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Algebra 2 Flashcards: Verify Functions Are Inverses

Study Verify Functions Are Inverses 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 Verify Functions Are Inverses, 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.

Algebra 2 Flashcards: Verify Functions Are Inverses

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QUESTION

What is g(f(x))g(f(x))g(f(x)) if f(x)=x2f(x)=x^2f(x)=x2 and g(x)=xg(x)=\sqrt{x}g(x)=x​ without restricting fff to x≥0x\ge 0x≥0?

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ANSWER

g(f(x))=∣x∣g(f(x))=|x|g(f(x))=∣x∣. Square root returns absolute value for all real inputs.

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Flashcard 1: What is g(f(x))g(f(x))g(f(x)) if f(x)=x2f(x)=x^2f(x)=x2 and g(x)=xg(x)=\sqrt{x}g(x)=x​ without restricting fff to x≥0x\ge 0x≥0?

Answer: g(f(x))=∣x∣g(f(x))=|x|g(f(x))=∣x∣. Square root returns absolute value for all real inputs.

Flashcard 2: What is g(f(x))g(f(x))g(f(x)) if f(x)=2x−5f(x)=2x-5f(x)=2x−5 and g(x)=x+52g(x)=\frac{x+5}{2}g(x)=2x+5​?

Answer: g(f(x))=xg(f(x))=xg(f(x))=x. Substituting: g(2x−5)=(2x−5)+52=xg(2x-5)=\frac{(2x-5)+5}{2}=xg(2x−5)=2(2x−5)+5​=x.

Flashcard 3: Identify the result of composing f(x)=x+7f(x)=x+7f(x)=x+7 and g(x)=x−7g(x)=x-7g(x)=x−7 as f(g(x))f(g(x))f(g(x)).

Answer: f(g(x))=xf(g(x))=xf(g(x))=x. Substituting: f(x−7)=(x−7)+7=xf(x-7)=(x-7)+7=xf(x−7)=(x−7)+7=x.

Flashcard 4: What is the inverse relationship between domain and range for inverse functions?

Answer: Dom(f)=Ran(g)\text{Dom}(f)=\text{Ran}(g)Dom(f)=Ran(g) and Ran(f)=Dom(g)\text{Ran}(f)=\text{Dom}(g)Ran(f)=Dom(g). Domain and range swap between inverse functions.

Flashcard 5: Identify f(g(x))f(g(x))f(g(x)) if f(x)=1xf(x)=\frac{1}{x}f(x)=x1​ and g(x)=1xg(x)=\frac{1}{x}g(x)=x1​ with x≠0x\ne 0x=0.

Answer: f(g(x))=xf(g(x))=xf(g(x))=x for x≠0x\ne 0x=0. Substituting: f(1x)=11x=xf(\frac{1}{x})=\frac{1}{\frac{1}{x}}=xf(x1​)=x1​1​=x.

Flashcard 6: What coordinate swap describes the inverse relationship between points on fff and f−1f^{-1}f−1?

Answer: If (a,b)(a,b)(a,b) is on fff, then (b,a)(b,a)(b,a) is on f−1f^{-1}f−1. Coordinates swap between function and inverse.

Flashcard 7: What is g(f(x))g(f(x))g(f(x)) if f(x)=3xf(x)=3xf(x)=3x and g(x)=x3g(x)=\frac{x}{3}g(x)=3x​?

Answer: g(f(x))=xg(f(x))=xg(f(x))=x. Substituting: g(3x)=3x3=xg(3x)=\frac{3x}{3}=xg(3x)=33x​=x.

Flashcard 8: What must be checked about domains when verifying inverses by composition?

Answer: Each composition must equal xxx for all xxx in its stated domain. Compositions must work on their proper domains.

Flashcard 9: What test is commonly used to decide whether a function is one-to-one before finding an inverse?

Answer: The horizontal line test. Checks if every horizontal line intersects once.

Flashcard 10: Identify the result of composing f(x)=x+7f(x)=x+7f(x)=x+7 and g(x)=x−7g(x)=x-7g(x)=x−7 as g(f(x))g(f(x))g(f(x)).

Answer: g(f(x))=xg(f(x))=xg(f(x))=x. Substituting: g(x+7)=(x+7)−7=xg(x+7)=(x+7)-7=xg(x+7)=(x+7)−7=x.

