Algebra 2 Quiz: Restrict Domain To Make Invertible
20 questions · exam conditions
0:00
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}.
← Back to quizzes

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.

What this quiz covers

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.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

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

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 3

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 4

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 5

The function g(x)=∣2x−6∣+1g(x) = |2x - 6| + 1 is not one-to-one on its natural domain. If we restrict the domain to make gg invertible, and we want the resulting function to include the point (1,5)(1, 5), which domain restriction is appropriate?

  1. x≥3x \geq 3
  2. x≤3x \leq 3 (correct answer)
  3. x≥1x \geq 1
  4. x≤1x \leq 1
Explanation: First, verify (1,5) is on the function: g(1) = |2(1) - 6| + 1 = |-4| + 1 = 5 ✓. The function g(x) = |2x - 6| + 1 has vertex at x = 3 (where 2x - 6 = 0). Since we need the function to include point (1,5) and x = 1 < 3, we must choose the left side of the vertex: x ≤ 3. Choice D (x ≤ 1) would exclude points near the vertex and create a smaller domain than necessary.

Question 6

A function h(x)=−x2+8x−7h(x) = -x^2 + 8x - 7 is defined on the interval [0,8][0, 8]. To create an invertible function, Maria restricts the domain to [0,4][0, 4] and Jose restricts it to [4,8][4, 8]. Which statement about their restricted functions is correct?

  1. Both functions have the same range, but Maria's function is decreasing while Jose's is increasing
  2. Both functions have the same range, but Maria's function is increasing while Jose's is decreasing (correct answer)
  3. Maria's function has a different range than Jose's function, making one restriction invalid
  4. Jose's restriction is invalid because it doesn't include the vertex of the parabola
Explanation: The function h(x) = -x² + 8x - 7 has vertex at x = 4, giving h(4) = 9. Since the parabola opens downward, the function increases on [0,4] and decreases on [4,8]. Both restrictions give the same range [-7, 9]. Maria's restriction [0,4] captures the increasing portion, while Jose's [4,8] captures the decreasing portion. Both create valid invertible functions with the same range.

Question 7

A function f(x)=∣x2−9∣f(x) = |x^2 - 9| is defined for all real numbers. To make this function invertible by restricting to an interval where f(x)≥7f(x) \geq 7, which domain restriction works?

  1. −4≤x≤4-4 \leq x \leq 4
  2. x≥3x \geq 3 or x≤−3x \leq -3
  3. −3≤x≤3-3 \leq x \leq 3
  4. x≥4x \geq 4 or x≤−4x \leq -4 (correct answer)
Explanation: When making a function invertible, you need to restrict its domain so the function becomes one-to-one (passes the horizontal line test). This means finding where f(x)=∣x2−9∣≥7f(x) = |x^2 - 9| \geq 7 and ensuring each y-value corresponds to exactly one x-value. First, solve ∣x2−9∣≥7|x^2 - 9| \geq 7. This absolute value inequality splits into two cases: x2−9≥7x^2 - 9 \geq 7 or x2−9≤−7x^2 - 9 \leq -7. The first gives x2≥16x^2 \geq 16, so x≥4x \geq 4 or x≤−4x \leq -4. The second gives x2≤2x^2 \leq 2, so −2≤x≤2-\sqrt{2} \leq x \leq \sqrt{2}. However, we need f(x)≥7f(x) \geq 7, and checking values in [−2,2][-\sqrt{2}, \sqrt{2}] shows f(x)=9−x2<7f(x) = 9 - x^2 < 7 there. So our constraint reduces to x≥4x \geq 4 or x≤−4x \leq -4. Now, on the intervals x≥4x \geq 4 and x≤−4x \leq -4, the function f(x)=x2−9f(x) = x^2 - 9 is strictly increasing on [4,∞)[4, \infty) and strictly decreasing on (−∞,−4](-\infty, -4]. Each interval is one-to-one, making the function invertible when restricted to either. Choice A (−4≤x≤4-4 \leq x \leq 4) includes the interval where f(x)<7f(x) < 7. Choice B (x≥3x \geq 3 or x≤−3x \leq -3) includes points where f(x)<7f(x) < 7 (like x=3x = 3). Choice C (−3≤x≤3-3 \leq x \leq 3) covers the region where f(x)<7f(x) < 7 entirely. Remember: for absolute value functions, always check both the constraint and the one-to-one requirement separately, then find their intersection.

