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 $f(x) = (x-1)^2$ restricted to $x \ge 1$, it's increasing from 0 to $\infty$. To find the inverse, set $y = (x-1)^2$ ($x \ge 1$), swap to $x = (y-1)^2$, solve $y-1 = \sqrt{x}$ (since $y \ge 1$ implies $y-1 \ge 0$), so $y = 1 + \sqrt{x}$. This matches the restricted range. For even-degree polynomials like $f(x) = x^2$ or $f(x) = (x - 3)^2 + 1$, standard restrictions are to the right or left of the vertex: restrict to $x \ge h$ or $x \le h$ where $h$ is the vertex's x-coordinate. This makes the function monotonic (always increasing or always decreasing), which guarantees one-to-one. For $(x-1)^2$, restricting to $x \ge 1$ gives the positive branch. Choice C correctly finds $f^{-1}(x) = 1 + \sqrt{x}$. 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 $x^2$, use $x \ge 0$; for $(x - h)^2 + k$, use $x \ge h$ or $x \le h$; for $x^4$, use $x \ge 0$. The restricted domain should be an interval $[a, \infty)$ or $(-\infty, a]$ where the function is one-to-one. After restricting, you can find the inverse: with $f(x) = (x-1)^2$ restricted to $x \ge 1$, the positive root ensures it works. Fantastic progress on finding inverses!

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!

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!

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) = 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 function fails without restriction because the inverse relation would map y to two x's, like 4 to ±2, which isn't a function. Choice B correctly explains that it fails the horizontal line test because some lines intersect more than once, so the inverse isn't a function. Choice A is incorrect because intersecting exactly twice means it fails, not passes, the test. 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 mastering the horizontal line test—keep it up!

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 $f(x) = x^2$) 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 $f(x) = x^2$ restricted to $x \geq 0$, it passes the test as it's strictly increasing from 0 to $\infty$. To find the inverse, set $y = x^2$ ($x \geq 0$), swap to $x = y^2$, and solve $y = \sqrt{x}$ (principal square root, since $y \geq 0$ matches the range). This gives a proper function inverse. For even-degree polynomials like $f(x) = x^2$ or $f(x) = (x - 3)^2 + 1$, standard restrictions are to the right or left of the vertex: restrict to $x \geq h$ or $x \leq h$ where $h$ is the vertex's x-coordinate. This makes the function monotonic (always increasing or always decreasing), which guarantees one-to-one. For $x^2$, restricting to $x \geq 0$ gives the 'standard' inverse $f^{-1}(x) = \sqrt{x}$ (principal square root). Restricting to $x \leq 0$ would give $f^{-1}(x) = -\sqrt{x}$ instead. Choice B correctly finds the inverse as $f^{-1}(x) = \sqrt{x}$ after the restriction. Choice A fails because $\pm \sqrt{x}$ is not a single function—it maps to two values, violating the definition of a 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 $x^2$, use $x \geq 0$; for $(x - h)^2 + k$, use $x \geq h$ or $x \leq h$; for $x^4$, use $x \geq 0$. The restricted domain should be an interval $[a, \infty)$ or $(-\infty, a]$ where the function is one-to-one. After restricting, you can find the inverse: with $f(x) = x^2$ restricted to $x \geq 0$, swap and solve: $y = x^2$, swap to $x = y^2$, solve for $y = \sqrt{x}$ (taking positive root because we restricted to $x \geq 0$). The domain restriction affects which branch of the inverse you get. No restriction means you can't choose between $\sqrt{x}$ and $-\sqrt{x}$—both would be needed, but that's not a function. Restriction lets you pick one branch! Excellent work on inverses—keep going!

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) = (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!

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 $f(x) = x^2$, standard restrictions are to the right or left of the vertex: restrict to $x \ge h$ or $x \le h$ where $h$ is the vertex's x-coordinate, making the function monotonic and one-to-one. For $x^2$ restricted to $x \ge 0$, this gives the 'standard' inverse $f^{-1}(x) = \sqrt{x}$ (principal square root), while restricting to $x \le 0$ would give $f^{-1}(x) = -\sqrt{x}$ instead. To find the inverse, set $y = x^2$ with $x \ge 0$, swap to $x = y^2$, and solve for $y = \sqrt{x}$, taking the non-negative root because the restricted domain ensures $y \ge 0$. Choice A correctly gives $f^{-1}(x) = \sqrt{x}$, matching the positive branch selected by the $x \ge 0$ 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 $x \ge 0$, 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!