Algebra 2 Flashcards: Restrict Domain To Make Invertible

What this deck covers

This deck focuses on Restrict Domain To Make Invertible, giving you a quick way to review the definitions, rules, and examples that matter most for Algebra 2.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What is $f^{-1}(x)$ for $f(x)=x^2-4$ with restricted domain $x\ge 0$?

Answer: $f^{-1}(x)=\sqrt{x+4}$. Solve $y=x^2-4$ for $x$ using the positive square root.

Flashcard 2: What is the vertex $x$-value used to split a parabola $f(x)=a(x-h)^2+k$ into one-to-one halves?

Answer: $x=h$. The vertex divides the parabola into symmetric halves.

Flashcard 3: What is $f^{-1}(x)$ for $f(x)=x^2-4$ with restricted domain $x\le 0$?

Answer: $f^{-1}(x)=-\sqrt{x+4}$. Solve $y=x^2-4$ for $x$ using the negative square root.

Flashcard 4: What is the range of $f(x)=-(x+2)^2$ when the domain is restricted to $x\ge -2$?

Answer: Range is $y\le 0$. The parabola opens downward with vertex at $(-2,0)$.

Flashcard 5: Which domain restriction makes $f(x)=x^2$ invertible and uses the negative branch?

Answer: Restrict to $x\le 0$. This restriction uses the left branch of the parabola.

Flashcard 6: What condition must a function satisfy to have an inverse function (as a function)?

Answer: $f$ must be one-to-one on its domain. Without this condition, multiple inputs would map to the same output.

Flashcard 7: What is $f^{-1}(x)$ for $f(x)=(x-3)^2$ with restricted domain $x\ge 3$?

Answer: $f^{-1}(x)=3+\sqrt{x}$. Solve $y=(x-3)^2$ for $x$ using the positive square root.

Flashcard 8: What restriction makes $f(x)=(x-3)^2$ invertible to match the negative square root branch?

Answer: Restrict to $x\le 3$. The vertex is at $x=3$, so restrict to the left side.

Flashcard 9: What restriction makes $f(x)=x^2$ one-to-one on an interval centered at $0$?

Answer: Use $x\ge 0$ or $x\le 0$ (one side only). Split at the vertex to create monotonic intervals.

Flashcard 10: What is $f^{-1}(x)$ for $f(x)=-(x-1)^2+7$ with restricted domain $x\le 1$?

Answer: $f^{-1}(x)=1-\sqrt{7-x}$. Solve $y=-(x-1)^2+7$ for $x$ using the left branch.

Flashcard 11: What is $f^{-1}(x)$ for $f(x)=-(x+2)^2$ with restricted domain $x\le -2$?

Answer: $f^{-1}(x)=-2-\sqrt{-x}$. Solve $y=-(x+2)^2$ for $x$ using the negative branch.

Flashcard 12: Identify the correct inverse for restricted $f(x)=(x-4)^2$ with domain $x\le 4$.

Answer: $f^{-1}(x)=4-\sqrt{x}$. For $x\le 4$, use the negative square root.

Flashcard 13: What is $f^{-1}(x)$ for $f(x)=(x+1)^2+5$ with restricted domain $x\le -1$?

Answer: $f^{-1}(x)=-1-\sqrt{x-5}$. Solve $y=(x+1)^2+5$ for $x$ using the negative branch.

Flashcard 14: Identify the restricted domain that makes $f(x)=-(x-1)^2+7$ invertible using the right branch.

Answer: Restrict to $x\ge 1$. The vertex is at $x=1$, so restrict to the right side.

Flashcard 15: What test determines whether a function is one-to-one using horizontal lines?

Answer: Horizontal line test: each $y$ hits graph once. If any horizontal line crosses the graph more than once, the function is not one-to-one.

Flashcard 16: What restriction makes $f(x)=x^2-4$ invertible to match the principal square root branch?

Answer: Restrict to $x\ge 0$. The vertex is at $x=0$, so restrict to the right side.

Flashcard 17: What restriction makes $f(x)=(x-h)^2+k$ one-to-one using the left branch?

