How to find the domain of a function - SAT Math

The purpose of this question is to understand when x values will yield y values. The term inside of a square root can be positive or equal to zero in order to yield a value. This means that x can be equal to all real numbers that are -9 or higher, which shows that the domain of the function is all real numbers that are at least -9.

Because x and y must be negative integers greater than negative five, then x and y can only be equal to the following values:

x can equal -4, -3, -2, or -1

y can equal -4, -3, -2, or -1

Now we can try all of the combinations of x and y, and see what x # y would equal. It is helpful to note that x # y is the same as y # x because 2yx + y + x = 2xy + x + y. This means that the order of x and y doesn't matter.

-4 # -4 = 2(-4)(-4) + -4 + -4 = 24

-4 # -3 = 2(-4)(-3) + -4 + -3 = 17

-4 # -2 = 10

-4 # -1 = 3

-3 # -3 = 12

-3 # -2 = 7

-3 # -1 = 2

-2 # -2 = 4

-2 # -1 = 1

-1 # -1 = 0

We don't need to find -3 # -4, -2 # -4, etc, because x # y = y # x .

The smallest value of x # y must be 0.

Using our properties of exponents, we could rewrite as

This means that when we input , we first subtract 2, then take this to the fourth power, then take the fifth root, and then add three. We want to look at these steps individually and see whether there are any values that wouldn’t work at each step. In other words, we want to know which values we can put into our function at each step without encountering any problems.

The first step is to subtract 2 from . The second step is to take that result and raise it to the fourth power. We can subtract two from any number, and we can take any number to the fourth power, which means that these steps don't put any restrictions on .

Then we must take the fifth root of a value. The trick to this problem is recognizing that we can take the fifth root of any number, positive or negative, because the function is defined for any value of ; thus the fact that has a fifth root in it doesn't put any restrictions on , because we can add three to any number; therefore, the domain for is all real values of .

If a value of x makes the denominator of a equation zero, that value is not part of the domain. This is true, even here where the denominator can be "cancelled" by factoring the numerator into

and then cancelling the from the numerator and the denominator.

This new expression, is the equation of the function, but it will have a hole at the point where the denominator originally would have been zero. Thus, this graph will look like the line with a hole where , which is

.

Thus the domain of the function is all values such that

The domain means what real number can you plug in that would still make the function work. For this case, we have to worry about the denominator so that it does not equal 0, so we solve the following. 7x – 1 = 0, 7x = 1, x = 1/7, so when x ≠ 1/7 the function will work.

The domain of f(x) is defined as all of the values of x for which f(x) is defined.

The first value we have to consider is 1. Let's find f(1):

f(1) = (13 – 3(1)2 +2(1))–1

= (1 – 3(1) + 2)–1

= (1 – 3 + 2)–1

= (0)–1.

Remember that, in general, a–1 = 1/a. Thus, 0–1 = 1/0. However, it is impossible to have 0 in the denominator of a fraction, because it is impossible to divide anything by zero. Thus, 0–1 is undefined. Since f(1) is undefined, 1 cannot belong to the domain of f(x).

Now let's find f(2):

f(2) = (23 – 3(2)2 +2(2))–1

= (8 – 3(4) + 4)–1

=(8 – 12 + 4)–1

= 0–1

Because f(2) is undefined, 2 is not in the domain of f(x).

Finally, we can look at f(–1):

f(–1) = ((–1)3 – 3(–1)2 +2(–1))–1

= (–1 – 3(1) – 2)–1

= (–1 – 3 – 2)–1 = (–6)–1 = –1/6.

f(–1) is defined, so –1 belongs to the domain of f(x).

Therefore only III is in the domain of f(x).

The domain of the function is all real numbers except x = –3. When x = –3, f(–3) is undefined.

When x = 0 or x = 3, the function is undefined due to its denominator.

Thus the domain is all real numbers x, such that x is not equal to 0 or 3.

The domain of a function or relation is the set of all possible x-values. Here that is simply a list of the x-coordinates of all of the coordinate pairs. So the domain is {–2, –3, 2, 4}.

The domain of a function includes all of the values of x for which that function is defined. In other words, the domain is all of the real values of x that will produce a real number. Let's look at the domains of each function one at a time.

First, let's examine

In general, when we are examining the domain of a function, we want to find places where we end up with zeros in the denominators or square-roots of negative numbers. For example, in the function , if we let x = 4, then we would be forced to evaluate 1/0, which isn't possible. We can never divide by zero. Thus, this function is not defined over all real values of x. We can eliminate it from the answer choices.

Next, let's look at . Let's set the denominator equal to zero to see if there are any values of x which might give us a zero in the denominator.

Subtract one from both sides.

Take the cube root of both sides.

Thus, if x = –1, then f(x) will be equal to 1/0, which isn't defined, because we can't divide by zero. Therefore, we can eliminate this answer choice.

Now, let's analyze .

We can never take the square root of a negative number. Thus, if , then f(x) won't be defined. For example, if x = 4, then , which would produce an imaginary number. Therefore, this function can't be the correct answer.

Next, let's look at . It will help us to rewrite f(x) in a form using square roots. In general, . As a result, we can rewrite f(x) as follows:

. In this form, we can see that if is negative, then f(x) won't be defined. Thus, if x = –2, for example, we would be forced to find the square root of –8, which produces an imaginary result. So, this function isn't the correct answer either.

By process of elimination the answer must be . The reasons that this function is defined for all values of x is because the denominator can never be zero. Thus, we can pick any value of x from negative to positive infinity, and we will get a real value for f(x).

The answer is

For the square root of a function to be defined on the real numbers, the radicand must be nonnegative. Therefore,

, or .

Any nonnegative number can be the radicand, so has no lower bound. This makes the natural domain

The key here is to factor the denominator, bearing in mind that once we do, we can find the values for which the denominator will be and therefore the values for which the function will not be valid.

We can thus deduce from those three factors that the function will not be valid when .

The domain of a function is all values that you can put in for x without breaking any rules. When first approaching this problem, you must realize that when dealing with a fraction, the denominator can never be 0. Thus, any x value that makes the denominator 0 must be removed from our domain set. Thus,

Since x=3 will make our denominator 0, it must be removed. All other values are permitted, so our answer is

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.

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 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 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 .

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 .