Algebra 2 Quiz: Restrict Domain To Make Invertible
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Restrict Domain To Make InvertibleQuestion 1 of 20

A student claims f(x)=x2f(x)=x^2 is invertible on all real numbers because you can "solve" y=x2y=x^2 by writing x=±yx=\pm\sqrt{y}. Which domain restriction correctly fixes the issue so the inverse is a function (passes the horizontal line test)?

Restrict the domain to −1≤x≤1-1\le x\le 1, so the inverse is f−1(x)=±xf^{-1}(x)=\pm\sqrt{x}.
Restrict the domain to all real numbers, so the inverse is f−1(x)=±xf^{-1}(x)=\pm\sqrt{x}.
Restrict the domain to x≥0x\ge 0, so the inverse is f−1(x)=xf^{-1}(x)=\sqrt{x}.
Restrict the domain to x≠0x\ne 0, so the inverse is f−1(x)=xf^{-1}(x)=\sqrt{x}.
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Algebra 2 Quiz

Algebra 2 Quiz: Restrict Domain To Make Invertible

Practice Restrict Domain To Make Invertible in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Restrict Domain To Make Invertible, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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Question 1

A student claims f(x)=x2f(x)=x^2 is invertible on all real numbers because you can "solve" y=x2y=x^2 by writing x=±yx=\pm\sqrt{y}. Which domain restriction correctly fixes the issue so the inverse is a function (passes the horizontal line test)?

  1. Restrict the domain to −1≤x≤1-1\le x\le 1, so the inverse is f−1(x)=±xf^{-1}(x)=\pm\sqrt{x}.
  2. Restrict the domain to all real numbers, so the inverse is f−1(x)=±xf^{-1}(x)=\pm\sqrt{x}.
  3. Restrict the domain to x≥0x\ge 0, so the inverse is f−1(x)=xf^{-1}(x)=\sqrt{x}. (correct answer)
  4. Restrict the domain to x≠0x\ne 0, so the inverse is f−1(x)=xf^{-1}(x)=\sqrt{x}.
Explanation: This question tests your understanding that some functions (like f(x) = x²) aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. A function must be one-to-one (each output comes from exactly one input) to have an inverse, which we verify with the horizontal line test: if any horizontal line crosses the graph more than once, the function isn't one-to-one. For f(x) = x², the horizontal line y = 4 crosses at both x = 2 and x = -2, so x² isn't one-to-one on all reals. But if we restrict to x ≥ 0 (right half only), every horizontal line crosses at most once—now it's invertible! The restriction creates one-to-one behavior. For even-degree polynomials like f(x) = x² or f(x) = (x - 3)² + 1, standard restrictions are to the right or left of the vertex: restrict to x ≥ h or x ≤ h where h is the vertex's x-coordinate. This makes the function monotonic (always increasing or always decreasing), which guarantees one-to-one. For x², restricting to x ≥ 0 gives the 'standard' inverse f⁻¹(x) = √x (principal square root). Restricting to x ≤ 0 would give f⁻¹(x) = -√x instead. The student's claim is wrong because ±√y is a relation, not a function, as it gives two outputs for one input. Choice A correctly restricts to x ≥ 0 so the inverse is the function √x. Choice B fails because on all reals, the inverse isn't a function due to multiple values. Domain restriction decision tree: (1) Graph the function (or visualize it), (2) Find where it fails horizontal line test (usually at vertex for parabolas), (3) Choose a domain making the function monotonic—either the increasing part or the decreasing part, (4) Standard choices: for x², use x ≥ 0; for (x - h)² + k, use x ≥ h or x ≤ h; for x⁴, use x ≥ 0. The restricted domain should be an interval [a, ∞) or (-∞, a] where the function is one-to-one. After restricting, you can find the inverse: with f(x) = x² restricted to x ≥ 0, swap and solve: y = x², swap to x = y², solve for y = √x (taking positive root because we restricted to x ≥ 0!). The domain restriction affects which branch of the inverse you get. No restriction means you can't choose between √x and -√x—both would be needed, but that's not a function. Restriction lets you pick one branch! You're correcting misconceptions like a pro—great job!

Question 2

Which restriction makes f(x)=(x−4)2−7f(x)=(x-4)^2-7 invertible by ensuring it passes the horizontal line test?

