To find the domain, you must find all the values you can put in for x. Thus, you must figure out what values would "break" your function and give you something unable to be computed. 

First we know that the number inside a square root must be positive. Thus, we can set the inside greater than or equal to 0 and solve.

We also know that the denominator of a fraction must never be 0. Thus, if we find out when it is 0, we can exclude that x value from our domain.

If we combine both of these for x, we can create an interval for our domain.

Consider the function . Then 

.

 is an absolute value of a linear function. Since the value of this function cannot be less than 0, its graph changes direction at the value of  at which . This value can be found by setting 

and solving for :

. Also, , so . Since the absolute value of an expression cannot be negative, 0 is the minimum end of the range of .

Also, if 

, 

then, using the properties of inequality,

Therefore, if , then . A function of the form  is a linear function and is either constantly increasing or constantly decreasing; since  in this case,  increases, and as  goes to infinity, so does . Therefore, the range of  has no upper bound, and the correct choice is .

Since the piecewise-defined function  is defined two different ways, one for nonpositive numbers and one for positive numbers, examine both definitions and determine each partial range separately;  the union of the partial ranges will be the overall range.

If , then 

Since 

,

applying the properties of inequality,

Therefore, on the portion of the domain comprising nonpositive numbers, the partial range of  is the set .

If , then 

Since 

,

applying the properties of inequality,

Therefore, on the portion of the domain comprising positive numbers, the partial range of  is the set .

The overall range is the union of these partial ranges, which is .

A quadratic function has a parabola as its graph; this graph changes direction (downward to upward, or vice versa) at a given point called the vertex.

 exists on a given domain interval if and only if there does not exist  and  on this domain such that , but . This will happen if the graph changes direction on the domain interval. The key is therefore to identify the interval that contains the vertex.

The -coordinate of the vertex of the parabola of the function

is .

The -coordinate of the vertex of the parabola of  can be found by setting :

.

Of the five intervals in the choices, 

,

so  cannot exist if the domain of  is restricted to this interval. This is the correct choice.

A quadratic function has a parabola as its graph; this graph decreases, then increases (or vice versa), with a vertex at which the change takes place. 

The -coordinate of the vertex of the parabola of the function

is .

The -coordinate of the vertex of the parabola of  can be found by setting :

.

, so the vertex is on the domain.  The maximum and the minimum of  must occur at the vertex and one endpoint, so evaluate , , and .

The minimum and maximum values of  are  and 12, respectively, so the correct range is .

A quadratic function has a parabola as its graph; this graph decreases, then increases (or vice versa), with a vertex at which the change takes place. 

The -coordinate of the vertex of the parabola of the function

is .

The -coordinate of the vertex of the parabola of  can be found by setting :

.

, so the vertex is not within the interval to which the domain is restricted. Therefore,  increases or decreases constantly on , and its maximum and minimum on this interval will be found on the endpoints. These values are  and , which can be evaluated using substitution:

The range is .

A quadratic function has a parabola as its graph; this graph changes direction at a vertex. 

The -coordinate of the vertex of the parabola of the function

is .

The -coordinate of the vertex of the parabola of  can be found by setting :

, so the vertex is not on the domain.  The maximum and the minimum of  must occur at the endpoints, so evaluate  and .

The minimum and maximum values of  are  and 40, respectively, so the correct range is .

A quadratic function has a parabola as its graph; this graph changes direction at a vertex. 

The -coordinate of the vertex of the parabola of the function

is .

The -coordinate of the vertex of the parabola of  can be found by setting :

, so the vertex is on the domain.  The maximum and the minimum of  must occur at the vertex and one endpoint, so evaluate , , and .

The minimum and maximum values of  are  and 17, respectively, so the correct range is .

First, we must determine whether  exists.

A quadratic function has a parabola as its graph; this graph changes direction at its vertex. 

 exists if and only if, if , then - or, equivalently, if there does not exist  and  such that , but . This will happen on any interval on which the graph of  constantly increases or constantly decreases, but if the graph changes direction on an interval, there will be  such that  on this interval. The key is therefore to determine whether the interval to which the domain is restricted contains the vertex.

The -coordinate of the vertex of the parabola of the function

is .

The -coordinate of the vertex of the parabola of  can be found by setting :

.

The vertex of the graph of  without its domain restriction is at the point with -coordinate . Since , the vertex is not in the interior of the domain; as a consequence,  exists on .

To determine the inverse of , first, rewrite  in vertex form

, the same as  in the standard form.

The graph of , if unrestricted, would have -coordinate , and -coordinate 

Therefore, .

The vertex form of  is therefore

Replace  with :

Switch  and :

Solve for . First, add 46 to both sides:

Multiply both sides by 2:

Take the square root of both sides:

Subtract 8 from both sides

Replace  with :

Either  or 

The domain of  is the set of nonnegative numbers; this is consequently the range of .  can only have negative values, so the only possible choice for  is .

A quadratic function has a parabola as its graph; this graph changes direction at its vertex.

The -coordinate of the vertex of the parabola of the function

is .

The -coordinate of the vertex of the parabola of  can be found by setting  and :

.

Since , the vertex falls within the domain of . 

Since, in , the quadratic coefficient is positive, the parabola curves upward. On the set of all nonnegative numbers, the function has no maximum. The minimum occurs at the vertex, which is in the domain; to calculate it, evaluate :

The range of the function given the domain restriction is .