All flashcards Flashcard 1: What is g ( f ( x ) ) g(f(x)) g ( f ( x )) if f ( x ) = x 2 f(x)=x^2 f ( x ) = x 2 and g ( x ) = x g(x)=\sqrt{x} g ( x ) = x without restricting f f f to x ≥ 0 x\ge 0 x ≥ 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 ) = 2 x − 5 f(x)=2x-5 f ( x ) = 2 x − 5 and g ( x ) = x + 5 2 g(x)=\frac{x+5}{2} g ( x ) = 2 x + 5 ? Answer: g ( f ( x ) ) = x g(f(x))=x g ( f ( x )) = x . Substituting: g ( 2 x − 5 ) = ( 2 x − 5 ) + 5 2 = x g(2x-5)=\frac{(2x-5)+5}{2}=x g ( 2 x − 5 ) = 2 ( 2 x − 5 ) + 5 = x .
Flashcard 3: Identify the result of composing f ( x ) = x + 7 f(x)=x+7 f ( x ) = x + 7 and g ( x ) = x − 7 g(x)=x-7 g ( x ) = x − 7 as f ( g ( x ) ) f(g(x)) f ( g ( x )) . Answer: f ( g ( x ) ) = x f(g(x))=x f ( g ( x )) = x . Substituting: f ( x − 7 ) = ( x − 7 ) + 7 = x f(x-7)=(x-7)+7=x f ( 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 ) = 1 x f(x)=\frac{1}{x} f ( x ) = x 1 and g ( x ) = 1 x g(x)=\frac{1}{x} g ( x ) = x 1 with x ≠ 0 x\ne 0 x = 0 . Answer: f ( g ( x ) ) = x f(g(x))=x f ( g ( x )) = x for x ≠ 0 x\ne 0 x = 0 . Substituting: f ( 1 x ) = 1 1 x = x f(\frac{1}{x})=\frac{1}{\frac{1}{x}}=x f ( x 1 ) = x 1 1 = x .
Flashcard 6: What coordinate swap describes the inverse relationship between points on f f f and f − 1 f^{-1} f − 1 ? Answer: If ( a , b ) (a,b) ( a , b ) is on f f f , then ( b , a ) (b,a) ( b , a ) is on f − 1 f^{-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 ) = 3 x f(x)=3x f ( x ) = 3 x and g ( x ) = x 3 g(x)=\frac{x}{3} g ( x ) = 3 x ? Answer: g ( f ( x ) ) = x g(f(x))=x g ( f ( x )) = x . Substituting: g ( 3 x ) = 3 x 3 = x g(3x)=\frac{3x}{3}=x g ( 3 x ) = 3 3 x = x .
Flashcard 8: What must be checked about domains when verifying inverses by composition? Answer: Each composition must equal x x x for all x x x 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 + 7 f(x)=x+7 f ( x ) = x + 7 and g ( x ) = x − 7 g(x)=x-7 g ( x ) = x − 7 as g ( f ( x ) ) g(f(x)) g ( f ( x )) . Answer: g ( f ( x ) ) = x g(f(x))=x g ( f ( x )) = x . Substituting: g ( x + 7 ) = ( x + 7 ) − 7 = x g(x+7)=(x+7)-7=x g ( x + 7 ) = ( x + 7 ) − 7 = x .
Flashcard 11: What is g ( f ( x ) ) g(f(x)) g ( f ( x )) if f ( x ) = 2 x − 5 f(x)=2x-5 f ( x ) = 2 x − 5 and g ( x ) = x + 5 2 g(x)=\frac{x+5}{2} g ( x ) = 2 x + 5 ? Answer: g ( f ( x ) ) = x g(f(x))=x g ( f ( x )) = x . Substituting: g ( 2 x − 5 ) = ( 2 x − 5 ) + 5 2 = x g(2x-5)=\frac{(2x-5)+5}{2}=x g ( 2 x − 5 ) = 2 ( 2 x − 5 ) + 5 = x .
Flashcard 12: What is g ( f ( x ) ) g(f(x)) g ( f ( x )) if f ( x ) = x − 4 5 f(x)=\frac{x-4}{5} f ( x ) = 5 x − 4 and g ( x ) = 5 x + 4 g(x)=5x+4 g ( x ) = 5 x + 4 ? Answer: g ( f ( x ) ) = x g(f(x))=x g ( f ( x )) = x . Substituting: g ( x − 4 5 ) = 5 ⋅ x − 4 5 + 4 = x g(\frac{x-4}{5})=5\cdot\frac{x-4}{5}+4=x g ( 5 x − 4 ) = 5 ⋅ 5 x − 4 + 4 = x .
Flashcard 13: What is g ( f ( x ) ) g(f(x)) g ( f ( x )) if f ( x ) = x 3 f(x)=x^3 f ( x ) = x 3 and g ( x ) = x 3 g(x)=\sqrt[3]{x} g ( x ) = 3 x ? Answer: g ( f ( x ) ) = x g(f(x))=x g ( f ( x )) = x . Substituting: g ( x 3 ) = x 3 3 = x g(x^3)=\sqrt[3]{x^3}=x g ( x 3 ) = 3 x 3 = 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 x x x . The composition should reduce to the identity.