Flashcard 11: What is g(f(x))g(f(x))g(f(x)) if f(x)=2x−5f(x)=2x-5f(x)=2x−5 and g(x)=x+52g(x)=\frac{x+5}{2}g(x)=2x+5​?

Answer: g(f(x))=xg(f(x))=xg(f(x))=x. Substituting: g(2x−5)=(2x−5)+52=xg(2x-5)=\frac{(2x-5)+5}{2}=xg(2x−5)=2(2x−5)+5​=x.

Flashcard 12: What is g(f(x))g(f(x))g(f(x)) if f(x)=x−45f(x)=\frac{x-4}{5}f(x)=5x−4​ and g(x)=5x+4g(x)=5x+4g(x)=5x+4?

Answer: g(f(x))=xg(f(x))=xg(f(x))=x. Substituting: g(x−45)=5⋅x−45+4=xg(\frac{x-4}{5})=5\cdot\frac{x-4}{5}+4=xg(5x−4​)=5⋅5x−4​+4=x.

Flashcard 13: What is g(f(x))g(f(x))g(f(x)) if f(x)=x3f(x)=x^3f(x)=x3 and g(x)=x3g(x)=\sqrt[3]{x}g(x)=3x​?

Answer: g(f(x))=xg(f(x))=xg(f(x))=x. Substituting: g(x3)=x33=xg(x^3)=\sqrt[3]{x^3}=xg(x3)=3x3​=x.

Flashcard 14: What is the key simplification goal when checking f(g(x))f(g(x))f(g(x)) to verify inverses?

Answer: Simplify until the result is exactly xxx. The composition should reduce to the identity.

Flashcard 15: What is g(f(x))g(f(x))g(f(x)) if f(x)=4xf(x)=\frac{4}{x}f(x)=x4​ and g(x)=4xg(x)=\frac{4}{x}g(x)=x4​ with x≠0x\ne 0x=0?

Answer: g(f(x))=xg(f(x))=xg(f(x))=x for x≠0x\ne 0x=0. Substituting: g(4x)=44x=xg(\frac{4}{x})=\frac{4}{\frac{4}{x}}=xg(x4​)=x4​4​=x.

Flashcard 16: Identify f(g(x))f(g(x))f(g(x)) if f(x)=ln⁡(x)f(x)=\ln(x)f(x)=ln(x) and g(x)=exg(x)=e^xg(x)=ex.

Answer: f(g(x))=xf(g(x))=xf(g(x))=x. Natural log and exponential are inverses.

Flashcard 17: Identify f(g(x))f(g(x))f(g(x)) if f(x)=exf(x)=e^xf(x)=ex and g(x)=ln⁡(x)g(x)=\ln(x)g(x)=ln(x).

Answer: f(g(x))=xf(g(x))=xf(g(x))=x for x>0x>0x>0. Domain restriction needed for ln⁡(x)\ln(x)ln(x).

Flashcard 18: Identify g(f(x))g(f(x))g(f(x)) if f(x)=exf(x)=e^xf(x)=ex and g(x)=ln⁡(x)g(x)=\ln(x)g(x)=ln(x).

Answer: g(f(x))=xg(f(x))=xg(f(x))=x. Exponential and natural log are inverses.

Flashcard 19: What is f(g(x))f(g(x))f(g(x)) if f(x)=2xf(x)=2^xf(x)=2x and g(x)=log⁡2(x)g(x)=\log_2(x)g(x)=log2​(x)?

Answer: f(g(x))=xf(g(x))=xf(g(x))=x for x>0x>0x>0. Domain restriction needed for log⁡2(x)\log_2(x)log2​(x).

Flashcard 20: What is g(f(x))g(f(x))g(f(x)) if f(x)=2xf(x)=2^xf(x)=2x and g(x)=log⁡2(x)g(x)=\log_2(x)g(x)=log2​(x)?

Answer: g(f(x))=xg(f(x))=xg(f(x))=x. Exponential and logarithm base 2 are inverses.

Flashcard 21: What is f(g(x))f(g(x))f(g(x)) if f(x)=x3+1f(x)=x^3+1f(x)=x3+1 and g(x)=x−13g(x)=\sqrt[3]{x-1}g(x)=3x−1​?

Answer: f(g(x))=xf(g(x))=xf(g(x))=x. Substituting: f(x−13)=(x−13)3+1=xf(\sqrt[3]{x-1})=(\sqrt[3]{x-1})^3+1=xf(3x−1​)=(3x−1​)3+1=x.

