SAT Math - Range and Domain

Which of the following is NOT a function?

$x=\left | y\right |$

$y = x^{3} -x +5$

$y = \sqrt{x-9} +3$

$y = -\sqrt{x^{2}-16}$

Explanation

A function has to pass the vertical line test, which means that a vertical line can only cross the function one time. To put it another way, for any given value of , there can only be one value of . For the function $x=\left | y\right |$ , there is one value for two possible values. For instance, if , then . But if , as well. This function fails the vertical line test. The other functions listed are a line,, the top half of a right facing parabola, $y = \sqrt{x-9} +3$ , a cubic equation, $y = x^{3} -x +5$ , and a semicircle, $y = -\sqrt{x^{2}-16}$ . These will all pass the vertical line test.

What is the domain of the following function? Please use interval notation.

$y=\left | -x\right |$

$(-\infty ,\infty )$

$(0,\infty )$

$(-\infty ,0)$

Explanation

A basic knowledge of absolute value and its functions is valuable for this problem. However, if you do not know what the typical shape of an absoluate value function looks like, one can always plug in values and plot points.

Upon doing so, we learn that the -values (domain) are not restricted on either end of the function, creating a domain of negative infinity to postive infinity.

If we plug in -100000 for , we get 100000 for .

If we plug in 100000 for , we get 100000 for .

Additionally, if we plug in any value for , we will see that we always get a real, defined value for .

**Extra Note: Due to the absolute value notation, the negative (-) next to the is not important, in that it will always be made positive by the absolute value, making this function the same as $y=\left | x\right |$ . If the negative (-) was outside of the absolute value, this would flip the function, making all corresponding -values negative. However, this knowledge is most important for range, rather than domain.

Give the domain of the function below.

$f(x) = \frac{1}{\sqrt{x^{2}+1}}$

$\mathbb{R}$

$\left { x | x eq 1\right }$

$\left { x | x eq -1\right }$

$\left { x | x< -1; or ; x> 1\right }$

$\left {x| -1<x<1 \right }$

Explanation

The domain is the set of possible value for the variable. We can find the impossible values of by setting the denominator of the fractional function equal to zero, as this would yield an impossible equation.

$\sqrt{x^2+1}=0$

Now we can solve for .

There is no real value of that will fit this equation; any real value squared will be a positive number.

The radicand is always positive, and $f(x) = \frac{1}{\sqrt{x^{2}+1}}$ is defined for all real values of . This makes the domain of the set of all real numbers.

What is the range of the equation $y=\frac{1}{5}$ ?

$[\frac{1}{5}]$

$\textup{There is no range.}$

$[\frac{1}{5},\infty]$

$(-\infty, \infty)$

$[-\frac{1}{5}, \frac{1}{5}]$

Explanation

The equation given represents a horizontal line. This means that every y-value on the domain is equal to $\frac{1}{5}$ .

The answer is: $[\frac{1}{5}]$

What is the range of the function $f(x)=\frac{2}{x}-3$ ?

All real numbers except .

All real numbers.

All real numbers except .

Explanation

Start by considering the term $\frac{2}{x}$ . $\frac{2}{x}$ will hold for all values of , except when . Thus, $\frac{2}{x}-3$ must be defined by all values except since the equation is just shifted down by .

If , which of these values of is NOT in the domain of this equation?

$y = x^2 + \frac{7}{x}$

$\frac{1}{2}$

$- \sqrt{7}$

Explanation

Using as the input () value for this equation generates an output () value that contradicts the stated condition of .

Therefore is not a valid value for and not in the equation's domain:

$y = (-1)^2 + \frac{7}{(-1)} \rightarrow y = 1 - 7 \rightarrow y= -6$

What is the domain of the function

$(-\infty ,\infty)$

$[3,\infty)$

$[-\infty ,\infty]$

$(3,\infty)$

$(-3,\infty )$

Explanation

The domain of a function is all the x-values that in that function. The function is a upward facing parabola with a vertex as (0,3). The parabola keeps getting wider and is not bounded by any x-values so it will continue forever. Parenthesis are used because infinity is not a definable number and so it can not be included.

What is the domain of the function? $y=\sqrt{-10x+3}$

$x\leq \frac{3}{10}$

$x\geq \frac{3}{10}$

$-\frac{3}{10}\leq x\leq \frac{3}{10}$

$\textup{All real numbers.}$

$x\leq- \frac{3}{10}$

Explanation

Notice this function resembles the parent function $y=\sqrt {x}$ . The value of must be zero or greater.

Set up an inequality to determine the domain of .

$-10x+3\geq 0$

Subtract three from both sides.

$-10x+3-3\geq 0-3$

$-10x\geq -3$

Divide by negative ten on both sides. The sign will switch.

$\frac{-10x}{-10}\geq \frac{-3}{-10}$

The domain is: $x\leq \frac{3}{10}$

is a sine curve. What are the domain and range of this function?

Domain: All real numbers

Range: $-1\leq f(x)\leq1$

Domain: All real numbers

Range:

Explanation

The domain includes the values that go into a function (the x-values) and the range are the values that come out (the or y-values). A sine curve represent a wave the repeats at a regular frequency. Based upon this graph, the maximum is equal to 1, while the minimum is equal to –1. The x-values span all real numbers, as there is no limit to the input fo a sine function. The domain of the function is all real numbers and the range is $-1\leq f(x)\leq1$ .