Question 8

The function g(x)=x4−8x2+16g(x) = x^4 - 8x^2 + 16 can be written as g(x)=(x2−4)2g(x) = (x^2 - 4)^2. To make this function invertible by domain restriction while ensuring the resulting function passes through the point (3,1)(\sqrt{3}, 1), which restriction is most appropriate?

  1. x≥2x \geq 2
  2. x≤−2x \leq -2
  3. 0≤x≤20 \leq x \leq 2 (correct answer)
  4. −2≤x≤0-2 \leq x \leq 0
Explanation: When you're asked to make a function invertible through domain restriction, you need to find an interval where the function is one-to-one (passes the horizontal line test) and includes the specified point. Let's analyze g(x)=(x2−4)2g(x) = (x^2 - 4)^2. Since this is a composition of even functions, it's symmetric about the y-axis and has multiple x-values producing the same y-value. To make it invertible, you must restrict the domain to eliminate this many-to-one behavior. First, find where the function equals zero: (x2−4)2=0(x^2 - 4)^2 = 0 when x2=4x^2 = 4, so x=±2x = \pm 2. The function has its minimum value of 0 at these points. Between x=−2x = -2 and x=2x = 2, the function decreases from x=−2x = -2 to x=0x = 0, then increases from x=0x = 0 to x=2x = 2. Since we need the function to pass through (3,1)(\sqrt{3}, 1), our domain must include x=3≈1.73x = \sqrt{3} \approx 1.73. Looking at the intervals, only options C and D could include this point. Option A (x≥2x \geq 2) doesn't include 3\sqrt{3} since 3<2\sqrt{3} < 2. Option B (x≤−2x \leq -2) also excludes 3\sqrt{3} since it's positive. Option D (−2≤x≤0-2 \leq x \leq 0) excludes 3\sqrt{3} since 3>0\sqrt{3} > 0. Option C (0≤x≤20 \leq x \leq 2) includes 3\sqrt{3} and creates a one-to-one function on this interval since g(x)g(x) is strictly increasing from x=0x = 0 to x=2x = 2. Study tip: For domain restrictions that make functions invertible, always check that your chosen interval both includes required points and eliminates the function's symmetry.

Question 9

Consider the piecewise function $$f(x) = \begin{cases} x^2 & \text{if } x \leq 0 \ 2x + 3 & \text{if } x > 0 \end{cases}

  1. The function is already one-to-one and requires no further domain restriction to be invertible (correct answer)
  2. Restrict to x≥0x \geq 0 only, since the quadratic piece creates the non-invertible behavior
  3. Restrict to x≤0x \leq 0 only, since the linear piece creates the non-invertible behavior
  4. The function cannot be made invertible by domain restriction because the pieces don't connect continuously
Explanation: Analyzing each piece: For x ≤ 0, f(x) = x² is decreasing (since we're on the left side of the parabola), giving range [0, ∞). For x > 0, f(x) = 2x + 3 is increasing, giving range (3, ∞). Since the ranges don't overlap and each piece is monotonic on its domain, the function is already one-to-one. At x = 0: f(0) = 0, and as x approaches 0 from the right, f(x) approaches 3, so there's no range overlap.

Question 10

A function f(x)=16−x2f(x) = \sqrt{16 - x^2} represents the upper semicircle of radius 4 centered at the origin. To create an invertible function that includes the point (22,22)(2\sqrt{2}, 2\sqrt{2}), which domain restriction should be used?

  1. −4≤x≤0-4 \leq x \leq 0
  2. −22≤x≤22-2\sqrt{2} \leq x \leq 2\sqrt{2}
  3. 0≤x≤40 \leq x \leq 4 (correct answer)
  4. 22≤x≤42\sqrt{2} \leq x \leq 4
Explanation: When working with function inverses, you need to understand that a function must be one-to-one (each y-value corresponds to exactly one x-value) to have an inverse. The semicircle function f(x)=16−x2f(x) = \sqrt{16 - x^2} fails the horizontal line test on its full domain [−4,4][-4, 4] because most y-values correspond to two different x-values. To find the correct domain restriction, first verify that the point (22,22)(2\sqrt{2}, 2\sqrt{2}) is actually on the semicircle: f(22)=16−(22)2=16−8=8=22f(2\sqrt{2}) = \sqrt{16 - (2\sqrt{2})^2} = \sqrt{16 - 8} = \sqrt{8} = 2\sqrt{2} ✓ Now examine each restriction for invertibility. Choice C, 0≤x≤40 \leq x \leq 4, restricts the function to the right half of the semicircle. On this interval, the function is strictly decreasing (as x increases from 0 to 4, y decreases from 4 to 0), making it one-to-one and therefore invertible. This domain also includes our required point (22,22)(2\sqrt{2}, 2\sqrt{2}). Choice A (−4≤x≤0-4 \leq x \leq 0) would also create an invertible function, but it's the left half of the semicircle and doesn't include the point (22,22)(2\sqrt{2}, 2\sqrt{2}) since 22>02\sqrt{2} > 0. Choice B (−22≤x≤22-2\sqrt{2} \leq x \leq 2\sqrt{2}) includes the required point but spans both sides of the y-axis, so the function isn't one-to-one on this interval. Choice D (22≤x≤42\sqrt{2} \leq x \leq 4) includes the point and creates invertibility, but it's unnecessarily restrictive. Remember: when restricting domains for inverse functions, choose the largest possible interval that maintains one-to-one behavior while satisfying any given constraints.

