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Algebra Help: Understanding Functions Domain And Range

Review real example questions for Understanding Functions Domain And Range in Algebra.

Question 1 / 10

0 of 10 answered

Let $g(x)=\dfrac{1}{x-5}$ . To keep $g$ a function with one output for each input, $g(x)$ must be defined for the input. For what values of $x$ is $g(x)$ defined? (Answer in words or interval notation.)

All questions

Question 1

All real numbers except $x=5$ (correct answer)
All real numbers except $x=0$
$[5,\infty)$
All real numbers

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For g(x) = 1/(x-5), we need to avoid division by zero, so the denominator x-5 cannot equal 0, which means x cannot equal 5. Choice A is correct because it states 'all real numbers except x=5', which excludes only the problematic value while allowing all others. Choice B incorrectly excludes x=0, but plugging in x=0 gives g(0)=1/(-5)=-1/5, which is perfectly defined. For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'

Question 2

A function $f$ is defined by $f(x)=\sqrt{x-2}$ . Because a function assigns exactly one output to each input in its domain, we must restrict inputs so the output is real. What is the domain of $f$ ? (Answer in interval notation.)

$(-\infty,2]$
$[2,\infty)$ (correct answer)
$(2,\infty)$
All real numbers

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For f(x) = √(x-2), we need the expression under the square root to be non-negative, so x-2 ≥ 0, which means x ≥ 2. Choice B is correct because [2,∞) represents all real numbers greater than or equal to 2, using a square bracket at 2 to show it's included. Choice A would give negative values under the square root, making the output imaginary rather than real. For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'

Question 3

Consider the piecewise function $f(x) = \begin{cases} 2x + 1 & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases}$ . What is $f(-2) + f(2)$ ?

1 (correct answer)
3
7
9

Explanation: Since -2 < 0, use the first piece: f(-2) = 2(-2) + 1 = -4 + 1 = -3. Since 2 ≥ 0, use the second piece: f(2) = 2² = 4. Therefore f(-2) + f(2) = -3 + 4 = 1. Choice B might result from calculation errors. Choice C might come from using wrong pieces. Choice D uses f(2) + f(2) + 1 incorrectly.

Question 4

Function $h$ is defined by the equation $h(x) = \sqrt{x + 3}$ . Which statement about the domain of $h$ is correct?

The domain is all real numbers because square roots can be taken of any number
The domain is $x > -3$ because the expression under the square root must be strictly positive
The domain is $x \geq 3$ because we cannot have negative values inside the square root function
The domain is $x \geq -3$ because the expression under the square root must be non-negative (correct answer)

Explanation: When you encounter a function involving a square root, you need to determine what values make the function undefined. The key principle is that square roots of negative numbers are not defined in the real number system. For $h(x) = \sqrt{x + 3}$ , the expression under the square root is $x + 3$ . This expression must be non-negative (greater than or equal to zero) for the function to produce real outputs. Setting up the inequality: $x + 3 \geq 0$ . Solving for $x$ : $x \geq -3$ . This means the domain includes $x = -3$ and all values greater than $-3$ . Let's check why each wrong answer misses the mark. Choice A incorrectly claims square roots can be taken of any number. While this is true for complex numbers, in Algebra 1 we work with real numbers only, where square roots of negative numbers are undefined. Choice B states the domain is $x > -3$ , incorrectly excluding $x = -3$ . However, when $x = -3$ , we get $\sqrt{-3 + 3} = \sqrt{0} = 0$ , which is perfectly valid. Choice C gives $x \geq 3$ , which stems from confusing the sign inside the square root—the expression is $x + 3$ , not $x - 3$ . Choice D correctly identifies that the domain is $x \geq -3$ because the radicand (expression under the square root) must be non-negative. Study tip: For square root functions, always set the expression inside the radical ≥ 0 and solve. Remember that zero is allowed under square roots, so use "greater than or equal to" in your final domain.

Question 5

If $g(x) = 3x - 2$ and $g(a) = 10$ , what is the value of $g(a + 1)$ ?

11
13 (correct answer)
31
28

Explanation: First find a: if g(a) = 10, then 3a - 2 = 10, so 3a = 12, thus a = 4. Then g(a + 1) = g(5) = 3(5) - 2 = 13. Choice A incorrectly adds 1 to g(a). Choice C uses g(3a) instead of g(a+1). Choice D uses g(a) + 3a incorrectly.

