How to find excluded values - PSAT Math

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Question

If the average (arithmetic mean) of , , and is , what is the average of , , and ?

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Answer

If we can find the sum of $\dpi{100} \small x+2$ , $\dpi{100} \small y-6$ , and 10, we can determine their average. There is not enough information to solve for $\dpi{100} \small x$ or $\dpi{100} \small y$ individually, but we can find their sum, $\dpi{100} \small x+y$ .

Write out the average formula for the original three quantities. Remember, adding together and dividing by the number of quantities gives the average: $\frac{x + y + 9}{3} = 12$

Isolate $\dpi{100} \small x+y$ :

$x + y + 9 = 36$

$x + y = 27$

Write out the average formula for the new three quantities:

$\frac{x + 2 + y - 6 + 10}{3} = ?$

Combine the integers in the numerator:

$\frac{x + y + 6}{3} = ?$

Replace $\dpi{100} \small x+y$ with 27:

$\frac{27+ 6}{3} = \frac{33}{3} = 11$

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Question

Which of the following provides the complete solution set for ?

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Answer

The absolute value will always be positive or 0, therefore all values of z will create a true statement as long as $2-z\neq 0$ . Thus all values except for 2 will work.

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Question

Define the function as follows:

$g(x) = \frac{x+3}{x-2} - \frac{x+2}{x-3}$

Give the domain of .

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Answer

The domain of is restricted by two different denominators, neither of which can be equal to 0, so the excluded values are:

$x-2 = 0 \Rightarrow x = 2$

$x-3 = 0 \Rightarrow x =3$

The correct response is therefore $\left ( -\infty, 2 \right ) \cup (2,3) \cup (3, \infty)$ .

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Question

Given the expression, $\frac{x^{2}-4}{x^2-x-2}$ , which value CANNOT be equal to ?

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Answer

cannot equal a value which would make the denominator equal to 0. In order to figure out what that value is, we must first simplify this fraction, then set each factor of the denominator equal to 0. As follows:

$\frac{x^{2}-4}{x^2-x-2}$

First, simplify by finding the original binomial multiples:

$\frac{(x-2)(x+2)}{(x-2)(x+1)}$

Now, set each factor of the denominator equal to 0

If OR , the denominator will equal 0. is the only choice provided by the answers.

Note: Even though you can cancel out from the numerator and denominator, still cannot be equal to two. The graph would have a hole at and an aymptote at .

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Question

Define the function on the set of real numbers as follows:

$g(x) = \frac{\sqrt{x+ 3}}{x- 4}$

Give the domain of .

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Answer

The domain of is restricted by two things.

First, the expression within the radical in the numerator must be nonnegative. We therefore solve for in the inequality

$x + 3 \geq 0$

$x \geq -3$ ,

or, in interval notation, $\left [-3, \infty \right )$

Second, the expression in the denominator must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded value(s):

We exclude 4 from $\left [-3, \infty \right )$ , so the correct response is

$\left [-3, 4\right ) \cup \left (4, \infty \right )$

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Question

Define the function as follows:

$f (x) = \frac{x^{2}- 1}{3x- 2}$

Give the domain of .

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Answer

The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the polynomial in the denominator, which must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded values:

$x= \frac{2}{3}$

Therefore, the domain of is the set of all real numbers except $\frac{2}{3}$ - that is,

$\left ( - \infty, \frac{2}{3}\right ) \cup \left ( \frac{2}{3}, \infty \right )$

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Question

Define the function on the set of real numbers as follows:

$h (x) =\frac{ \sqrt{x-1}}{x^{2}-3x-4}$

Give the domain of .

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Answer

There are two things restricting the domain of .

One is the radical symbol in the numerator. The expression inside the radical must be nonnegative, so solve the inequality:

$x - 1 \geq 0$

$x \geq 1$ ,

or, in interval notation, $[1, \infty)$

The other is the denominator, which must be equal to 0, so set, and solve for in, the equation:

$x^{2}- 3x - 4 = 0$

$x - 4 = 0 \Rightarrow x = 4$

$x+ 1 = 0 \Rightarrow x = -1$

is already excluded from the domain; we exclude 4, so the domain is

$\left [1, 4 \right ) \cup \left ( 4 ,\infty)$ .

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Question

Define the function on the real numbers as follows:

$f(x) = \frac{x- 7}{\sqrt{10-x}}$

Give the domain of .

