After restricting f(x)=x2 to the domain x≥0 so that it passes the horizontal line test, what is the inverse function f−1(x)?
Answer: f−1(x)=x
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!
After restricting the domain of f(x)=x2 to x≥0 so it passes the horizontal line test, what is the inverse function f−1(x)?
Answer: f−1(x)=x
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)=x2, 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, making the function monotonic and one-to-one. For x2 restricted to x≥0, this gives the 'standard' inverse f−1(x)=x (principal square root), while restricting to x≤0 would give f−1(x)=−x instead. To find the inverse, set y=x2 with x≥0, swap to x=y2, and solve for y=x, taking the non-negative root because the restricted domain ensures y≥0. Choice A correctly gives f−1(x)=x, matching the positive branch selected by the x≥0 restriction. Choice B includes ±, 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≥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!
Which restriction(s) make f(x)=x2 pass the horizontal line test? (Choose one option.)
Answer: Either restrict to x≥0 or restrict to 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!
Explain why f(x)=x2 needs a domain restriction to have an inverse function. Which statement is correct?
Answer: A restriction is needed because some horizontal lines intersect the graph twice, so one output can come from two different inputs.
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, consider the horizontal line y = 4: it intersects the parabola at both x = 2 and x = -2, meaning f(2) = f(-2) = 4. This shows that one output (4) comes from two different inputs (2 and -2), violating the one-to-one requirement for invertibility. Choice B correctly explains that a restriction is needed because some horizontal lines intersect the graph twice, so one output can come from two different inputs. Choice A is incorrect because many horizontal lines (like y = 4) intersect the graph twice, not just once. Choice C confuses the horizontal line test with the vertical line test (which checks if something is a function). Choice D is incorrect because x² is defined for all real numbers, including negative x. The horizontal line test reveals why quadratic functions need domain restrictions: without restriction, you can't uniquely determine which input produced a given output, making inversion impossible.
Explain why f(x)=x2 needs a domain restriction to have an inverse function. Which statement best connects this to the horizontal line test?
Answer: It needs a restriction because some horizontal lines intersect the graph more than once, so different inputs can share the same output.
You're doing wonderfully understanding the need for restrictions—let's clarify why! For f(x) = x², it's not one-to-one on all reals, failing the horizontal line test as lines like y=4 hit at x=2 and x=-2, meaning different inputs share outputs. We restrict to pass the test, ensuring no horizontal line intersects more than once. The horizontal line test directly checks one-to-one: multiple intersections mean not invertible. Choice B correctly explains this connection to horizontal lines and shared outputs. Choice A mentions vertical lines, which test if it's a function, not one-to-one; D is wrong as quadratics need restrictions. Strategy: visualize graph, apply horizontal line test at vertex; restrict to one side. This sets up inverse branch selection—impressive insight!
The function f(x)=−x2+10 is not invertible on all real numbers because it fails the horizontal line test. Which domain restriction makes f one-to-one?
Answer: Domain: 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 + 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.
Restrict the domain of f(x)=−x2+10 (domain: all reals) so that it is invertible. Which restriction works?
Answer: Domain: [0,∞)
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.
A function h(x)=−x2+8x−7 is defined on the interval [0,8]. To create an invertible function, Maria restricts the domain to [0,4] and Jose restricts it to [4,8]. Which statement about their restricted functions is correct?
Answer: Both functions have the same range, but Maria's function is increasing while Jose's is decreasing
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.
The function f(x)=x4 (domain: all reals) fails the horizontal line test because f(2)=f(−2). Which domain restriction makes f invertible?
Answer: Domain: [0,∞)
This question tests your understanding that some functions (like f(x) = x⁴) 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) = x⁴, the function is even (symmetric about the y-axis), so f(2) = f(-2) = 16, failing the one-to-one requirement. The graph looks like a flatter version of x², with the same U-shape but steeper sides. Choice B correctly restricts to [0,∞), which includes only the right half where the function is increasing and one-to-one—every horizontal line crosses at most once on this domain. Choice A incorrectly suggests no restriction is needed, but we've established that f(x) = f(-x) for all x, 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 pattern holds for all even powers!
After restricting f(x)=(x−4)2 to the domain x≤4 so that it is one-to-one, what is the inverse function f−1(x)?
Answer: f−1(x)=4−x
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.
The function g(x)=∣2x−6∣+1 is not one-to-one on its natural domain. If we restrict the domain to make g invertible, and we want the resulting function to include the point (1,5), which domain restriction is appropriate?
