbers only.

This means that every square root functions produce only positive outputs, we do not consider the negative outputs.

Thus, for the above function, $h(x)=\sqrt{x–5}$, outputs must be only positive values.

To find out other output values that we might have to exclude from the range, we go ahead to solve for inverse of the function;

$y=\sqrt{x–5}$

$y^2=x–5$

$x=y^2+5$

This implies that no value needs to be excluded, since y would accommodate all real values,

Range is then shown to be:

Interval Notation: $\text{rng}(h)=[0,\mathrm{\infty })$

Set Notation: $\text{rng}(h)=\{y\in \mathbb{R}:y\ge 0\}$

G.

$h(x)=\frac{7x–\sqrt{x^2–4}}{x}$

Solving for inverse of the function;

$y=\frac{7x–\sqrt{x^2–4}}{x}$

$xy=7x–\sqrt{x^2–4}$

$\sqrt{x^2–4}=7x–xy$

$x^2–4=(7x–xy{)}^2$

$x^2–4=49x^2–14x^{2}y+x^2{y}^2$

$x^2–49x^2+14x^{2}y–x^2{y}^2=4$

$–48x^2+14x^{2}y–x^2{y}^2=4$

$x^2(–48+14y–y^2)=4$

$x^2=\frac{4}{–48+14y–y^2}$

$x=\sqrt{\frac{4}{–48+14y–y^2}}$

$x=\frac{2}{\sqrt{–48+14y–y^2}}$

Values that can enter into this inverse function are those such that $\sqrt{–y^2+14y–48}\ne 0$ and $–y^2+14y–48\ge 0$; 

since $–y^2+14y–48$ must not give a square root that is equal to zero, we have $–y^2+14y–48>0$

Evaluating this inequality by using quadratic formula;

$y>\frac{–14\pm \sqrt{{14}^2–4\times –1\times –48}}{2\times –1}$

$y>\frac{–14\pm \sqrt{196–192}}{–2}$

$y>\frac{–14\pm \sqrt{4}}{–2}$

$y>\frac{–14\pm 2}{–2}$

$y>\frac{–14+2}{–2}$ and $y<\frac{–14–2}{–2}$

$y>\frac{–12}{–2}$ and $y<\frac{–16}{–2}$

$y>6$ and $y<8$

It is therefore seen that $6<y<8$

Range is therefore:

Interval Notation: $\text{rng}(h)=(6,8)$

Set Notation: $\text{rng}(h)=\{y\in \mathbb{R}:6<y<8\}$

H.

$h(x)=\log (x+2)+\frac{x}{x+2}$

As a beginner, you can only understand the range of this type of function by plotting its graph;

But We can boycott having to plot the graph of the above function by detaching $\log (x+2)$ and $\frac{x}{x+2}$

We have $y=\log (x+2)$ and $y=\frac{x}{x+2}$

We then find the inverses of both individual (detached) equations;

$y=\log (x+2)$ ${10}^y=x+2$ $x={10}^y–2$

$y=\frac{x}{x+2}$ $xy+2y=x$ $x–xy=2y$ $x(1–y)=2y$ $x=\frac{2y}{1–y}$

We conclude that $\log (x+2)$ will yield the set of all real numbers; whilst $\frac{x}{x+2}$ will have an asymptote (explained in next blog post) at 1; since we will end up having to add these two outputs, we see that we have an output of all real numbers for the function $h(x)=\log (x+2)+\frac{x}{x+2}$

Range:

Interval Notation: $\text{rng}(h)=(–\mathrm{\infty },\mathrm{\infty })$

Set Notation: $\text{rng}(h)=\{y:y\in \mathbb{R}\}$

I.

$f(x)=e^x$

We can quickly get the range of the above function by solving for the inverse function;

$y=e^x$

Taking the natural log of both sides:

${\log }_{e}y={\log }_e{e}^x$

${\log }_{e}y=x{\log }_{e}e$

${\log }_{e}y=x$

$x=\ln y$

$y$ can only take values greater than 0, hence, range is:

Interval Notation: $\text{rng}(f)=(0,\mathrm{\infty })$

Set Notation: $\text{rng}(f)=\{y\in \mathbb{R}:y>0\}$

J.

Prove that $f(x)\ge 1$ in $f(x)=\frac{1}{2}(3^x+3^{–x})$

What this question is trying to require of us is that we should check if all $f(x)$ for $f(x)=\frac{1}{2}(3^x+3^{–x})$ is greater than 1; this, in other words, mean that we should check if the range of the function is greater than 1.