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Answer

The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the expression in the denominator, which must be nonzero. Furthermore, the radicand must be nonnegative. Combined, these facts mean that the radicand must be positive, and that the following inequality be solved:

or, equivalently, .

In interval form, this is $(- \infty, 10 )$

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Question

Define the function as follows:

$h(x) = \frac{x-6}{x^{2}- 9}$

Give the domain of .

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Answer

The numerator, being a polynomial, does not restrict our domain. The denominator, however, does restrict it to the values for which it is not equal to 0. We set the denominator equal to 0 to find the excluded values:

$x^{2} - 9 = 0$

$x- 3 = 0 \Rightarrow x = 3$

$x+3 = 0 \Rightarrow x = -3$

The domain, in interval notation, is therefore

$\left ( -\infty, -3 \right ) \cup (-3,3) \cup (3, \infty)$ .

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Question

Define the functions and as follows:

$f(x) = \frac{5}{x-4}$ and $g(x) = \frac{2x-8}{x-7}$

Give the domain of the function .

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Answer

The domain of the product of two functions is the intersection of the domains of the individual functions.

The domain of is restricted to all values of that yield a nonzero denominator. Since this means that

$x- 4 \ne 0$ ,

then, subsequently,

$x \ne 4$ ,

so the domain is the set of all real numbers except 4.

Similarly, the domain of is restricted to all values of that yield a nonzero denominator. This set is found to be the set of all real numbers except 7.

The intersection of these sets is the set of all real numbers except 4 and 7, or

$\left (-\infty, 4 \right ) \cup \left ( 4,7\right ) \cup \left ( 7, \infty \right )$ .

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Question

Define the function as follows:

$f(x) = \frac{x}{4- \frac{2}{x-2}}$

Give the domain of .

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Answer

The definition of has two denominators of two fractions (one within the other), so we must exclude the values of that make either denominator equal to zero.

One denominator is . Since it cannot be zero, we have

$x- 2 \ne 0$ ,

and, subsequently,

$x \ne 2$ .

The value 2 is excluded from the domain.

The other denominator is $4- \frac{2}{x-2}$ . If it is equal to zero, then

$4- \frac{2}{x-2} = 0$

$4= \frac{2}{x-2}$

$4\cdot \frac{x-2}{4} = \frac{2}{x-2} \cdot \frac{x-2}{4}$

$x-2 = \frac{1}{2}$

$x=2 \frac{1}{2}$

Therefore, this value is also excluded from the domain.

The correct domain is the set $\left ( -\infty , 2 \right ) \cup \left ( 2, 2\frac{1}{2}\right ) \cup \left ( 2\frac{1}{2}, \infty \right )$ .

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Question

Define the function as follows:

$f(x)=\frac{x-4}{1- \frac{3}{\sqrt x}}$

Give the domain of .

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Answer

The numerator , being a polynomial, does not restrict the domain.

appears as a radicand of a square root in the definition of , so must be restricted to nonnegative numbers. Also, $\sqrt{x}$ is a denominator, which means 0 must also be excluded. Therefore, we are restricted so far to positive numbers.

There is one more denominator, which is $1- \frac{3}{\sqrt x}$ ; it must be nonzero. We set this equal to zero to determine any additional value(s) that must be excluded:

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

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

$1 \cdot \sqrt{x}= \frac{3}{\sqrt x} \cdot \sqrt{x}$

$\sqrt{x} = 3$

Therefore, the domain is the set of all positive numbers except 9 - or

$\left ( 0, 9\right ) \cup \left ( 9, \infty \right )$ .

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Question

Which of the following are answers to the equation below?

$\frac{x^2-4}{x^2+5x+6}=0$

I. -3

II. -2

III. 2

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Answer

Given a fractional algebraic equation with variables in the numerator and denominator of one side and the other side equal to zero, we rely on a simple concept. Zero divided by anything equals zero. That means we can focus in on what values make the numerator (the top part of the fraction) zero, or in other words,

The expression is a difference of squares that can be factored as

Solving this for gives either or . That means either of these values will make our numerator equal zero. We might be tempted to conclude that both are valid answers. However, our statement earlier that zero divided by anything is zero has one caveat. We can never divide by zero itself. That means that any values that make our denominator zero must be rejected. Therefore we must also look at the denominator.

The left side factors as follows

This means that if is or , we end up dividing by zero. That means that cannot be a valid solution, leaving as the only valid answer. Therefore only #3 is correct.