Flashcard 22: What is g(f(x))g(f(x))g(f(x)) if f(x)=x3+1f(x)=x^3+1f(x)=x3+1 and g(x)=x−13g(x)=\sqrt[3]{x-1}g(x)=3x−1​?

Answer: g(f(x))=xg(f(x))=xg(f(x))=x. Substituting: g(x3+1)=(x3+1)−13=xg(x^3+1)=\sqrt[3]{(x^3+1)-1}=xg(x3+1)=3(x3+1)−1​=x.

Flashcard 23: What is g(f(x))g(f(x))g(f(x)) if f(x)=3xf(x)=3xf(x)=3x and g(x)=x3g(x)=\frac{x}{3}g(x)=3x​?

Answer: g(f(x))=xg(f(x))=xg(f(x))=x. Substituting: g(3x)=3x3=xg(3x)=\frac{3x}{3}=xg(3x)=33x​=x.

Flashcard 24: What is f(g(x))f(g(x))f(g(x)) if f(x)=x−12f(x)=\frac{x-1}{2}f(x)=2x−1​ and g(x)=2x+1g(x)=2x+1g(x)=2x+1?

Answer: f(g(x))=xf(g(x))=xf(g(x))=x. Substituting: f(2x+1)=(2x+1)−12=xf(2x+1)=\frac{(2x+1)-1}{2}=xf(2x+1)=2(2x+1)−1​=x.

Flashcard 25: What is g(f(x))g(f(x))g(f(x)) if f(x)=x−12f(x)=\frac{x-1}{2}f(x)=2x−1​ and g(x)=2x+1g(x)=2x+1g(x)=2x+1?

Answer: g(f(x))=xg(f(x))=xg(f(x))=x. Substituting: g(x−12)=2⋅x−12+1=xg(\frac{x-1}{2})=2\cdot\frac{x-1}{2}+1=xg(2x−1​)=2⋅2x−1​+1=x.

Flashcard 26: Identify f(g(x))f(g(x))f(g(x)) if f(x)=exf(x)=e^xf(x)=ex and g(x)=ln⁡(x)g(x)=\ln(x)g(x)=ln(x).

Answer: f(g(x))=xf(g(x))=xf(g(x))=x for x>0x>0x>0. Domain restriction needed for ln⁡(x)\ln(x)ln(x).

Flashcard 27: What is f(g(x))f(g(x))f(g(x)) if f(x)=xf(x)=\sqrt{x}f(x)=x​ and g(x)=x2g(x)=x^2g(x)=x2 with domain x≥0x\ge 0x≥0 for ggg?

Answer: f(g(x))=xf(g(x))=xf(g(x))=x for x≥0x\ge 0x≥0. Substituting: f(x2)=x2=xf(x^2)=\sqrt{x^2}=xf(x2)=x2​=x when x≥0x\ge 0x≥0.

Flashcard 28: What is g(f(x))g(f(x))g(f(x)) if f(x)=xf(x)=\sqrt{x}f(x)=x​ and g(x)=x2g(x)=x^2g(x)=x2 with domain x≥0x\ge 0x≥0 for fff?

Answer: g(f(x))=xg(f(x))=xg(f(x))=x for x≥0x\ge 0x≥0. Substituting: g(x)=(x)2=xg(\sqrt{x})=(\sqrt{x})^2=xg(x​)=(x​)2=x when x≥0x\ge 0x≥0.

Flashcard 29: What is f(g(x))f(g(x))f(g(x)) if f(x)=x2f(x)=x^2f(x)=x2 and g(x)=xg(x)=\sqrt{x}g(x)=x​ with domain restriction x≥0x\ge 0x≥0 for fff?

Answer: f(g(x))=xf(g(x))=xf(g(x))=x for x≥0x\ge 0x≥0. Substituting: f(x)=(x)2=xf(\sqrt{x})=(\sqrt{x})^2=xf(x​)=(x​)2=x when x≥0x\ge 0x≥0.

Flashcard 30: What is g(f(x))g(f(x))g(f(x)) if f(x)=x2f(x)=x^2f(x)=x2 and g(x)=xg(x)=\sqrt{x}g(x)=x​ with domain restriction x≥0x\ge 0x≥0 for fff?

Answer: g(f(x))=xg(f(x))=xg(f(x))=x for x≥0x\ge 0x≥0. Substituting: g(x2)=x2=xg(x^2)=\sqrt{x^2}=xg(x2)=x2​=x when x≥0x\ge 0x≥0.