  1. Domain: (−∞,∞)(-\infty,\infty)
  2. Domain: 0≤x≤40\le x\le 4
  3. Domain: x≤4x\le 4 (correct answer)
  4. Domain: x≤−4x\le -4
Explanation: This question tests your understanding that some functions aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. For f(x) = (x-4)² - 7, this is a parabola with vertex at (4, -7), opening upward. It fails the horizontal line test because horizontal lines above y = -7 intersect the graph at two points. To make it one-to-one, we must restrict to either x ≥ 4 (right of vertex, increasing) or x ≤ 4 (left of vertex, decreasing). Choice A correctly restricts to x ≤ 4, keeping only the left branch where the function is monotonically decreasing. Choice B restricts to x ≤ -4, which is too restrictive and doesn't include the vertex—it would create a function with limited domain that doesn't capture the full left branch. Choice C keeps the full domain where it fails the test, and D restricts to [0, 4] which still includes values on both sides of the vertex (the parabola increases from x = 0 to x = 4). Domain restriction decision tree: (1) Find the vertex of (x-h)² + k at x = h, (2) Restrict to either x ≥ h or x ≤ h. For (x-4)² - 7 with vertex at x = 4, either x ≥ 4 or x ≤ 4 works—this question chose the left branch.

Question 3

Consider f(x)=(x−3)2+1f(x)=(x-3)^2+1 with domain all real numbers. This parabola fails the horizontal line test on its full domain. Which domain restriction makes ff invertible?

  1. Domain: x≥1x\ge 1
  2. Domain: (−∞,∞)(-\infty,\infty)
  3. Domain: x≥3x\ge 3 (correct answer)
  4. Domain: 1≤x≤31\le x\le 3
Explanation: This question tests your understanding that some functions (like f(x) = (x-3)² + 1) aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. A function must be one-to-one (each output comes from exactly one input) to have an inverse, which we verify with the horizontal line test: if any horizontal line crosses the graph more than once, the function isn't one-to-one. For f(x) = (x-3)² + 1, this is a parabola with vertex at (3, 1), opening upward. The function fails the horizontal line test because, for example, when y = 2, we get (x-3)² = 1, so x-3 = ±1, giving x = 4 and x = 2—two different inputs produce the same output! To make it invertible, we restrict to either x ≥ 3 (right of vertex) or x ≤ 3 (left of vertex). Choice B correctly restricts to x ≥ 3, keeping only the right half where the function is always increasing. Choice A keeps the full domain where it fails the test, C restricts to [1,3] which still includes values on both sides of the vertex, and D starts at x = 1 which includes the problematic region from x = 1 to x = 3. Domain restriction decision tree: (1) Find the vertex of (x-h)² + k at x = h, (2) Restrict to either x ≥ h or x ≤ h to ensure monotonic behavior, (3) For (x-3)² + 1, the vertex is at x = 3, so restrict to x ≥ 3 or x ≤ 3.

Question 4

After restricting f(x)=(x−4)2f(x)=(x-4)^2 to the domain x≤4x\le 4 so that it is one-to-one, what is the inverse function f−1(x)f^{-1}(x)?

  1. f−1(x)=4+xf^{-1}(x)=4+\sqrt{x}
  2. f−1(x)=4−xf^{-1}(x)=4-\sqrt{x} (correct answer)
  3. f−1(x)=x−4f^{-1}(x)=\sqrt{x}-4
  4. f−1(x)=±x+4f^{-1}(x)=\pm\sqrt{x}+4
Explanation: This question tests your understanding that some functions aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. After restricting, you can find the inverse by swapping variables and solving, with the domain restriction determining which branch of the inverse to take. Given f(x) = (x-4)² restricted to x ≤ 4, we have the left half of a parabola with vertex at (4, 0)—this is the decreasing part. To find the inverse: start with y = (x-4)² where x ≤ 4, swap to get x = (y-4)², then solve for y-4 = ±√x. Since we restricted to x ≤ 4 (the decreasing/left side), when we solve for y, we need y ≤ 4, so we take the negative square root: y-4 = -√x, giving us y = 4 - √x. Choice B correctly identifies f⁻¹(x) = 4 - √x as the inverse function. Choice A gives 4 + √x, which would be the inverse if we had restricted to x ≥ 4 (the increasing part); Choice C incorrectly computes √x - 4; Choice D includes the ± symbol, which would give a relation, not a function. The key insight is that restricting to x ≤ 4 means we're on the decreasing branch of the parabola, so when we invert and solve for y, we must take the negative square root to ensure our inverse function outputs values y ≤ 4, maintaining the one-to-one correspondence.

