Range and Domain

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SAT Subject Test in Math II › Range and Domain

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

All real numbers except .

CORRECT

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 .

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

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

CORRECT

$\textup{There is no range.}$

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

$(-\infty, \infty)$

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

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 domain of the function

$(-\infty ,\infty)$

CORRECT

$[3,\infty)$

$[-\infty ,\infty]$

$(3,\infty)$

$(-3,\infty )$

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? $y=\sqrt{-10x+3}$

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

CORRECT

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

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

$\textup{All real numbers.}$

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

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

Define $f(x)= \frac{x- 4}{x-7}$ .

Give the range of .

The correct range is not among the other responses.

CORRECT

$(-\infty, 7) \cup (7, \infty)$

$(-\infty, 4) \cup (4, \infty)$

$(-\infty, 4) \cup (4, 7) \cup (7, \infty)$

$(-\infty, 4) \cup (7, \infty)$

Explanation

The function can be rewritten as follows:

$f(x)= \frac{x- 4}{x-7}$

$f(x)= \frac{x- 7 + 3}{x-7}$

$f(x)= \frac{x- 7 }{x-7}+ \frac{ 3}{x-7}$

$f(x)=1+ \frac{ 3}{x-7}$

The expression $\frac{ 3}{x-7}$ can assume any value except for 0, so the expression $f(x)=1+ \frac{ 3}{x-7}$ can assume any value except for 1. The range is therefore the set of all real numbers except for 1, or

$(-\infty, 1) \cup (1, \infty)$ .

This choice is not among the responses.

Define $f(x)= \frac{x- 4}{x-7}$ .

Give the domain of .

$(-\infty, 7) \cup (7, \infty)$

CORRECT

$(-\infty, 4) \cup (4, \infty)$

$(-\infty, 4) \cup (4, 7) \cup (7, \infty)$

$(-\infty, 4) \cup (7, \infty)$

Explanation

In a rational function, the domain excludes exactly the value(s) of the variable which make the denominator equal to 0. Set the denominator to find these values:

The domain is the set of all real numbers except 7 - that is, $(-\infty, 7) \cup (7, \infty)$ .

Define $f(x) = 2- 4\cos x$

Give the range of .

CORRECT

$\left ( -\infty, \infty\right )$

Explanation

$-1 \leq \cos x \leq 1$ for any real value of .

Therefore,

$-4(-1) \cdot \geq -4 \cdot \cos x \geq -4 \cdot 1$

$4 \geq -4 \cos x \geq -4$

$4 + 2 \geq -4 \cos x + 2 \geq -4 + 2$

$6 \geq 2-4 \cos x \geq -2$

$-2 \leq f(x) \leq 6$

The range is .

Define $f(x) = \sqrt[3]{100 - x^{2}}$

Give the domain of .

$(-\infty, \infty)$

CORRECT

$\left [-10, 10 \right ]$

$\left [-10, \infty)$

$(-\infty,10]$

$\left [0, 10 \right ]$

Explanation

Every real number has one real cube root, so there are no restrictions on the radicand of a cube root expression. The domain is the set of all real numbers.

Define $f(x) = 2- \cos 4x$

Give the range of .

$\left [ 1,3\right ]$

CORRECT

$\left ( -\infty, \infty\right )$

$\left [ 1,2\right ]$

Explanation

$-1 \leq \cos \theta \leq 1$ for any real value of $\theta$ , so

$-1 \leq \cos 4x \leq 1$

$-1 \cdot \left (-1 \right ) \geq -1 \cdot \cos 4x \geq -1 \cdot 1$

$1 \geq - \cos 4x \geq -1$

$1+ 2 \geq - \cos 4x + 2 \geq -1 + 2$

$3 \geq 2 - \cos 4x \geq 1$

$1 \leq f(x) \leq 3$ ,

making the range $\left [ 1,3\right ]$ .

Define $f (x) = \sqrt{16-x^{2}}$ .

Give the range of .

CORRECT

$(-\infty, 4]$

Explanation

The radicand within a square root symbol must be nonnegative, so

$16 - x^{2} \geq 0$

$16 \geq x^{2}$

$x^{2} \leq 16$

This happens if and only if $-4 \leq x \leq 4$ , so the domain of is .

$f (x) = \sqrt{16-x^{2}}$ assumes its greatest value when $16-x^{2}$ , which is the point on where $x^{2}$ is least - this is at .

$f (0) = \sqrt{16-0^{2}} = \sqrt{16} = 4$

Similarly, $f (x) = \sqrt{16-x^{2}}$ assumes its least value when $16-x^{2}$ , which is the point on where $x^{2}$ is greatest - this is at $x = \pm 4$ .