Answer: x≤3
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.
Explain why f(x)=x2 (domain: all real numbers) needs a domain restriction to have an inverse. Which statement correctly uses the horizontal line test idea?
Answer: It fails the horizontal line test because some horizontal lines intersect the graph more than once, so the inverse would not be a function.
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!
The function f(x)=x2 (domain: all real numbers) is not invertible because it fails the horizontal line test. Which domain restriction makes f one-to-one (and therefore invertible)?
Answer: Domain: [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 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 fails because it keeps the full domain where f(2) = f(-2) = 4, violating one-to-one; Choice C incorrectly claims every function has an inverse; Choice D creates a strange split domain that still includes both positive and negative values for most outputs. 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.
The function f(x)=−x2+10 (domain: all real numbers) fails the horizontal line test. Which domain restriction makes f one-to-one and invertible?
Answer: Domain: x≥0
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.
The function f(x)=x2 (domain: all real numbers) is not invertible because it fails the horizontal line test (for example, f(2)=f(−2)=4). Which domain restriction makes f one-to-one and therefore invertible?
Answer: Domain: [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 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!
After restricting f(x)=x2−4x+3 to the domain x≥2 so it passes the horizontal line test, find f−1(x).
Answer: f−1(x)=2+x+1
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. With f(x) = x² - 4x + 3 = (x - 2)² - 1 restricted to x ≥ 2, we find the inverse by setting y = (x - 2)² - 1 and solving for x: y + 1 = (x - 2)², so √(y + 1) = x - 2 (taking positive root since x ≥ 2 means x - 2 ≥ 0), giving x = 2 + √(y + 1). Swapping variables gives f⁻¹(x) = 2 + √(x + 1). Choice A correctly gives f⁻¹(x) = 2 + √(x + 1), which undoes the transformation: shift right 2, square, then shift down 1. Choice B would correspond to restricting to x ≤ 2 (left branch), C has the wrong transformation, and D isn't a function. 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.
The function f(x)=x2−4x+3 has domain all real numbers and fails the horizontal line test. What domain restriction makes f invertible?
Answer: Domain: [2,∞)
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.
Restrict the domain of f(x)=(x−1)2 to x≥1 so that f is invertible. What is f−1(x)?
Answer: f−1(x)=1+x (domain x≥0)
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-1)² with domain restricted to x ≥ 1, we keep only the right branch of the parabola (vertex at x = 1), making it one-to-one and invertible. To find the inverse, start with y = (x-1)², swap variables to get x = (y-1)², then solve for y: taking the square root gives y-1 = ±√x, but since we restricted to x ≥ 1 (the increasing branch), we take the positive root: y-1 = √x, so y = 1 + √x. The domain of f⁻¹ is the range of the restricted f, which is [0, ∞). Choice C correctly gives f⁻¹(x) = 1 + √x with domain x ≥ 0. Choice A has the wrong sign (would be for the left branch restriction), B subtracts 1 instead of adding, and D incorrectly includes ± which would give two outputs per input. After restricting, you can find the inverse: with f(x) = (x-1)² restricted to x ≥ 1, the function maps [1, ∞) to [0, ∞) bijectively. The domain restriction to x ≥ 1 ensures we take the positive square root when inverting, giving us the unique inverse function.
For f(x)=(x+5)2 with domain restricted to x≤−5, what is the inverse function f−1(x)?
Answer: f−1(x)=−x−5 (domain x≥0)
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+5)² with domain restricted to x ≤ -5, we keep only the left branch of the parabola (vertex at x = -5), making it one-to-one. On this branch, the function is decreasing: as x increases toward -5, f(x) decreases toward 0. To find the inverse, start with y = (x+5)², swap to get x = (y+5)², solve for y: taking the square root gives y+5 = ±√x, but since we restricted to x ≤ -5 (the decreasing branch), we must take the negative root: y+5 = -√x, so y = -√x - 5. Choice B correctly gives f⁻¹(x) = -√x - 5 with domain x ≥ 0 (which is the range of the restricted f). Choice A would be correct for the right branch restriction (x ≥ -5), C incorrectly includes ±, and D has the signs wrong. The key insight: when you restrict to the left/decreasing branch of a parabola, you take the negative square root when finding the inverse; for the right/increasing branch, you take the positive square root. The domain restriction determines which branch of the square root to use!
Which restriction(s) make f(x)=x2−4x pass the horizontal line test and therefore have an inverse? (Choose the best answer.)
Answer: Domain: [2,∞)
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!