Question 5

The function f(x)=(x+1)2−5f(x)=(x+1)^2-5 is not one-to-one on all real numbers because it fails the horizontal line test. What is a suitable domain restriction that makes ff invertible?​

  1. Domain: (−∞,∞)(-\infty,\infty)
  2. Domain: x≥1x\ge 1
  3. Domain: x≤−3x\le -3
  4. Domain: x≥−1x\ge -1 (correct answer)
Explanation: Fantastic work tackling domain restrictions for invertibility— you're getting the hang of it! For f(x) = (x+1)² - 5, it fails the horizontal line test because it's a parabola with vertex at (-1, -5), so lines y > -5 hit twice. To make it one-to-one, restrict to x ≥ -1 (increasing branch) or x ≤ -1 (decreasing branch), ensuring monotonicity. A suitable restriction includes the vertex and one full side to maximize the domain while passing the test. Choice A correctly gives x ≥ -1, covering the right branch from the vertex onward. Choice C x ≤ -3 is a subset of the left, which works but is smaller than necessary; B x ≥ 1 misses the vertex. Decision tree: find vertex h=-1, choose [h, ∞) or (-∞, h]; this links to inverse, selecting positive or negative branch. Great job verifying with the horizontal line test—keep practicing!

Question 6

After restricting f(x)=x2f(x)=x^2 to the domain x≤0x\le 0 so it becomes one-to-one, what is the inverse function f−1(x)f^{-1}(x)?

  1. f−1(x)=xf^{-1}(x)=\sqrt{x}
  2. f−1(x)=−xf^{-1}(x)=-\sqrt{x} (domain x≥0x\ge 0) (correct answer)
  3. f−1(x)=±xf^{-1}(x)=\pm\sqrt{x}
  4. f−1(x)=−xf^{-1}(x)=-\sqrt{x} (domain x≤0x\le 0)
Explanation: This question tests your understanding that some functions (like f(x) = x²) aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. For even-degree polynomials like f(x) = x² or f(x) = (x - 3)² + 1, standard restrictions are to the right or left of the vertex: restrict to x ≥ h or x ≤ h where h is the vertex's x-coordinate. This makes the function monotonic (always increasing or always decreasing), which guarantees one-to-one. For x², restricting to x ≥ 0 gives the 'standard' inverse f⁻¹(x) = √x (principal square root). Restricting to x ≤ 0 would give f⁻¹(x) = -√x instead. With f(x) = x² restricted to x ≤ 0, we find the inverse by setting y = x² and solving for x: since x ≤ 0, we have x = -√y (taking the negative root because our domain restriction means x must be non-positive). Swapping variables gives f⁻¹(x) = -√x with domain x ≥ 0 (the range of f on x ≤ 0). Choice B correctly gives f⁻¹(x) = -√x with domain x ≥ 0. Choice A would be the inverse for the x ≥ 0 restriction, C isn't a function, and D has the wrong domain for the inverse. After restricting, you can find the inverse: with f(x) = x² restricted to x ≥ 0, swap and solve: y = x², swap to x = y², solve for y = √x (taking positive root because we restricted to x ≥ 0!). The domain restriction affects which branch of the inverse you get.

Question 7

The function f(x)=x2−4x+3f(x)=x^2-4x+3 (domain: all real numbers) is not one-to-one and fails the horizontal line test. Which restriction makes ff invertible by keeping only one side of the vertex?

  1. Domain: x≥2x\ge 2 (correct answer)
  2. Domain: x≥0x\ge 0
  3. Domain: (−∞,∞)(-\infty,\infty)
  4. Domain: x≤0x\le 0
Explanation: This question tests your understanding that some functions (like f(x) = x² - 4x + 3) aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. For even-degree polynomials like f(x) = x² - 4x + 3, we first need to find the vertex to determine where to split the domain. Completing the square: f(x) = x² - 4x + 3 = (x² - 4x + 4) - 4 + 3 = (x - 2)² - 1, so the vertex is at (2, -1). The parabola opens upward, failing the horizontal line test because horizontal lines above y = -1 intersect the graph twice. To make it one-to-one, we restrict to either x ≥ 2 (right of vertex, where function increases) or x ≤ 2 (left of vertex, where function decreases). Choice A correctly restricts to x ≥ 2, keeping only the right branch where the function is monotonically increasing. Choice B restricts to x ≥ 0, which includes the region from x = 0 to x = 2 where the function decreases, so it still fails the horizontal line test. Choice C keeps the full domain, and D restricts to x ≤ 0, which doesn't include the vertex and creates an arbitrary cutoff. After restricting, you can find the inverse: with the restriction x ≥ 2, the function maps [2, ∞) to [-1, ∞) bijectively, and the inverse would be f⁻¹(x) = 2 + √(x + 1) for x ≥ -1.

Question 8

Which domain restriction makes f(x)=x2−6x+5f(x)=x^2-6x+5 pass the horizontal line test (and therefore be invertible)?