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Question

Find the excluded values of the following algebraic fraction

$\frac{2x^2-5x-12}{x^2-11x+28}$

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Answer

To find the excluded values of a algebraic fraction you need to find when the denominator is zero. To find when the denominator is zero you need to factor it. This denominator factors into

so this is zero when x=4,7 so our answer is

$x\neq4,7$

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Question

If the average (arithmetic mean) of , , and is , what is the average of , , and ?

Tap to reveal answer

Answer

If we can find the sum of $\dpi{100} \small x+2$ , $\dpi{100} \small y-6$ , and 10, we can determine their average. There is not enough information to solve for $\dpi{100} \small x$ or $\dpi{100} \small y$ individually, but we can find their sum, $\dpi{100} \small x+y$ .

Write out the average formula for the original three quantities. Remember, adding together and dividing by the number of quantities gives the average: $\frac{x + y + 9}{3} = 12$

Isolate $\dpi{100} \small x+y$ :

$x + y + 9 = 36$

$x + y = 27$

Write out the average formula for the new three quantities:

$\frac{x + 2 + y - 6 + 10}{3} = ?$

Combine the integers in the numerator:

$\frac{x + y + 6}{3} = ?$

Replace $\dpi{100} \small x+y$ with 27:

$\frac{27+ 6}{3} = \frac{33}{3} = 11$

← Didn't Know|Knew It →

Question

Which of the following provides the complete solution set for ?

Tap to reveal answer

Answer

The absolute value will always be positive or 0, therefore all values of z will create a true statement as long as $2-z\neq 0$ . Thus all values except for 2 will work.

← Didn't Know|Knew It →

Question

Given the expression, $\frac{x^{2}-4}{x^2-x-2}$ , which value CANNOT be equal to ?

Tap to reveal answer

Answer

cannot equal a value which would make the denominator equal to 0. In order to figure out what that value is, we must first simplify this fraction, then set each factor of the denominator equal to 0. As follows:

$\frac{x^{2}-4}{x^2-x-2}$

First, simplify by finding the original binomial multiples:

$\frac{(x-2)(x+2)}{(x-2)(x+1)}$

Now, set each factor of the denominator equal to 0

If OR , the denominator will equal 0. is the only choice provided by the answers.

Note: Even though you can cancel out from the numerator and denominator, still cannot be equal to two. The graph would have a hole at and an aymptote at .

← Didn't Know|Knew It →

Question

Define the function on the set of real numbers as follows:

$g(x) = \frac{\sqrt{x+ 3}}{x- 4}$

Give the domain of .

Tap to reveal answer

Answer

The domain of is restricted by two things.

First, the expression within the radical in the numerator must be nonnegative. We therefore solve for in the inequality

$x + 3 \geq 0$

$x \geq -3$ ,

or, in interval notation, $\left [-3, \infty \right )$

Second, the expression in the denominator must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded value(s):

We exclude 4 from $\left [-3, \infty \right )$ , so the correct response is

$\left [-3, 4\right ) \cup \left (4, \infty \right )$

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Question

Define the function as follows:

$f (x) = \frac{x^{2}- 1}{3x- 2}$

Give the domain of .

Tap to reveal answer

Answer

The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the polynomial in the denominator, which must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded values:

$x= \frac{2}{3}$

Therefore, the domain of is the set of all real numbers except $\frac{2}{3}$ - that is,

$\left ( - \infty, \frac{2}{3}\right ) \cup \left ( \frac{2}{3}, \infty \right )$

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Question

Define the function on the set of real numbers as follows:

$h (x) =\frac{ \sqrt{x-1}}{x^{2}-3x-4}$

Give the domain of .

Tap to reveal answer

Answer

There are two things restricting the domain of .

One is the radical symbol in the numerator. The expression inside the radical must be nonnegative, so solve the inequality:

$x - 1 \geq 0$

$x \geq 1$ ,

or, in interval notation, $[1, \infty)$

The other is the denominator, which must be equal to 0, so set, and solve for in, the equation:

$x^{2}- 3x - 4 = 0$

$x - 4 = 0 \Rightarrow x = 4$

$x+ 1 = 0 \Rightarrow x = -1$

is already excluded from the domain; we exclude 4, so the domain is

$\left [1, 4 \right ) \cup \left ( 4 ,\infty)$ .

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