Solving for the inverse of the function;

$y=\frac{1}{2}(3^x+3^{–x})$

$2y=3^x+3^{–x}$

$2y=3^x+\frac{1}{3^x}$

$2y=\frac{3^{2x}+1}{3^x}$

$2y{3}^x=3^{2x}+1$

$3^{2x}–2y{3}^x+1=0$

Let $3^x=a$

$a^2–2ay+1=0$

Using quadratic formula to simplify $a$;

$a=\frac{–(–2y)\pm \sqrt{(–2y{)}^2–4\times 1\times 1}}{2\times 1}$

$a=\frac{2y\pm \sqrt{4y^2–4}}{2}$

$a=\frac{2y\pm \sqrt{4(y^2–1)}}{2}$

$a=\frac{2y\pm 2\sqrt{y^2–1}}{2}$

$a=y\pm \sqrt{y^2–1}$

Recall, $3^x=a$

$3^x=y\pm \sqrt{y^2–1}$

$\log 3^x=\log (y\pm \sqrt{y^2–1})$

$x\log 3=\log (y\pm \sqrt{y^2–1})$

$x=\frac{\log (y\pm \sqrt{y^2–1})}{\log 3}$

From the above inverse function, the radicand $y^2–1$ must not be negative;

$y^2–1\ge 0$

$y^2\ge 1$

$y\ge \sqrt{1}$

$y\ge 1$ or $y\le –1$

 For the logarithmic expression, $y\pm \sqrt{y^2–1}$ must be greater than $0$;

This implies $y$ must take only positive values,

Range is therefore:

$\text{rng}(f)=\{y\in \mathbb{R}:y\ge 1\}$

It is therefore, seen that $f(x)\ge 1$

Domain and Range on a Graph

As you would normally know, on a Graph, the horizontal axis is the $x$–axis, while the vertical axis is the $y$–axis.

$x$–axis is the axis for independent variables, that is, inputs; while $y$–axis is the axis for dependent variables, that is, outputs.

This means that the domain of a function is shown along the $x$–axis of a graph, while range are shown along the $y$–axis.

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Summary

📌 The domain of a function is the set of all input values that will make a function defined.

📌 The Codomain of a function is the set of all possible output values that may come out of a function.

📌 The Range of a function is the set of all actual output values that come out of a function.

📌 For a function to have a domain that will make it defined:

• Denominator must not be equal to zero.
• Radicand inside a square root must be non-negative.
• Logarithmic expression must be greater than 0, etc.

Practice Questions

Question 1

Which of these is the Domain for $f(x)=x^3–7x^2+5x–2$

A. $\text{dom}(f):\{x|x\in \mathbb{R},x\ne 2\}$

B. $\text{dom}(f):\{x|x\in \mathbb{R}\}$

C. $\text{dom}(f):\{x|x\in \mathbb{R},x\ge 2\}$

D. $\text{dom}(f):\{x|x\in \mathbb{R},x>2\}$

Question 2

Calculate domain and range of $f(x)=\sin x$

A. $\text{dom}(f)=\{x\in \mathbb{R}\}$
       $\text{rng}(f)=\{y\in \mathbb{R}:–1\le y\le 1\}$

B. $\text{dom}(f)=\{x\in \mathbb{R}\}$
       $\text{rng}(f)=\{y\in \mathbb{R}\}$

C. $\text{dom}(f)=\{x\in \mathbb{R}:0\le x\le 360\}$
       $\text{rng}(f)=\{y\in \mathbb{R}:–1\le y\le 1\}$

D. $\text{dom}(f)=\{x\in \mathbb{R}\}$
       $\text{rng}(f)=\{y\in \mathbb{R}:y\ge 0\}$

Question 3

Find domain and range for $2x+3y=4$

A. $\text{dom}(f)=\{x\in \mathbb{R}:x\ge 1\}$
       $\text{rng}(f)=\{y\in \mathbb{R}\}$

B. $\text{dom}(f)=\{x\in \mathbb{R}\}$
       $\text{rng}(f)=\{y\in \mathbb{R}\}$

C. $\text{dom}(f)=\{x\in \mathbb{R}:x\ne 3\}$
       $\text{rng}(f)=\{y\in \mathbb{R}:y\ne 2\}$

D. $\text{dom}(f)=\{x\in \mathbb{R}:x\ge 3\}$
       $\text{rng}(f)=\{y\in \mathbb{R}:y\ge 2\}$

Question 4

Find domain and range for $f(x)=3x^2+2x–5$

A. $\text{dom}(f)=\{x:x\in \mathbb{R}\}$
        $\text{rng}(f)=\{y:y\in \mathbb{R}\}$

B. $\text{dom}(f)=\{x:x\in \mathbb{R}\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:y\le \frac{16}{3}\}$

C. $\text{dom}(f)=\{x:x\in \mathbb{R}\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:y\ne \frac{16}{3}\}$

D. $\text{dom}(f)=\{x:x\in \mathbb{R}\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:y\ge \frac{16}{3}\}$

Question 5

Find domain and range for $f(x)=\frac{x^2–x}{x–2}$

A. $\text{dom}(f)=\{x\in \mathbb{R}:x\ne 2\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:y\ge 3–2\surd 2,y\le 3+2\surd 2\}$