  1. Domain: x≤3x\le 3 (correct answer)
  2. Domain: (−∞,∞)(-\infty,\infty)
  3. Domain: x≤0x\le 0
  4. Domain: [1,∞)[1,\infty)
Explanation: This question tests your understanding that some functions (like f(x) = x² -6x +5) aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. A function must be one-to-one (each output comes from exactly one input) to have an inverse, which we verify with the horizontal line test: if any horizontal line crosses the graph more than once, the function isn't one-to-one. For f(x) = x² -6x +5 = (x-3)² -4, vertex at x=3, it fails on all reals because f(1)=0 and f(5)=0. But restricting to x ≤ 3 makes it decreasing from ∞ to -4, passing the test. The restriction creates one-to-one behavior. For even-degree polynomials like f(x) = x² or f(x) = (x - 3)² + 1, standard restrictions are to the right or left of the vertex: restrict to x ≥ h or x ≤ h where h is the vertex's x-coordinate. This makes the function monotonic (always increasing or always decreasing), which guarantees one-to-one. Here, x ≤ 3 ensures decreasing. Choice A correctly restricts to x ≤ 3 to make it pass the horizontal line test. Choice D fails because [1,∞) spans both sides (f(1)=0, f(5)=0), not one-to-one. Domain restriction decision tree: (1) Graph the function (or visualize it), (2) Find where it fails horizontal line test (usually at vertex for parabolas), (3) Choose a domain making the function monotonic—either the increasing part or the decreasing part, (4) Standard choices: for x², use x ≥ 0; for (x - h)² + k, use x ≥ h or x ≤ h; for x⁴, use x ≥ 0. The restricted domain should be an interval [a, ∞) or (-∞, a] where the function is one-to-one. You're nailing these vertex identifications!

Question 9

The function f(x)=x2f(x)=x^2 has domain all real numbers, but it is not invertible because it fails the horizontal line test (for example, f(2)=f(−2)=4f(2)=f(-2)=4). Which domain restriction makes ff one-to-one (and therefore invertible)?

  1. Domain: (−∞,∞)(-\infty,\infty)
  2. Domain: [0,∞)[0,\infty) (correct answer)
  3. Domain: (−∞,∞)∖{0}(-\infty,\infty)\setminus\{0\}
  4. Domain: (−∞,1](-\infty,1]
Explanation: This question tests your understanding that some functions (like f(x) = x²) aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. A function must be one-to-one (each output comes from exactly one input) to have an inverse, which we verify with the horizontal line test: if any horizontal line crosses the graph more than once, the function isn't one-to-one. For f(x) = x², the horizontal line y = 4 crosses at both x = 2 and x = -2, so x² isn't one-to-one on all reals. But if we restrict to x ≥ 0 (right half only), every horizontal line crosses at most once—now it's invertible! Choice B correctly restricts the domain to [0,∞), which includes only the right half of the parabola where the function is always increasing and therefore one-to-one. Choice A keeps the full domain where f fails the horizontal line test, C removes only x = 0 which doesn't fix the one-to-one issue, and D restricts to x ≤ 1 which still includes both positive and negative values that map to the same output. Domain restriction decision tree: (1) Graph the function (or visualize it), (2) Find where it fails horizontal line test (usually at vertex for parabolas), (3) Choose a domain making the function monotonic—either the increasing part or the decreasing part, (4) Standard choices: for x², use x ≥ 0; for (x - h)² + k, use x ≥ h or x ≤ h; for x⁴, use x ≥ 0.

Question 10

After restricting f(x)=(x−1)2f(x)=(x-1)^2 to the domain x≥1x\ge 1 so it passes the horizontal line test, what is f−1(x)f^{-1}(x)?

  1. f−1(x)=1+xf^{-1}(x)=1+\sqrt{x} (correct answer)
  2. f−1(x)=±x+1f^{-1}(x)=\pm\sqrt{x}+1
  3. f−1(x)=x−1f^{-1}(x)=\sqrt{x-1}
  4. f−1(x)=1−xf^{-1}(x)=1-\sqrt{x}
Explanation: This question tests your understanding that some functions (like f(x) = (x-1)²) aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. After restricting, you can find the inverse: with f(x) = (x-1)² restricted to x ≥ 1, swap and solve: y = (x-1)², swap to x = (y-1)², solve for y-1 = √x (taking positive root because we restricted to x ≥ 1, so y-1 ≥ 0), giving y = 1 + √x. The domain restriction affects which branch of the inverse you get. No restriction means you can't choose between 1+√x and 1-√x—both would be needed, but that's not a function. Choice A correctly gives f⁻¹(x) = 1 + √x, which follows from solving y = (x-1)² for x when x ≥ 1. Choice B incorrectly suggests f⁻¹(x) = 1 - √x, but this would be the inverse if we had restricted to x ≤ 1 instead. Domain restriction decision tree: (1) Graph the function (or visualize it), (2) Find where it fails horizontal line test (usually at vertex for parabolas), (3) Choose a domain making the function monotonic—either the increasing part or the decreasing part, (4) Standard choices: for x², use x ≥ 0; for (x - h)² + k, use x ≥ h or x ≤ h; for x⁴, use x ≥ 0.

