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Algebra 2 Flashcards: Restrict Domain To Make Invertible

Study Restrict Domain To Make Invertible in Algebra 2 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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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.

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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.

Algebra 2 Flashcards: Restrict Domain To Make Invertible

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QUESTION

Which domain restriction makes f(x)=x2f(x)=x^2f(x)=x2 invertible and uses the negative branch?

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ANSWER

Restrict to x≤0x\le 0x≤0. This restriction uses the left branch of the parabola.

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Flashcard 1: Which domain restriction makes f(x)=x2f(x)=x^2f(x)=x2 invertible and uses the negative branch?

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

Flashcard 2: Which domain restriction makes f(x)=x2f(x)=x^2f(x)=x2 invertible and matches the principal square root?

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

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

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

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

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

Flashcard 5: What is f−1(x)f^{-1}(x)f−1(x) for f(x)=(x−3)2f(x)=(x-3)^2f(x)=(x−3)2 with restricted domain x≥3x\ge 3x≥3?

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

Flashcard 6: What restriction makes f(x)=(x−3)2f(x)=(x-3)^2f(x)=(x−3)2 invertible to match the negative square root branch?

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

Flashcard 7: What is f−1(x)f^{-1}(x)f−1(x) for f(x)=x2−4f(x)=x^2-4f(x)=x2−4 with restricted domain x≤0x\le 0x≤0?

Answer: f−1(x)=−x+4f^{-1}(x)=-\sqrt{x+4}f−1(x)=−x+4​. Solve y=x2−4y=x^2-4y=x2−4 for xxx using the negative square root.

Flashcard 8: What is the inverse of f(x)=x2f(x)=x^2f(x)=x2 if the restricted domain is [−2,0][-2,0][−2,0]?

Answer: f−1(x)=−xf^{-1}(x)=-\sqrt{x}f−1(x)=−x​ for 0≤x≤40\le x\le 40≤x≤4. The range [0,4][0,4][0,4] comes from squaring the domain [−2,0][-2,0][−2,0].

Flashcard 9: What restriction makes f(x)=(x−h)2+kf(x)=(x-h)^2+kf(x)=(x−h)2+k one-to-one using the left branch?

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

Flashcard 10: What is f−1(x)f^{-1}(x)f−1(x) for f(x)=(x−3)2f(x)=(x-3)^2f(x)=(x−3)2 with restricted domain x≤3x\le 3x≤3?

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

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

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

Flashcard 12: What restriction makes f(x)=(x−h)2+kf(x)=(x-h)^2+kf(x)=(x−h)2+k one-to-one using the right branch?

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

Flashcard 13: What restriction makes f(x)=−(x+2)2f(x)=-(x+2)^2f(x)=−(x+2)2 invertible using the right half of the parabola?

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

Flashcard 14: What is f−1(x)f^{-1}(x)f−1(x) for f(x)=x2−4f(x)=x^2-4f(x)=x2−4 with restricted domain x≤0x\le 0x≤0?

Answer: f−1(x)=−x+4f^{-1}(x)=-\sqrt{x+4}f−1(x)=−x+4​. Solve y=x2−4y=x^2-4y=x2−4 for xxx using the negative square root.

Flashcard 15: Identify the restricted domain that makes f(x)=x2f(x)=x^2f(x)=x2 invertible if the domain is [−2,2][-2,2][−2,2].

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

Flashcard 16: What is the range of f(x)=(x−3)2f(x)=(x-3)^2f(x)=(x−3)2 when the domain is restricted to x≥3x\ge 3x≥3?

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

Flashcard 17: What is the range of f−1(x)f^{-1}(x)f−1(x) if fff is restricted to the domain x≤3x\le 3x≤3?

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

Flashcard 18: What is the first algebraic step to find an inverse after restricting the domain?

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

Flashcard 19: What must you check after solving for yyy 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 20: Identify the correct inverse for restricted f(x)=(x−4)2f(x)=(x-4)^2f(x)=(x−4)2 with domain x≥4x\ge 4x≥4.

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

Flashcard 21: Identify the restricted domain that makes f(x)=(x+5)2f(x)=(x+5)^2f(x)=(x+5)2 invertible using the left branch.

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

Flashcard 22: Identify the restricted domain that makes f(x)=−(x−1)2+7f(x)=-(x-1)^2+7f(x)=−(x−1)2+7 invertible using the right branch.

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

Flashcard 23: What is f−1(x)f^{-1}(x)f−1(x) for f(x)=−(x−1)2+7f(x)=-(x-1)^2+7f(x)=−(x−1)2+7 with restricted domain x≥1x\ge 1x≥1?

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

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

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

Flashcard 25: Which restriction makes f(x)=x2+2xf(x)=x^2+2xf(x)=x2+2x invertible by using the increasing side of the parabola?

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

Flashcard 26: Which restriction makes f(x)=x2+2xf(x)=x^2+2xf(x)=x2+2x invertible by using the decreasing side of the parabola?

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

Flashcard 27: What happens to domain and range when taking an inverse f−1f^{-1}f−1?

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

Flashcard 28: What is the relationship between the graphs of fff and f−1f^{-1}f−1?

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

Flashcard 29: 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 30: 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.