Question 6

Consider the relation $R = \{(2,5), (3,7), (2,9), (4,1)\}$ . If we remove exactly one ordered pair to make this relation a function, what is the range of the resulting function?

$\{2, 3, 4\}$
$\{5, 7, 9, 1\}$
$\{5, 7, 1\}$ or $\{9, 7, 1\}$ (correct answer)
$\{2, 3, 4, 5, 7, 9, 1\}$

Explanation: The relation fails to be a function because input 2 maps to both 5 and 9. We must remove either (2,5) or (2,9). If we remove (2,5), the range is {9,7,1}. If we remove (2,9), the range is {5,7,1}. Choice A gives the domain. Choice B gives the original range before removal. Choice D combines domain and range incorrectly.

Question 7

The height of a toy rocket above the ground is modeled by $H(t)=-2t^2+8t+1$ , where $t$ is time in seconds. The model is only valid from launch until it hits the ground at $t=4$ . What is the domain of $H$ in this context?

$(0,4)$
$( -\infty, \infty)$
$[0,4]$ (correct answer)
$[4, \infty)$

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For the rocket height H(t) = -2t² + 8t + 1, the context limits t from launch at t=0 to when it hits the ground at t=4, so only those times are meaningful. Choice C is correct because [0,4] includes t=0 (launch) and t=4 (landing), capturing the valid interval. Choice A ignores the context and assumes all real t, but negative time or after landing doesn't make sense here—always factor in the real-world scenario! For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'

Question 8

A ball's height (in meters) after $t$ seconds is modeled by a function $H(t)$ , and the model is only intended to be used for $0\le t\le 6$ . What is the domain of $H$ in this context? (Use interval notation.)

$(0,6)$
$[0,6]$ (correct answer)
$[6,\infty)$
$( -\infty,6]$

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). The problem states the model is only intended for 0 ≤ t ≤ 6, meaning we're tracking the ball's height from t = 0 seconds (when thrown) to t = 6 seconds (perhaps when it hits the ground). Choice B is correct because [0,6] uses square brackets to show both endpoints are included—the model works at t = 0 and at t = 6. Choice A uses parentheses, incorrectly excluding the endpoints, but the problem clearly states t can equal 0 and 6. When a problem gives you explicit restrictions like "for 0 ≤ t ≤ 6," that's your domain! Square brackets [ ] mean "include this endpoint," while parentheses ( ) mean "get close but don't include."

Question 9

A function $T(n)$ gives the total number of minutes it takes to download $n$ files. In this context, $n$ must be a whole number (you can download $0$ files, $1$ file, $2$ files, etc.). Which set best represents the domain of $T$ ?

$\{n\mid n\text{ is any real number}\}$
$\{n\mid n\text{ is any integer}\}$
$(0,\infty)$
$\{0,1,2,3,\dots\}$ (correct answer)

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). Since n represents the number of files downloaded, it must be a whole number—you can't download 2.5 files or -3 files! The domain includes 0 (downloading no files), 1, 2, 3, and so on. Choice C is correct because {0,1,2,3,...} represents exactly the non-negative integers (whole numbers), which are the only sensible values for counting files. Choice B includes negative integers like -1, -2, which don't make sense for file counts. When dealing with counting problems, your domain is usually the non-negative integers {0,1,2,3,...}. This set has a special name: the whole numbers or natural numbers including zero!

Question 10

Consider the function $h(x)=x^2-4$ . What is the range of $h$ ? (Answer in interval notation.)

$(-\infty,\infty)$
$(-\infty,-4]$
$[-4,\infty)$ (correct answer)
$(-4,\infty)$

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The range is the set of all possible output values (y-values) the function can produce: for a quadratic that opens upward with vertex at (2, -3), the range is all y-values greater than or equal to -3 because the parabola goes up from that minimum point forever. For h(x) = x² - 4, this is a parabola opening upward (positive coefficient on x²) with vertex at (0, -4), found by noting the minimum occurs when x=0, giving h(0)=0-4=-4. Choice C is correct because [-4,∞) represents all y-values greater than or equal to -4, using a square bracket to show -4 is included as the minimum value. Choice B incorrectly suggests the range goes down to negative infinity, but parabolas opening upward have a minimum point, not a maximum. For finding range: linear functions usually have range = all real numbers. For quadratics, find the vertex first—if it opens up, range is y ≥ (vertex y-value); if it opens down, range is y ≤ (vertex y-value). For square roots, range is usually y ≥ 0. The function type tells you a lot!