Question 11

After restricting the domain of f(x)=x2f(x)=x^2 to x≥0x \ge 0 so it passes the horizontal line test, what is the inverse function f−1(x)f^{-1}(x)?

  1. f−1(x)=±xf^{-1}(x)=\pm\sqrt{x}
  2. f−1(x)=x2f^{-1}(x)=x^2
  3. f−1(x)=−xf^{-1}(x)=-\sqrt{x}
  4. f−1(x)=xf^{-1}(x)=\sqrt{x} (correct answer)
Explanation: This question builds on restricting domains to make functions invertible, focusing now on finding the inverse after the restriction—great job getting this far! For even-degree polynomials like f(x)=x2f(x) = x^2, 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, making the function monotonic and one-to-one. For x2x^2 restricted to x≥0x \ge 0, this gives the 'standard' inverse f−1(x)=xf^{-1}(x) = \sqrt{x} (principal square root), while restricting to x≤0x \le 0 would give f−1(x)=−xf^{-1}(x) = -\sqrt{x} instead. To find the inverse, set y=x2y = x^2 with x≥0x \ge 0, swap to x=y2x = y^2, and solve for y=xy = \sqrt{x}, taking the non-negative root because the restricted domain ensures y≥0y \ge 0. Choice A correctly gives f−1(x)=xf^{-1}(x) = \sqrt{x}, matching the positive branch selected by the x≥0x \ge 0 restriction. Choice B includes ±\pm, but the inverse must be a function, so we can't have both branches; C is the original function, and D takes the negative root, which wouldn't match the domain. Use this strategy: after restricting, swap variables and solve, ensuring the branch aligns with the restriction— for x≥0x \ge 0, it's the positive root. This restriction lets you pick one branch, turning a non-invertible function into one with a clear inverse—keep practicing, you're doing awesome!

Question 12

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: [0,∞)[0,\infty)
  3. Domain: [2,∞)[2,\infty) (correct answer)
  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 13

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!

Question 14

The function f(x)=−x2+10f(x)=-x^2+10 (domain: all real numbers) fails the horizontal line test. Which domain restriction makes ff one-to-one and invertible?

  1. Domain: x≥0x\ge 0 (correct answer)
  2. Domain: (−∞,∞)(-\infty,\infty)
  3. Domain: −1≤x≤1-1\le x\le 1
  4. Domain: x≥10x\ge 10
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. For f(x) = -x² + 10, this is a downward-opening parabola with vertex at (0, 10). The negative coefficient flips the parabola upside down compared to x². It fails the horizontal line test because, for example, when y = 6, we get -x² + 10 = 6, so x² = 4, giving x = ±2—two inputs with the same output! To make it invertible, we restrict to either x ≥ 0 (right of vertex) or x ≤ 0 (left of vertex). Choice A correctly restricts to x ≥ 0, keeping only the right branch where the function is monotonically decreasing (as x increases from 0, f(x) decreases from 10). Choice B keeps the full domain where it fails the test, C restricts to [-1, 1] which still includes both positive and negative values, and D restricts to x ≥ 10 which would give an empty or very limited function since the parabola's maximum is at y = 10. Domain restriction strategy: For -x² + k (vertex at x = 0), restrict to either x ≥ 0 or x ≤ 0. The downward opening doesn't change where to restrict, just the monotonic behavior on each branch.

Question 15

After restricting f(x)=x2f(x)=x^2 to the domain x≥0x\ge 0 so that it passes the horizontal line test, what is the inverse function f−1(x)f^{-1}(x)?

