Every function you ever came across with has its **domain** and **codomain**. For a function $f$ that takes every member from the set $X$ and relates it to some member of $Y$ set, we say that $X$ is its * domain* and $Y$ is its

*.*

**codomain**Every function is given by its *formula*. Because of that, we can precisely determine domain of every given function.

*If a function is given by its formula, the domain of the function is the set of all numbers for which that formula makes sense.*

The function domain is marked with$ D_f$.

There are few rules you need to know in order to correctly find the domain of a function.

**1. The denominator must never be equal to zero.**

Let’s take, for an example, the function$ f(x) = \frac{1}{x}$. Try to insert any number you can think of. You can find its value, no matter how little or how large the numbers are, but you know nothing about the value of this function in $ f(0)$. $ f(0) = \frac{1}{0}$ which is undefined. This means that this function is defined in the whole set of real numbers except for $0$.

“The **domain of the given function** is the set of all $x$ from the set of real numbers that are **different from zero**.”

If you take a look at a graph of this function, you see that it is not defined in $0$.

We can also represent the domain graphically.

**Example 1.** Find the domain of the following functions:

a) $ f(x) = \frac{5}{2x – 1}$

The numerator of a fraction can be anything, so we don’t have to worry about it. The main problem is denominator. Since it can’t be equal to zero, we’ll equalize that whole expression with it in order to find what it can’t be.

$ 2x – 1 \not= 0$

$ 2x \not= 1$

$ x \not= \frac{1}{2}$

$ D_f = \{ x \in \mathbb{R} : x \not= \frac{1}{2} \}$

b) $ f(x) = \frac{2x^3 + x^2 + 2}{x^2 – 4}$

Since the numerator is a polynomial, which is a continuous function, it is defined in a whole set of real numbers. Again, the only problem is the denominator.

$ x^2 – 4 \not= 0$

$ x^2 \not= 4$

$ x \not= \pm 2$

$ D_f = \{ x \in \mathbb{R} : x \not= \pm 2 \}$

**2. The expression under the even root must be greater than or equal to zero.**

When talking about functions we are thinking about real functions of real variable. This is why, for this area, the imaginary numbers can’t appear. Negative odd roots also aren’t a problem, for example $\sqrt[3]{-8} = \sqrt[3]{(-2)(-2)(-2)}$.

**Example 2**. Find the domain of the following functions:

a) $ f(x) = \sqrt{x^2 – 4x + 3}$

$ x^2 – 4x + 3 \ge 0$

$ x \ge 3, x \le 1$

$ D_f = \{ x \in \mathbb{R} : x \le 1 \} \{ x \in \mathbb{R} : x \ge 3 \}$

What if we have more than one condition? What if, for example, the even root appears in the fraction? For these cases you’ll have to watch the intersection between those two conditions.

**Example 3**. Find the domain of the following function:

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

$\sqrt[4]{x^2 – 4x + 3} > 0$, $x^2 – 4x + 3 > 0$

The intersection of these two conditions is $ x^2 – 4x + 3 > 0$.

$ D_f = \{ x \in \mathbb{R} : x < 1 \} \cup \{ x \in \mathbb{R} : x > 2 \}$

**3. The base and the argument of the logarithm must be greater than zero.**

$ log_a b$ $a > 0, b >0 $

When dealing with functions whose part is a logarithm, you have two conditions. First is the base and the second is argument. In order to find the domain of that function you have to find the intersection of these two conditions.

**Example 4**. Find the domain of the following function:

$ f(x) = log_{x + 2}(2x + 1)$

$ x + 2 > 0$, $ 2x + 1 > 0$

$ x > -2$, $ x > -\frac{1}{2}$

$ D_f = \{ x \in \mathbb{R} : x > -\frac{1}{2} \}$

**Example 5**. Find the domain of the following function:

$ f(x) = \sqrt{ln(2x + 3)}$

$ ln(2x + 3) > 0$

Remember that the function values of logarithm are greater than $0$ for arguments greater than $1$.

$ x > -1$

$ D_f = \{ x \in \mathbb{R} : x > -1 \}$

**4. Inverse trigonometric functions’ domain is the same as codomain of the restricted trigonometric functions.**

Since the restriction of sine is the function $ in: [-\frac{\pi}{2}, \frac{\pi}{2}] \rightarrow [-1, 1]$, its codomain is the segment $ [-1, 1]$. The inverse function, arcsine, is a function whose domain is a segment $ [-1, 1]$.

**Example 6**. Find the domain of the following function:

$ f(y) = arcsin(\frac{y}{2})$

$\frac{y}{2} \in [-1, 1] \rightarrow y \in [-2, 2]$

$ D_f = \{ y \in \mathbb{R} : – 2 \le y \le 2\}$

The same applies to $arccosine$.

**Example 7**. Find the domain of the following function:

$ f(x) = tan x$

$ f(x) = \frac{sin x}{cos x}$

The domain is the set of real numbers such that $ cos x \not= 0$.

$ D_f = \{ x \in \mathbb{R} : x \not= k\pi, k \in \mathbb{Z}$

## Domain of functions of multiple variables

**If a function of multiple variables is given by its formula, the domain of that function is the set of all points for which that formula makes sense.**

To determine the domain of a function of multiple variables you do the same you did with functions of one variable. The sketches of the domain will now be two dimensional.

**Example 8**. Find the domain of the following function:

$ f(x) = \sqrt{1 – x^2 – y^2}$

$1 – x^2 – y^2 \ge 0$

$ x^2 + y^2 \le 1$

$ D_f = \{ (x, y) \in \mathbb{R} : x^2 + y^2 \le 1 \}$

**Example 9**. Find the domain of the following function:

$ f(x, y) = ln (x^2 + y)$

$ x^2 + y > 0$

$ y > – x^2$

$ D_f = \{ (x, y) \in \mathbb{R} : y > -x^2 \}$

**Example 10**. Find the domain of the following function:

$ f(x, y) = \sqrt{-ln (y^2 + 2x + 1}$

$ – ln {(y^2 + 2x + 1}) \ge 0$

$ ln {(y^2 + 2x + 1)} \le 0$

$ y^2 + 2x + 1 \le 0$

$ y^2 + 2x \le 0$