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SAT II Math II : Range and Domain

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Example Question #1 :Range And Domain

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

Give the range of.

Possible Answers:

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

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

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

The correct range is not among the other responses.

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

Correct answer:

The correct range is not among the other responses.

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.

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Example Question #2 :Range And Domain

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

Give the domain of.

Possible Answers:

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

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

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

(4,7)

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

Correct answer:

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

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:

x- 7 =0

x = 7

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

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Example Question #3 :Range And Domain

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

Give the domain of.

Possible Answers:

$(-\infty,10]$

$\left [0, 10 \right ]$

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

$\left [-10, \infty)$

$(-\infty, \infty)$

Correct answer:

$(-\infty, \infty)$

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.

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Example Question #4 :Range And Domain

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

Give the range of.

Possible Answers:

[-6, 2]

[-6, -2]

[-2,6]

[2,6]

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

Correct answer:

[-2,6]

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 [-2,6] .

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Example Question #5 :Range And Domain

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

Give the range of.

Possible Answers:

[-2,6]

$\left [ 1,2\right ]$

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

[-2,2]

$\left [ 1,3\right ]$

Correct answer:

$\left [ 1,3\right ]$

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 $\left [ 1,3\right ]$ .

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Example Question #6 :Range And Domain

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

Give the range of.

Possible Answers:

[0, 16]

[0, 4]

[-16,16]

$(-\infty, 4]$

[-4,4]

Correct answer:

[0, 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 ofis [-4,4] .

$f (x) = \sqrt{16-x^{2}}$ assumes its greatest value when $16-x^{2}$ , which is the point on [-4,4] where $x^{2}$ is least - this is at x = 0 .

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

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

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

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

Therefore, the range ofis [0, 4] .

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Example Question #7 :Range And Domain

Define $f(x) = \frac{5}{\sin 4x}$

Give the range of.

Possible Answers:

$\left ( -\frac{5}{4}, \frac{5}{4} \right )$

$(- \infty, -5] \cup [5, \infty)$

$(- \infty, 0) \cup (0, \infty)$

$\left (- \infty, -\frac{5}{4} \right )\cup\left ( \frac{5}{4}, \infty \right )$

( -5 , 5 )

Correct answer:

$(- \infty, -5] \cup [5, \infty)$

Explanation:

$f(x) = \frac{5}{\sin 4x}$ can be rewritten as $f (x) = 5 \csc 4x$ .

For all real values of $\theta$ ,

$\csc \theta\leq -1$ or $\csc \theta\geq 1$ .

Therefore,

$\csc 4x \leq -1$ or $\csc 4x\geq 1$ and

$5 \csc 4x \leq -5$ or $5 \csc 4x\geq 5$ .

The range ofis $(- \infty, -5] \cup [5, \infty)$ .

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Example Question #1 :Properties Of Functions And Graphs

What is the domain of the function f(x)=x^2+3?

Possible Answers:

$(-\infty ,\infty)$

$[-\infty ,\infty]$

$(3,\infty)$

$[3,\infty)$

$(-3,\infty )$

Correct answer:

$(-\infty ,\infty)$

Explanation:

The domain of a function is all the x-values that in that function. The function f(x)=x^2+3 是一个向上的抛物线与一个顶点(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.

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Example Question #9 :Range And Domain

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

Possible Answers:

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

$\textup{All real numbers.}$

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

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

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

Correct answer:

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

Explanation:

Notice this function resembles the parent function $y=\sqrt {x}$ . The value ofmust 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: $x\leq \frac{3}{10}$

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Example Question #10 :Range And Domain

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

Possible Answers:

All real numbers except.

所有的实数。

All real numbers except.

Correct answer:

All real numbers except.

Explanation:

Start by considering the term $\frac{2}{x}$ . $\frac{2}{x}$ will hold for all values of, except when x=0 . Thus, $\frac{2}{x}-3$ must be defined by all values exceptsince the equation is just shifted down by.

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SAT II Math II : Range and Domain

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #1 :Range And Domain

Example Question #2 :Range And Domain

Example Question #3 :Range And Domain

Example Question #4 :Range And Domain

Example Question #5 :Range And Domain

Example Question #6 :Range And Domain

Example Question #7 :Range And Domain

Example Question #1 :Properties Of Functions And Graphs

Example Question #9 :Range And Domain

Example Question #10 :Range And Domain

All SAT II Math II Resources

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