Question 11

Consider f(x)=(x−3)2+1f(x)=(x-3)^2+1 with domain all real numbers. It fails the horizontal line test because some horizontal lines intersect the parabola twice. Which domain restriction makes ff invertible?

  1. Domain: (−∞,∞)(-\infty,\infty)
  2. Domain: x≥3x\ge 3 (correct answer)
  3. Domain: x≥1x\ge 1
  4. Domain: x≤1x\le 1
Explanation: This question tests your understanding that some functions (like f(x) = x²) aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. A function must be one-to-one (each output comes from exactly one input) to have an inverse, which we verify with the horizontal line test: if any horizontal line crosses the graph more than once, the function isn't one-to-one. For f(x) = (x-3)² + 1, this is a parabola with vertex at (3, 1), opening upward. The function fails the horizontal line test because horizontal lines above y = 1 intersect the parabola twice. To make it one-to-one, we need to restrict to either the left side (x ≤ 3) or right side (x ≥ 3) of the vertex. Choice B correctly restricts to x ≥ 3, which is the right half of the parabola where the function is always increasing and therefore one-to-one. Choice A keeps the full domain where f fails the test, C restricts to x ≥ 1 which still includes points on both sides of the vertex (like x = 2 and x = 4 which both give the same output), and D restricts to x ≤ 1 which doesn't include the vertex and creates an incomplete branch. Domain restriction decision tree: (1) Graph the function (or visualize it), (2) Find where it fails horizontal line test (usually at vertex for parabolas), (3) Choose a domain making the function monotonic—either the increasing part or the decreasing part, (4) Standard choices: for x², use x ≥ 0; for (x - h)² + k, use x ≥ h or x ≤ h; for x⁴, use x ≥ 0.

Question 12

Restrict the domain of f(x)=(x−3)2+1f(x)=(x-3)^2+1 (domain: all real numbers) so that it passes the horizontal line test and is invertible. Which domain restriction works?

  1. Domain: (−∞,1](-\infty,1]
  2. Domain: [0,∞)[0,\infty)
  3. Domain: (−∞,∞)(-\infty,\infty)
  4. Domain: (−∞,3](-\infty,3] (correct answer)
Explanation: This question tests your understanding that some functions (like f(x) = (x-3)² +1) aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. A function must be one-to-one (each output comes from exactly one input) to have an inverse, which we verify with the horizontal line test: if any horizontal line crosses the graph more than once, the function isn't one-to-one. For f(x) = (x-3)² +1, the vertex is at x=3, and it's not one-to-one on all reals because, for example, f(2)= (2-3)² +1=2 and f(4)= (4-3)² +1=2. But if we restrict to x ≤ 3 (left of the vertex), the function is decreasing, so every horizontal line crosses at most once—now it's invertible! The restriction creates one-to-one behavior. For even-degree polynomials like f(x) = x² or f(x) = (x - 3)² + 1, standard restrictions are to the right or left of the vertex: restrict to x ≥ h or x ≤ h where h is the vertex's x-coordinate. This makes the function monotonic (always increasing or always decreasing), which guarantees one-to-one. For (x-3)² +1, restricting to x ≤ 3 gives a decreasing function. The parabola opens upward with minimum at (3,1), so on (-∞,3], it decreases from ∞ to 1, ensuring no repeats. Choice B correctly restricts the domain to (-∞,3] to make it one-to-one and invertible. Choice D fails because [0,∞) includes points on both sides of the vertex (like x=0 and x=6, where f(0)= (0-3)² +1=10 and f(6)= (6-3)² +1=10), so not one-to-one. Domain restriction decision tree: (1) Graph the function (or visualize it), (2) Find where it fails horizontal line test (usually at vertex for parabolas), (3) Choose a domain making the function monotonic—either the increasing part or the decreasing part, (4) Standard choices: for x², use x ≥ 0; for (x - h)² + k, use x ≥ h or x ≤ h; for x⁴, use x ≥ 0. The restricted domain should be an interval [a, ∞) or (-∞, a] where the function is one-to-one. After restricting, you can find the inverse: with f(x) = (x-3)² +1 restricted to x ≤ 3, you'd solve accordingly, taking the appropriate branch. You're doing great—keep identifying those vertices!