Answer: Restrict to $x\le h$. The vertex $x=h$ is the axis of symmetry of the parabola.

Flashcard 18: What restriction makes $f(x)=(x-3)^2$ invertible to match the principal square root branch?

Answer: Restrict to $x\ge 3$. The vertex is at $x=3$, so restrict to the right side.

Flashcard 19: What is the inverse of $f(x)=|x|$ when restricted to $x\ge 0$?

Answer: $f^{-1}(x)=x$ for $x\ge 0$. When $x\ge 0$, $f(x) = |x| = x$, so $f^{-1}(x) = x$.

Flashcard 20: What is the range of $f(x)=(x-3)^2$ when the domain is restricted to $x\ge 3$?

Answer: Range is $y\ge 0$. The parabola opens upward with vertex at $(3,0)$.

Flashcard 21: Which domain restriction makes $f(x)=x^2$ invertible and matches the principal square root?

Answer: Restrict to $x\ge 0$. This restriction uses the right branch of the parabola.

Flashcard 22: What restriction makes $f(x)=-(x+2)^2$ invertible using the right half of the parabola?

Answer: Restrict to $x\ge -2$. The vertex is at $x=-2$, so restrict to the right side.

Flashcard 23: What is the range of $f^{-1}(x)$ if $f$ is restricted to the domain $x\le 3$?

Answer: Range of $f^{-1}$ is $x\le 3$. The range of the inverse equals the domain of the original function.

Flashcard 24: What is $f^{-1}(x)$ for $f(x)=-(x-1)^2+7$ with restricted domain $x\ge 1$?

Answer: $f^{-1}(x)=1+\sqrt{7-x}$. Solve $y=-(x-1)^2+7$ for $x$ using the right branch.

Flashcard 25: What is the goal of restricting a domain to make a function invertible?

Answer: Make the function one-to-one. Only one-to-one functions have inverses that are also functions.

Flashcard 26: What is $f^{-1}(x)$ for $f(x)=-(x+2)^2$ with restricted domain $x\ge -2$?

Answer: $f^{-1}(x)=-2+\sqrt{-x}$. Solve $y=-(x+2)^2$ for $x$ using the positive branch.

Flashcard 27: What is the range of $f(x)=x^2$ when the domain is restricted to $x\ge 0$?

Answer: Range is $y\ge 0$. Squaring non-negative inputs produces non-negative outputs.

Flashcard 28: What is $f^{-1}(x)$ for $f(x)=(x+1)^2+5$ with restricted domain $x\ge -1$?

Answer: $f^{-1}(x)=-1+\sqrt{x-5}$. Solve $y=(x+1)^2+5$ for $x$ using the positive branch.

Flashcard 29: Identify the restricted domain that makes $f(x)=(x+5)^2$ invertible using the left branch.

Answer: Restrict to $x\le -5$. The vertex is at $x=-5$, so restrict to the left side.

Flashcard 30: What must you check after solving for $y$ when finding an inverse of a restricted quadratic?

Answer: Choose the correct sign to match the restriction. The sign must correspond to the restricted domain interval.

Flashcard 31: What is $f^{-1}(x)$ for $f(x)=(x-3)^2$ with restricted domain $x\le 3$?

Answer: $f^{-1}(x)=3-\sqrt{x}$. Solve $y=(x-3)^2$ for $x$ using the negative square root.

Flashcard 32: What is the first algebraic step to find an inverse after restricting the domain?

Answer: Write $y=f(x)$ and swap $x$ and $y$. This sets up the equation to solve for the inverse.

Flashcard 33: What is the inverse of $f(x)=|x|$ when restricted to $x \geq 0$?

Answer: $f^{-1}(x)=x$ for $x \geq 0$. When $x \geq 0$, $f(x) = |x| = x$, so $f^{-1}(x) = x$.

Flashcard 34: What is the inverse of $f(x)=x^2$ when the domain is restricted to $x\le 0$?

Answer: $f^{-1}(x)=-\sqrt{x}$. The negative square root matches the left branch restriction.