  1. f−1(x)=x2f^{-1}(x)=x^2
  2. f−1(x)=xf^{-1}(x)=\sqrt{x} (correct answer)
  3. f−1(x)=±xf^{-1}(x)=\pm\sqrt{x}
  4. f−1(x)=1xf^{-1}(x)=\dfrac{1}{\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. 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. When we restrict f(x) = x² to domain x ≥ 0, we're keeping only the right half of the parabola. To find the inverse, we swap x and y, then solve for y: starting with y = x² (where x ≥ 0), swap to get x = y², then solve for y = ±√x. Since our original function had domain x ≥ 0 (non-negative inputs only), the range was also y ≥ 0 (non-negative outputs only), so when we invert, we need y ≥ 0, giving us y = √x (the positive square root). Choice B correctly identifies f⁻¹(x) = √x as the inverse function. Choice A incorrectly suggests f⁻¹(x) = x², which would be composing f with itself, not inverting; Choice C gives ±√x, which isn't a function because it assigns two values to each input; Choice D suggests 1/√x, which has no connection to inverting x². 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!

Question 16

The function f(x)=x2f(x)=x^2 (domain: all real numbers) 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: (−∞,∞)(-\infty,\infty) except x=0x=0
  4. Domain: [−1,1][-1,1]
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! 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 full domain fails because it's symmetric, so positive and negative inputs give the same output, like f(3) = f(-3) = 9, violating one-to-one. Choice B correctly restricts the domain to [0, ∞) to make it one-to-one and invertible. Choice D fails because [-1,1] still includes both positive and negative inputs, so f(1) = f(-1) = 1, 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) = 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! Keep practicing these restrictions, and you'll master invertibility in no time!

Question 17

The function f(x)=(x+1)2f(x)=(x+1)^2 is not one-to-one on all real numbers (it fails the horizontal line test). Which domain restriction makes it invertible?

  1. Domain: x≥−1x\ge -1 (correct answer)
  2. Domain: x≥1x\ge 1
  3. Domain: (−∞,∞)(-\infty,\infty)
  4. Domain: −2≤x≤2-2\le x\le 2
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. 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)², vertex at x=-1, it fails on all reals because f(-2)=1 and f(0)=1. But restricting to x ≥ -1 makes it increasing from 0 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, x ≥ -1 ensures monotonicity. Choice A correctly restricts to x ≥ -1 to make it invertible. Choice D fails because [-2,2] includes both sides (f(-2)=1, f(0)=1), 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 building strong skills here!

Question 18

Consider h(x)=2sin⁡(x)+1h(x) = 2\sin(x) + 1 on the interval [0,2π][0, 2\pi]. To create an invertible function that achieves the maximum value of h(x)h(x), which domain restriction should be used?

  1. [0,π2][0, \frac{\pi}{2}] (correct answer)
  2. [π2,3π2][\frac{\pi}{2}, \frac{3\pi}{2}]
  3. [π,2π][\pi, 2\pi]
  4. [3π2,2π][\frac{3\pi}{2}, 2\pi]
Explanation: The function h(x) = 2sin(x) + 1 has maximum value 3 when sin(x) = 1, which occurs at x = π/2. For invertibility, we need a monotonic interval that includes x = π/2. On [0, 2π], sin(x) increases on [0, π/2], decreases on [π/2, 3π/2], and increases on [3π/2, 2π]. To include the maximum at x = π/2 while maintaining monotonicity, we can use either [0, π/2] (increasing to maximum) or [π/2, π] (decreasing from maximum, but this isn't listed). Choice A gives the increasing portion that reaches the maximum.

Question 19

Restrict the domain of f(x)=−x2+10f(x)=-x^2+10 (domain: all reals) so that it is invertible. Which restriction works?

  1. Domain: [0,∞)[0,\infty) (correct answer)
  2. Domain: (−∞,5](-\infty,5]
  3. Domain: (−∞,∞)(-\infty,\infty)
  4. Domain: (−∞,∞)∖{0}(-\infty,\infty)\setminus\{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, this is a downward-opening parabola with vertex at (0, 10). The function fails the horizontal line test because horizontal lines below y = 10 cross the graph twice. To make it one-to-one, we need to restrict to either x ≥ 0 (right side, decreasing) or x ≤ 0 (left side, increasing). Choice A correctly restricts to [0,∞), which is the right half where the function is decreasing and one-to-one. Choice C suggests (-∞,5], but this doesn't align with the vertex location at x = 0 and would include parts where the function isn't monotonic. 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

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.