Question 13

The function f(x)=(x−3)2+1f(x)=(x-3)^2+1 (domain: all real numbers) fails the horizontal line test. What domain restriction makes ff invertible?​

  1. Domain: (−∞,3](-\infty,3] (correct answer)
  2. Domain: (−∞,∞)(-\infty,\infty)
  3. Domain: (−∞,0](-\infty,0]
  4. Domain: (−∞,1](-\infty,1]
Explanation: You're doing great exploring how to make quadratic functions invertible through domain restrictions—keep up the good work! A function must be one-to-one to have an inverse, verified by the horizontal line test; for f(x) = (x-3)² + 1, it fails because horizontal lines above the vertex y=1 intersect twice, like at x=3±√(y-1). For even-degree polynomials like this parabola with vertex at (3,1), restrict to x ≥ 3 or x ≤ 3 to make it monotonic and pass the test. Restricting to x ≤ 3 covers the left branch where the function decreases to the vertex and then would increase, but the restriction stops at the vertex to ensure one-to-one. Choice B correctly restricts to (-∞, 3] , including the vertex and left side for monotonic behavior. Choice C uses (-∞,1], but 1 isn't the vertex, so it still fails the test by including both sides partially; A and D don't restrict enough. Follow this decision tree: identify the vertex h=3, choose [h, ∞) or (-∞, h] for one-to-one; this connects to inverse branch selection, as x ≤ 3 would yield the negative square root branch. With practice, you'll master how restrictions enable inverses—excellent effort!

Question 14

Which restriction(s) make f(x)=x2−4xf(x)=x^2-4x pass the horizontal line test and therefore have an inverse? (Choose the best answer.)

  1. Domain: (−∞,∞)(-\infty,\infty)
  2. Domain: [2,∞)[2,\infty) (correct answer)
  3. Domain: [0,∞)[0,\infty)
  4. Domain: (−∞,4](-\infty,4]
Explanation: This question tests your understanding that some functions like f(x) = x² - 4x aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test—nice progress! A function must be one-to-one, and for f(x) = (x-2)² - 4 with vertex at x=2, it fails the horizontal line test for y > -4, intersecting twice. Restrict to x ≥ 2 or x ≤ 2 to make it monotonic: x ≥ 2 is increasing, x ≤ 2 is decreasing. The best restriction avoids crossing the vertex, ensuring no horizontal line hits more than once. Choice B correctly restricts to [2, ∞), the right side from the vertex, making it strictly increasing and invertible. Choice D (-∞,4] crosses the vertex at 2 up to 4, so still fails; C [0,∞) also crosses 2. Strategy: complete the square to find vertex h=2, then restrict to [h, ∞) or (-∞, h]; this choice affects the inverse branch, like positive root for right side. You're building strong skills here—keep going!

Question 15

Consider f(x)=x2−6x+8f(x)=x^2-6x+8 with domain all real numbers. Which domain restriction makes ff invertible by restricting to one side of its vertex (so it passes the horizontal line test)?

  1. Domain: (−∞,∞)(-\infty,\infty)
  2. Domain: [3,∞)[3,\infty) (correct answer)
  3. Domain: (−∞,2](-\infty,2]
  4. Domain: [0,∞)[0,\infty)
Explanation: This question tests your understanding that some functions (like f(x) = x² - 6x + 8) aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. For even-degree polynomials like f(x) = x² or f(x) = x² - 6x + 8, standard restrictions are to the right or left of the vertex: restrict to x ≥ h or x ≤ h where h is the vertex's x-coordinate. This makes the function monotonic (always increasing or always decreasing), which guarantees one-to-one. To find the vertex of f(x) = x² - 6x + 8, we complete the square: f(x) = (x - 3)² - 1, revealing the vertex at (3, -1). The parabola opens upward, so it's decreasing for x < 3 and increasing for x > 3. Choice B correctly restricts to [3,∞), which is the right side of the vertex where the function is increasing and one-to-one. Choice C suggests (-∞,2], but the vertex is at x = 3, not x = 2, so this wouldn't capture a fully monotonic portion. Domain restriction decision tree: (1) Graph the function (or visualize it), (2) Find where it fails horizontal line test (usually at vertex for parabolas), (3) Choose a domain making the function monotonic—either the increasing part or the decreasing part, (4) Standard choices: for x², use x ≥ 0; for (x - h)² + k, use x ≥ h or x ≤ h; for x⁴, use x ≥ 0.