Flashcard 35: Which restriction makes $f(x)=x^2+2x$ invertible by using the decreasing side of the parabola?

Answer: Restrict to $x\le -1$. Complete the square to find vertex at $x=-1$, then use left side.

Flashcard 36: What is the inverse of $f(x)=x^2$ if the restricted domain is $[-2,0]$?

Answer: $f^{-1}(x)=-\sqrt{x}$ for $0\le x\le 4$. The range $[0,4]$ comes from squaring the domain $[-2,0]$.

Flashcard 37: What happens to domain and range when taking an inverse $f^{-1}$?

Answer: Domain and range swap. The input and output sets exchange roles.

Flashcard 38: What is the inverse of $f(x)=|x|$ when restricted to $x\le 0$?

Answer: $f^{-1}(x)=-x$ for $x\ge 0$. When $x\le 0$, $f(x) = |x| = -x$, so $f^{-1}(x) = -x$.

Flashcard 39: What restriction makes $f(x)=-(x-h)^2+k$ one-to-one using the right branch?

Answer: Restrict to $x\ge h$. The vertex $x=h$ is the axis of symmetry of the parabola.

Flashcard 40: What restriction makes $f(x)=(x-h)^2+k$ one-to-one using the right branch?

Answer: Restrict to $x\ge h$. The vertex $x=h$ is the axis of symmetry of the parabola.

Flashcard 41: What is $f^{-1}(x)$ for $f(x)=(x+5)^2$ with restricted domain $x\le -5$?

Answer: $f^{-1}(x)=-5-\sqrt{x}$. Solve $y=(x+5)^2$ for $x$ using the negative branch.

Flashcard 42: What does restricting the domain of a function mean?

Answer: Limit inputs to a subset of the original domain. This creates a new function with a smaller domain.

Flashcard 43: What is the domain of $f^{-1}(x)$ if $f$ is restricted to be one-to-one with range $y\ge 5$?

Answer: Domain of $f^{-1}$ is $x\ge 5$. The domain of the inverse equals the range of the original function.

Flashcard 44: Identify the correct inverse for restricted $f(x)=(x-4)^2$ with domain $x\ge 4$.

Answer: $f^{-1}(x)=4+\sqrt{x}$. For $x\ge 4$, use the positive square root.

Flashcard 45: What restriction makes $f(x)=-(x-h)^2+k$ one-to-one using the left branch?

Answer: Restrict to $x\le h$. The vertex $x=h$ is the axis of symmetry of the parabola.

Flashcard 46: What restriction makes $f(x)=(x+1)^2+5$ invertible using the increasing side?

Answer: Restrict to $x\ge -1$. The vertex is at $x=-1$, so restrict to the right side.

Flashcard 47: Which restriction makes $f(x)=x^2+2x$ invertible by using the increasing side of the parabola?

Answer: Restrict to $x\ge -1$. Complete the square to find vertex at $x=-1$, then use right side.

Flashcard 48: Identify the restricted domain that makes $f(x)=x^2$ invertible if the domain is $[-2,2]$.

Answer: Restrict to $[0,2]$ or $[-2,0]$. Choose one monotonic piece from each side of the vertex at $x=0$.

Flashcard 49: What is the relationship between the graphs of $f$ and $f^{-1}$?

Answer: They reflect across the line $y=x$. The graphs are mirror images across the diagonal line $y=x$.

Flashcard 50: What does it mean for a function $f$ to be one-to-one on its domain?

Answer: $f(a)=f(b)\Rightarrow a=b$. Each input maps to exactly one output.

Flashcard 51: What is the inverse of $f(x)=x^2$ if the restricted domain is $[0,2]$?

Answer: $f^{-1}(x)=\sqrt{x}$ for $0\le x\le 4$. The range $[0,4]$ comes from squaring the domain $[0,2]$.

Flashcard 52: What is the inverse of $f(x)=x^2$ when the domain is restricted to $x\ge 0$?

Answer: $f^{-1}(x)=\sqrt{x}$. The positive square root matches the right branch restriction.