B. $\text{dom}(f)=\{x\in \mathbb{R}:x\ne 2\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:y\le 3–2\surd 2,y\ge 3+2\surd 2\}$

C. $\text{dom}(f)=\{x\in \mathbb{R}:x\ne 2\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:3–2\surd 2\le y\le 3+2\surd 2\}$

D. $\text{dom}(f)=\{x\in \mathbb{R}:x\ne 2\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:y\ne 3+2\surd 2\}$

Question 6

Find domain and range for $f(x)=x^5+4x^4–3x^2+7$

A. $\text{dom}(f)=\{x\in \mathbb{R}:x\ne 7\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:y\le 3\}$

B. $\text{dom}(f)=\{x\in \mathbb{R}\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:y\ge 3\}$

C. $\text{dom}(f)=\{x\in \mathbb{R}\}$
        $\text{rng}(f)=\{y\in \mathbb{R}\}$

D. $\text{dom}(f)=\{x\in \mathbb{R}\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:y\ne 3\}$

Question 7

Calculate domain and range for $f(x)=\frac{x}{x^2–9}$

A. $\text{dom}(f)=\{x\in \mathbb{R}:x\ne 3\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:y\ne 2\}$

B. $\text{dom}(f)=\{x\in \mathbb{R}:x\ne 3,x\ne –3\}$
        $\text{rng}(f)=\{y\in \mathbb{R}\}$

C. $\text{dom}(f)=\{x\in \mathbb{R}\}$
        $\text{rng}(f)=\{y\in \mathbb{R}\}$

D. $\text{dom}(f)=\{x\in \mathbb{R}:x\ne 9\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:y\ne 9\}$

Question 8

Find domain and range for $f(x)=\ln (x+2)$

A. $\text{dom}(f)=\{x\in \mathbb{R}\}$
        $\text{rng}(f)=\{y\in \mathbb{R}\}$

B. $\text{dom}(f)=\{x\in \mathbb{R}:x\ge –2\}$
        $\text{rng}(f)=\{y\in \mathbb{R}\}$

C. $\text{dom}(f)=\{x\in \mathbb{R}:x>–2\}$
        $\text{rng}(f)=\{y\in \mathbb{R}\}$

D. $\text{dom}(f)=\{x\in \mathbb{R}:x<–2\}$
        $\text{rng}(f)=\{y\in \mathbb{R}\}$

Question 9

Find domain and range for     $f(x)={\sin }^2(x–2)$

A. $\text{dom}(f)=\{x\in \mathbb{R}\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:0\le y\le 1\}$

B. $\text{dom}(f)=\{x\in \mathbb{R}:0\le x\le 1\}$
        $\text{rng}(f)=\{y\in \mathbb{R}\}$

C. $\text{dom}(f)=\{x\in \mathbb{R}\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:–1\le y\le 1\}$

D. $\text{dom}(f)=\{x\in \mathbb{R}\}$
        $\text{rng}(f)=\{y\in \mathbb{R}\}$

Question 10

Calculate domain and range for $f(x)=\frac{6}{4–\sqrt{6x–2}}$

A. $\text{dom}(f)=\{x\in \mathbb{R}:\frac{1}{3}\le x<3\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:y<0,y\ge \frac{3}{2}\}$

B. $\text{dom}(f)=\{x\in \mathbb{R}:x>3\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:y<0,y\ge \frac{3}{2}\}$

C. $\text{dom}(f)=\{x\in \mathbb{R}:\frac{1}{3}\le x<3,x>3\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:y<0,y\ge \frac{3}{2}\}$

D. $\text{dom}(f)=\{x\in \mathbb{R}:\frac{1}{3}\le x<3,x>3\}$
        $\text{rng}(f)=\{y\in \mathbb{R}:y<0\}$

Recommended Videos

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Domain And range of trigonometric functions

Test Questions

DOMAIN, RANGE

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DOMAIN, RANGE

Total Questions: 20

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Quiz Answers

1.

{x ∈ R : x ≠ 1, x ≠ –1}

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2.

{x ∈ R}

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3.

{x ∈ R : x ≠ 2}

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4.

{x ∈ R : x > ¹/₂}

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5.

{x ∈ R : x ≥ 0}

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6.

{x ∈ R : x > ¹/₂}

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7.

{x ∈ R : x ≥ –4}

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8.

{x ∈ R : x > 4}

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9.

{x ∈ R}

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10.

{x ∈ R}

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11.

{y ∈ R : y ≥ 0}

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12.

{y ∈ R : y > 5}

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13.

{y ∈ R : y ≥ 2 or y≤–2}

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14.

{y ∈ R}

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15.

{y ∈ R : y ≠ 1}

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16.

{y ∈ R : y ≠ log 2}

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17.

{y ∈ R}

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18.

{y ∈ R}

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19.

{y ∈ R : y ≠ 0}

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20.

{y ∈ R}

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