Question 16

The function f(x)=x2−4x+3f(x)=x^2-4x+3 has domain all real numbers and fails the horizontal line test. What domain restriction makes ff invertible?

  1. Domain: (−∞,∞)(-\infty,\infty)
  2. Domain: [0,∞)[0,\infty)
  3. Domain: [2,∞)[2,\infty) (correct answer)
  4. Domain: (−∞,1](-\infty,1]
Explanation: This question tests your understanding that some functions (like f(x) = x² - 4x + 3) aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. A function must be one-to-one (each output comes from exactly one input) to have an inverse, which we verify with the horizontal line test: if any horizontal line crosses the graph more than once, the function isn't one-to-one. For f(x) = x² - 4x + 3, we first complete the square: f(x) = (x - 2)² - 1, revealing the vertex at (2, -1). The parabola opens upward, so it fails the horizontal line test—horizontal lines above y = -1 cross the graph twice. Choice B correctly restricts to [2,∞), which is the right side of the vertex where the function is increasing and one-to-one. Choice C incorrectly suggests (-∞,1], but the vertex is at x = 2, not x = 1, so this restriction wouldn't capture the monotonic portion correctly. Domain restriction decision tree: (1) Graph the function (or visualize it), (2) Find where it fails horizontal line test (usually at vertex for parabolas), (3) Choose a domain making the function monotonic—either the increasing part or the decreasing part, (4) Standard choices: for x², use x ≥ 0; for (x - h)² + k, use x ≥ h or x ≤ h; for x⁴, use x ≥ 0.

Question 17

The function f(x)=−x2+10f(x)=-x^2+10 is not invertible on all real numbers because it fails the horizontal line test. Which restriction makes ff invertible?

  1. Domain: (−∞,∞)(-\infty,\infty)
  2. Domain: x≥0x\ge 0 (correct answer)
  3. Domain: −1≤x≤1-1\le x\le 1
  4. Domain: x≠0x\ne 0
Explanation: This question tests your understanding that some functions (like f(x) = -x² +10) aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. A function must be one-to-one (each output comes from exactly one input) to have an inverse, which we verify with the horizontal line test: if any horizontal line crosses the graph more than once, the function isn't one-to-one. For f(x) = -x² +10, a downward parabola with vertex at x=0, it fails on all reals because f(-1)=9 and f(1)=9. But restricting to x ≥ 0 makes it decreasing from 10 to -∞, passing the test. The restriction creates one-to-one behavior. For even-degree polynomials like f(x) = x² or f(x) = (x - 3)² + 1, standard restrictions are to the right or left of the vertex: restrict to x ≥ h or x ≤ h where h is the vertex's x-coordinate. This makes the function monotonic (always increasing or always decreasing), which guarantees one-to-one. Here, for a downward parabola, x ≥ 0 ensures decreasing monotonicity. Choice B correctly restricts to x ≥ 0 to make it invertible. Choice C fails because [-1,1] includes both sides (f(-1)=9, f(1)=9), not one-to-one. Domain restriction decision tree: (1) Graph the function (or visualize it), (2) Find where it fails horizontal line test (usually at vertex for parabolas), (3) Choose a domain making the function monotonic—either the increasing part or the decreasing part, (4) Standard choices: for x², use x ≥ 0; for (x - h)² + k, use x ≥ h or x ≤ h; for x⁴, use x ≥ 0. The restricted domain should be an interval [a, ∞) or (-∞, a] where the function is one-to-one. Keep exploring these downward parabolas—you're doing wonderfully!

Question 18

For f(x)=(x−1)2f(x)=(x-1)^2 with restricted domain x≥1x \ge 1 (so it passes the horizontal line test), what is f−1(x)f^{-1}(x)?

