Step 1: We need to determine what kind of function we have here. We have a rational function.

Step 2: Since we have a rational function, the denominator cannot be equal to $\displaystyle 0$. We will equal the denominator to $\displaystyle 0$ and find the values of $\displaystyle x$ that make the bottom zero. 

$\displaystyle \sqrt{4-x^2}=0$
Square both sides:
$\displaystyle (\sqrt {4-x^2})^2=0^2\rightarrow 4-x^2=0$
Add $\displaystyle x^2$ to both sides of the equal sign:

$\displaystyle 4-x^2+x^2=0+x^2$
Simplify:
$\displaystyle 4=x^2$
Take the square root of both sides
$\displaystyle \sqrt{4}=\sqrt {x^2}\rightarrow \pm2=x$

However, we said earlier that these two solutions cannot be values of x, so we must change the sign:

$\displaystyle \pm 2 \neq x$

Step 3: We need to write the solution in interval notation form. 

The smallest number we can plug in for $\displaystyle x$ is $\displaystyle -\infty$, and the biggest is $\displaystyle +\infty$. $\displaystyle x$ cannot be $\displaystyle -2$ and $\displaystyle 2$, but it can be anything else. So, we should have three intervals:

1) Between $\displaystyle -\infty$ and $\displaystyle -2$, which is written as $\displaystyle (-\infty,-2)$
2) Between $\displaystyle -2$ and $\displaystyle 2$, which is written as $\displaystyle (-2,2)$
3) Between $\displaystyle 2$ and $\displaystyle +\infty$, which is written as $\displaystyle (2,+\infty)$

The full solution in interval notation is $\displaystyle (-\infty,-2)\cup (-2,2)\cup (2,+\infty)$. 

Step 1: We have a square root here, where the inside of the radical must be greater or equal than $\displaystyle 0$, so we get a real answer.

Step 2: Set the inside equation greater or equal than zero...

$\displaystyle x+3 \ge 0$

Step 3: Solve for x...

$\displaystyle x \ge -3$... Write this in interval notation. $\displaystyle -3$ is the leftmost boundary, infinity is the right boundary. $\displaystyle -3$ has a bracket with it, infinity does not..

Final Answer: $\displaystyle [-3,+\infty)$

The domain is the set of all first elements of ordered pairs (x-coordinates). The range is the set of all second elements of ordered pairs (y-coordinates).

Given this set of ordered pairs $\displaystyle \left \{ (-4,3) (-3,3)(-2,3)(-1,3)\right \}$, the

domain is $\displaystyle \left \{ -4,-3,-2,-1\right \}$ and the range is $\displaystyle \left \{ 3\right \}$. 

Note: If a value of a coordinate is repeated, it only gets listed once in the set.

The surface area of a cube is the total outside area of that cube. The formula to find the Area of a cube is $\displaystyle A=6s^{2}$.

A cube has a side length of at least $\displaystyle 4$ inches. The surface area is a function of side length. What is the domain (side length) and the range (surface area)?

The independent variable is the length of the side.  The domain consists of numbers that represent the length of sides.  The problem states that the length of the sides is at least $\displaystyle 4$ inches. So, the domain is all numbers greater than or equal to four.

$\displaystyle D\left \{ s\mid s\geq 4\right \}$

The range consists of the numbers that corresponds with the chosen $\displaystyle x-$values in a function.

To find the range, find the Area of the cube with the length of the side measuring $\displaystyle 4$ inches; this is the smallest value that $\displaystyle s$ (or the side length measurement), can be equal to.

$\displaystyle A \geq 6 (4^{2})$

$\displaystyle A\geq6 (16)$

$\displaystyle A\geq 96$

$\displaystyle R\left \{ A\mid A\geq 96\right \}$

The correct answer is $\displaystyle D\left \{ s\mid s\geq 4\right \}$   ;   $\displaystyle R\left \{ A\mid A\geq 96\right \}$

For a function $\displaystyle f$ defined by an expression with variable  $\displaystyle x$  the domain of $\displaystyle f$ is the set of all real numbers that the variable $\displaystyle x$  can take such that the expression defining the function is real. 

The expression defining function contains a square root.

 $\displaystyle y=-\sqrt{-3x+6}$.  

The domain is the set of all first elements of ordered pairs or $\displaystyle x-coordinates.$

Because there cannot be a negative inside the radical sign, set the inside of the radical  $\displaystyle \geq 0$  and solve.   The expression under the radical has to satisfy the condition $\displaystyle -3x+6\geq 0$  for the function to take real values.  

Solve:

$\displaystyle -3x+6\geq 0$

Subtract $\displaystyle 6$ from both sides of the equation.

