The domain of a rational function is the set of all values of $\displaystyle x$ for which the denominator is not equal to 0 (the value of the numerator is irrelevant), so we set the denominator to 0 and solve for $\displaystyle x$ to find the excluded values.

$\displaystyle x^{2}-6x+8 =0$

This is a quadratic function, so we factor the expression as $\displaystyle (x+?)(x+?)$, replacing the question marks with two numbers whose product is 8 and whose sum is $\displaystyle -6$. These numbers are $\displaystyle -4,-2$, so

$\displaystyle x^{2}-6x+8 =0$ 

becomes

$\displaystyle (x-2)(x-4) = 0$

So either 

$\displaystyle x-2 = 0$, in which case $\displaystyle x = 2$, 

or $\displaystyle x-4 = 0$, in which case $\displaystyle x = 4$.

Therefore, 2 and 4 are the only numbers excluded from the domain.

In order for the function to be real, the value inside of the square root must be greater than or equal to zero. The domain refers to the possible values of the independent variable (x-value) that allow this to be true.

$\displaystyle \small \sqrt{x}$

For this term to be real, $\displaystyle \small x$ must be greater than or equal to zero.

$\displaystyle x\geq0$

$\displaystyle \left ( -\infty , -3 \right ) = \left \{x| -\infty < x< -3 \right \}$

and $\displaystyle \cup$ stands for OR in Set Builder Notation

The graph will open to the left.  The contents inside the square root cannot be negative.  

Set the inside equal to zero.

$\displaystyle \frac{1}{3}-x=0$

$\displaystyle x=\frac{1}{3}$

Any number greater than one-third will be invalid, but any number below will be the domain of this function.

The domain is:   

$\displaystyle \left(-\infty,\frac{1}{3}\right]$

At  $\displaystyle x=0$, there is a hole in this function:

$\displaystyle f(0)= \frac{0^{2}-2(0)}{0}=\frac{0}{0}$

At all other values of x, both positive and negative, this function will be defined.

To find the range, we first need to find the domain of the function. Then we will determine the range by finding the output values based on the domain.

 $\displaystyle f(x) =\frac{\sqrt{x^{3}}+21}{x^{2}-x}$ 

First, we can factor out x from the denominator:

$\displaystyle f(x) =\frac{\sqrt{x^{3}}+21}{x(x-1)}$

Since the function will be undefined when the denominator equals 0, we know that x=0 and x=1 are not in the domain of this function. We also know that this function is undefined for all negative number since the function includes a root of an even degree. The exponent under the square root does not change this, since cubing a negative number will still result in a negative number, as with any odd degreed exponent. 

So, this function is defined at all non-negative values except for 0 and 1.

$\displaystyle D: (0,1)\cup(1,\infty )$

The domain is defined as the input values or x values of a set.

So domain: 4,3,7

The range is defined as the output values or y values of a set.

So range: 2,7,3

In order for the set to be a function, each input value must have only one corresponding output value. In this example, the input value 4 has two output values: 2 and 3. The set is not a function.

The domain of a set of ordered pairs is the $\displaystyle x$ values.

The $\displaystyle x$ values are the first number in each set of coordinates.

The $\displaystyle x$ values are:

$\displaystyle (1, 3, 6, 7, 9)$

The domain consists of all values that the input $\displaystyle (x)$ can be without making the output $\displaystyle (f(x))$ unreasonable. In our problem, the only condition that would dissatisfy the equations parameters is a negative inside the square root. However, having $\displaystyle x^{2}$ inside the square root makes this a bit tricky, because we have to consider that squaring this value will always yield something positive. Thus, we cannot have any values of $\displaystyle x$ whose squares are strictly less than $\displaystyle 9$. Thus, the domain must be all values of $\displaystyle x$ that are greater than or equal to $\displaystyle 3$ and less than or equal to $\displaystyle -3$.

To solve this equation you must look at the denominator since the denominator can never equal zero.

You need to set the denominator equal to $\displaystyle 0$ then solve for $\displaystyle x$.

$\displaystyle x^2=0$,  then square root both sides to get

$\displaystyle x=0$, the value that $\displaystyle x$ cannot be, therefore, is $\displaystyle 0$.