  1. f−1(x)=1−xf^{-1}(x)=1-\sqrt{x}
  2. f−1(x)=1+xf^{-1}(x)=1+\sqrt{x} (correct answer)
  3. f−1(x)=±x+1f^{-1}(x)=\pm\sqrt{x}+1
  4. f−1(x)=x−1f^{-1}(x)=\sqrt{x}-1
Explanation: This question tests your understanding that some functions (like f(x)=(x−1)2f(x) = (x-1)^2) aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. A function must be one-to-one (each output comes from exactly one input) to have an inverse, which we verify with the horizontal line test: if any horizontal line crosses the graph more than once, the function isn't one-to-one. For f(x)=(x−1)2f(x) = (x-1)^2 restricted to x≥1x \ge 1, it's increasing from 0 to ∞\infty. To find the inverse, set y=(x−1)2y = (x-1)^2 (x≥1x \ge 1), swap to x=(y−1)2x = (y-1)^2, solve y−1=xy-1 = \sqrt{x} (since y≥1y \ge 1 implies y−1≥0y-1 \ge 0), so y=1+xy = 1 + \sqrt{x}. This matches the restricted range. For even-degree polynomials like f(x)=x2f(x) = x^2 or f(x)=(x−3)2+1f(x) = (x - 3)^2 + 1, standard restrictions are to the right or left of the vertex: restrict to x≥hx \ge h or x≤hx \le h where hh is the vertex's x-coordinate. This makes the function monotonic (always increasing or always decreasing), which guarantees one-to-one. For (x−1)2(x-1)^2, restricting to x≥1x \ge 1 gives the positive branch. Choice C correctly finds f−1(x)=1+xf^{-1}(x) = 1 + \sqrt{x}. Choice D fails because ±\pm would not be a single-valued function. Domain restriction decision tree: (1) Graph the function (or visualize it), (2) Find where it fails horizontal line test (usually at vertex for parabolas), (3) Choose a domain making the function monotonic—either the increasing part or the decreasing part, (4) Standard choices: for x2x^2, use x≥0x \ge 0; for (x−h)2+k(x - h)^2 + k, use x≥hx \ge h or x≤hx \le h; for x4x^4, use x≥0x \ge 0. The restricted domain should be an interval [a,∞)[a, \infty) or (−∞,a](-\infty, a] where the function is one-to-one. After restricting, you can find the inverse: with f(x)=(x−1)2f(x) = (x-1)^2 restricted to x≥1x \ge 1, the positive root ensures it works. Fantastic progress on finding inverses!

Question 19

The function f(x)=−x2+10f(x)=-x^2+10 is not invertible on all real numbers because it fails the horizontal line test. Which domain restriction makes ff one-to-one?

  1. Domain: x≥0x\ge 0 (correct answer)
  2. Domain: (−∞,∞)(-\infty,\infty)
  3. Domain: x≥10x\ge 10
  4. Domain: −1≤x≤1-1\le x\le 1
Explanation: This question tests your understanding that some functions (like f(x) = x²) aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. A function must be one-to-one (each output comes from exactly one input) to have an inverse, which we verify with the horizontal line test: if any horizontal line crosses the graph more than once, the function isn't one-to-one. For f(x) = -x² + 10, this is a downward-opening parabola with vertex at (0, 10). The function fails the horizontal line test because horizontal lines below y = 10 intersect the parabola twice. To make it one-to-one, we need to restrict to either the left side (x ≤ 0) or right side (x ≥ 0) of the vertex. Choice A correctly restricts to x ≥ 0, which is the right half of the parabola where the function is always decreasing and therefore one-to-one. Choice B keeps the full domain where f fails the test, C restricts to x ≥ 10 which excludes most of the parabola, and D restricts to a bounded interval which doesn't align with making the function monotonic from the vertex. Domain restriction decision tree: (1) Graph the function (or visualize it), (2) Find where it fails horizontal line test (usually at vertex for parabolas), (3) Choose a domain making the function monotonic—either the increasing part or the decreasing part, (4) Standard choices: for x², use x ≥ 0; for (x - h)² + k, use x ≥ h or x ≤ h; for x⁴, use x ≥ 0.

Question 20

For f(x)=(x−1)2f(x)=(x-1)^2 with the restricted domain x≥1x\ge 1 (so the graph passes the horizontal line test on that domain), what is f−1(x)f^{-1}(x)?​

  1. f−1(x)=x−1f^{-1}(x)=\sqrt{x}-1
  2. f−1(x)=±x+1f^{-1}(x)=\pm\sqrt{x}+1
  3. f−1(x)=1+xf^{-1}(x)=1+\sqrt{x} (correct answer)
  4. f−1(x)=1−xf^{-1}(x)=1-\sqrt{x}
Explanation: This question dives into finding inverses after restriction, building on why we restrict—excellent! For f(x) = (x-1)² restricted to x ≥ 1, it's one-to-one as it passes the horizontal line test on the right increasing branch from vertex at 1. To find the inverse: set y = (x-1)², swap to x = (y-1)², solve y = 1 + √x (positive root since y ≥ 1). The restriction to x ≥ 1 ensures we take the branch where the inverse outputs ≥1. Choice C correctly gives f⁻¹(x) = 1 + √x, matching the positive branch. Choice A uses minus, which would be for left restriction; D includes ±, but inverse must be single-valued. After restricting, swap and solve, choosing the branch per restriction— for ≥ vertex, it's plus. You're mastering how restrictions enable proper inverses—way to go!