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!

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.

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, which is (x-2)² -1 with vertex at x=2, it's not one-to-one on all reals because f(1)=0 and f(3)=0. But if we restrict to x ≥ 2 (right of the vertex), the function is increasing from -1 to ∞, so every horizontal line above -1 crosses 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. Here, [2,∞) ensures monotonic increase. Choice A correctly restricts the domain to [2,∞) to make it one-to-one and invertible. Choice C fails because [0,∞) spans both sides of the vertex (f(0)=3, f(4)=3), violating 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 by solving y = (x-2)² -1 for x in terms of y, taking the positive branch since x ≥ 2. Great job spotting the vertex—you've got this!

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!

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). With f(x) = x² restricted to x ≥ 0, we find the inverse by swapping variables and solving: y = x² becomes x = y², so y = √x (we take the positive root because our restricted domain x ≥ 0 means we're on the right branch). Choice A correctly gives f⁻¹(x) = √x with domain x ≥ 0, which is the range of the restricted f. Choice B with ±√x isn't a function (one input gives two outputs), C reverses nothing (x² is the original function), and D has the wrong domain restriction. 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!

This question tests your understanding that some functions (like f(x) = (x+2)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. For f(x) = (x+2)², this is a parabola with vertex at (-2, 0), opening upward. It fails the horizontal line test because, for example, when y = 4, we get (x+2)² = 4, so x+2 = ±2, giving x = 0 and x = -4—two inputs with the same output! To make it invertible, we must restrict to either x ≥ -2 (right of vertex) or x ≤ -2 (left of vertex). Choice A correctly restricts to x ≤ -2, keeping only the left branch where the function is monotonically decreasing (as x increases from -∞ to -2, f(x) decreases from ∞ to 0). Choice B keeps the full domain where it fails the test, C restricts to [-3, 3] which includes both sides of the vertex, and D restricts to x ≤ 2 which includes the entire parabola and more. Domain restriction strategy: For (x - h)² + k, the vertex is at x = h; restrict to either x ≥ h (for the increasing branch) or x ≤ h (for the decreasing branch). Here, with vertex at x = -2, either x ≥ -2 or x ≤ -2 works—the question chose the left branch.

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.

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.

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 parabola has its vertex at (0, 0) and opens upward. It's decreasing for x < 0 and increasing for x > 0. To make it pass the horizontal line test, we need to keep only one "arm" of the parabola—either the left side where it's decreasing or the right side where it's increasing. Both restrictions work equally well: restricting to x ≥ 0 gives the right half (increasing), while restricting to x ≤ 0 gives the left half (decreasing). Choice C correctly states that either restriction works: we can restrict to x ≥ 0 OR restrict to x ≤ 0, and both create one-to-one functions that pass the horizontal line test. Choices A and B each give only one valid restriction, missing that both work; Choice D incorrectly claims no restriction is needed, when in fact f(2) = f(-2) = 4 shows the unrestricted function isn't one-to-one. After restricting, you get different inverse functions: restricting to x ≥ 0 gives f⁻¹(x) = √x, while restricting to x ≤ 0 gives f⁻¹(x) = -√x. Both are valid inverses on their respective restricted domains!

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!