How to find the domain of a function - PSAT Math

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

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

f(x) has no square roots or denominators, so it will always be defined; there are no restrictions on x because any and all values will lead to a real result.

Therefore, the domain is the set of all real numbers.

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}.