$\displaystyle -3x - 6 + 6 \geq 0 + -6$

$\displaystyle -3x \geq -6$

Divide both sides of the equation by $\displaystyle -3$.

$\displaystyle \frac{-3x}{-3} \geq \frac{-6}{-3}$

Because you are dividing both sides of the equation by a negative integer  $\displaystyle (-3)$this will change the sign from  $\displaystyle \geq$   to  $\displaystyle \leq$.

$\displaystyle x \leq 2$

The domain or d is all  $\displaystyle x \leq 2.$

The domain is the set of all first elements of ordered pairs (x-coordinates). The range is the set of all second elements of ordered pairs (y-coordinates).

The domain, or all values of  $\displaystyle x$ $\displaystyle \left \{ 2,3,4\right \}$.

In order to get the range or all values of  $\displaystyle y,$  insert the given values of  $\displaystyle x$ (the domain) into the equation $\displaystyle y=x+8$ and solve for $\displaystyle y$ (the range).

$\displaystyle y = 2^{2} +8 = 4+8 = 12$

$\displaystyle y =3^{2} +8 = 9+8 = 17$

$\displaystyle y = 4^{2} + 8 = 16+ 8 =24$

The range for the function  with the domain of $\displaystyle \left \{ 2,3,4\right \}$ is

$\displaystyle \left \{ 12,17,24\right \}$.

The ordered pairs represented on the graph are:$\displaystyle \left \{ (0,2), (1,3), (2,4), (-1,1), (-2,0)\right \}$

The domain is the set of all first elements of ordered pairs (x-coordinates). The range is the set of all second elements of ordered pairs (y-coordinates).

The domain is $\displaystyle \left \{-2,-1, 0, 1, 2\right \}$; the range is $\displaystyle \left \{0,1,2,3,4\right \}$.

For a function $\displaystyle f$ defined by an expression with variable $\displaystyle x$ the domain of $\displaystyle f$ is the set of all real numbers that the variable $\displaystyle x$  can take such that the expression defining the function is real. 

The expression defining function $\displaystyle f(x) = \sqrt{2x-10}$ contains a square root.       The expression under the radical has to satisfy the condition $\displaystyle 2x-10\geq 0$  for the function to take real values.

Solve  $\displaystyle 2x-10\geq 0$

Add $\displaystyle 10$ to both sides of the equation.

$\displaystyle 2x-10 +10\geq 0+10$

$\displaystyle 2x\geq 10$

Divide both sides by $\displaystyle 2$.

$\displaystyle 2x\div2 \geq10\div2$

The domain is all $\displaystyle x\geq5.$

To find the domain and the range of the inverse of a relation with this set of coordinates, first find the domain and range of the set of coordinates given.

The domain is all of the $\displaystyle x$-coordinates and the range is all of the $\displaystyle y$-coordinates. Remember, if a coordinate is the same, it is only listed once.

The domain of $\displaystyle \left \{ (0,1) (2,3) (-2,-3) (4,1) \right \}$ is

 $\displaystyle \left \{ -2,0,2,4\right \}$  and the range is $\displaystyle \left \{ -3,1,3 \right \}$.

However, the question is asking for the domain and the range of the inverse.  So, the values of $\displaystyle x$ (the domain) will now become the values of $\displaystyle y$, (the range) and the values of $\displaystyle y$ (the range) will become the values of $\displaystyle x$, or the domain.

The domain for the inverse is $\displaystyle \left \{ -3,1,3\right \}$ and the range is $\displaystyle \left \{-2, 0,2,4\right \}$.

The domain is the set of all first elements of ordered pairs (x-coordinates). The range is the set of all second elements of ordered pairs (y-coordinates).

$\displaystyle f(x) = x^{2} -3$   domain or values of $\displaystyle x$ $\displaystyle \left \{ -3,-2,-1,0,1,2,3\right \}$

Plug all of the values of $\displaystyle x$ that have been given into the equation

$\displaystyle y=x^{2} -3$

$\displaystyle y=-3^{2} -3 = 9-3 =6$

$\displaystyle y = -2^{2} -3 = 4-3=1$

$\displaystyle y =-1^{2} -3 = 1-3 = -2$

$\displaystyle y = 0^{2} -3 = 0-3 = -3$

$\displaystyle y = 1^{2} - 3 = 1-3=-2$

$\displaystyle y= 2^{2} -3 = 4-3=1$

$\displaystyle y = 3^{2} - 3 = 9-3=6$

Even though some of the values of $\displaystyle y$ are repeated, they are only listed once in the set..

The range, or all values of $\displaystyle y$ for the function     are

$\displaystyle \left \{-3,-2,1,6\right \}$.