Flashcard 15: What is g ( f ( x ) ) g(f(x)) g ( f ( x )) if f ( x ) = 4 x f(x)=\frac{4}{x} f ( x ) = x 4 and g ( x ) = 4 x g(x)=\frac{4}{x} g ( x ) = x 4 with x ≠ 0 x\ne 0 x = 0 ? Answer: g ( f ( x ) ) = x g(f(x))=x g ( f ( x )) = x for x ≠ 0 x\ne 0 x = 0 . Substituting: g ( 4 x ) = 4 4 x = x g(\frac{4}{x})=\frac{4}{\frac{4}{x}}=x g ( x 4 ) = x 4 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 ) = e x g(x)=e^x g ( x ) = e x . Answer: f ( g ( x ) ) = x f(g(x))=x f ( g ( x )) = x . Natural log and exponential are inverses.
Flashcard 17: Identify f ( g ( x ) ) f(g(x)) f ( g ( x )) if f ( x ) = e x f(x)=e^x f ( x ) = e x and g ( x ) = ln ( x ) g(x)=\ln(x) g ( x ) = ln ( x ) . Answer: f ( g ( x ) ) = x f(g(x))=x f ( g ( x )) = x for x > 0 x>0 x > 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 ) = e x f(x)=e^x f ( x ) = e x and g ( x ) = ln ( x ) g(x)=\ln(x) g ( x ) = ln ( x ) . Answer: g ( f ( x ) ) = x g(f(x))=x g ( 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 ) = 2 x f(x)=2^x f ( x ) = 2 x and g ( x ) = log 2 ( x ) g(x)=\log_2(x) g ( x ) = log 2 ( x ) ? Answer: f ( g ( x ) ) = x f(g(x))=x f ( g ( x )) = x for x > 0 x>0 x > 0 . Domain restriction needed for log 2 ( x ) \log_2(x) log 2 ( x ) .
Flashcard 20: What is g ( f ( x ) ) g(f(x)) g ( f ( x )) if f ( x ) = 2 x f(x)=2^x f ( x ) = 2 x and g ( x ) = log 2 ( x ) g(x)=\log_2(x) g ( x ) = log 2 ( x ) ? Answer: g ( f ( x ) ) = x g(f(x))=x g ( 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 ) = x 3 + 1 f(x)=x^3+1 f ( x ) = x 3 + 1 and g ( x ) = x − 1 3 g(x)=\sqrt[3]{x-1} g ( x ) = 3 x − 1 ? Answer: f ( g ( x ) ) = x f(g(x))=x f ( g ( x )) = x . Substituting: f ( x − 1 3 ) = ( x − 1 3 ) 3 + 1 = x f(\sqrt[3]{x-1})=(\sqrt[3]{x-1})^3+1=x f ( 3 x − 1 ) = ( 3 x − 1 ) 3 + 1 = x .
Flashcard 22: What is g ( f ( x ) ) g(f(x)) g ( f ( x )) if f ( x ) = x 3 + 1 f(x)=x^3+1 f ( x ) = x 3 + 1 and g ( x ) = x − 1 3 g(x)=\sqrt[3]{x-1} g ( x ) = 3 x − 1 ? Answer: g ( f ( x ) ) = x g(f(x))=x g ( f ( x )) = x . Substituting: g ( x 3 + 1 ) = ( x 3 + 1 ) − 1 3 = x g(x^3+1)=\sqrt[3]{(x^3+1)-1}=x g ( x 3 + 1 ) = 3 ( x 3 + 1 ) − 1 = x .
Flashcard 23: What is g ( f ( x ) ) g(f(x)) g ( f ( x )) if f ( x ) = 3 x f(x)=3x f ( x ) = 3 x and g ( x ) = x 3 g(x)=\frac{x}{3} g ( x ) = 3 x ? Answer: g ( f ( x ) ) = x g(f(x))=x g ( f ( x )) = x . Substituting: g ( 3 x ) = 3 x 3 = x g(3x)=\frac{3x}{3}=x g ( 3 x ) = 3 3 x = x .
Flashcard 24: What is f ( g ( x ) ) f(g(x)) f ( g ( x )) if f ( x ) = x − 1 2 f(x)=\frac{x-1}{2} f ( x ) = 2 x − 1 and g ( x ) = 2 x + 1 g(x)=2x+1 g ( x ) = 2 x + 1 ? Answer: f ( g ( x ) ) = x f(g(x))=x f ( g ( x )) = x . Substituting: f ( 2 x + 1 ) = ( 2 x + 1 ) − 1 2 = x f(2x+1)=\frac{(2x+1)-1}{2}=x f ( 2 x + 1 ) = 2 ( 2 x + 1 ) − 1 = x .
Flashcard 25: What is g ( f ( x ) ) g(f(x)) g ( f ( x )) if f ( x ) = x − 1 2 f(x)=\frac{x-1}{2} f ( x ) = 2 x − 1 and g ( x ) = 2 x + 1 g(x)=2x+1 g ( x ) = 2 x + 1 ? Answer: g ( f ( x ) ) = x g(f(x))=x g ( f ( x )) = x . Substituting: g ( x − 1 2 ) = 2 ⋅ x − 1 2 + 1 = x g(\frac{x-1}{2})=2\cdot\frac{x-1}{2}+1=x g ( 2 x − 1 ) = 2 ⋅ 2 x − 1 + 1 = x .
Flashcard 26: Identify f ( g ( x ) ) f(g(x)) f ( g ( x )) if f ( x ) = e x f(x)=e^x f ( x ) = e x and g ( x ) = ln ( x ) g(x)=\ln(x) g ( x ) = ln ( x ) . Answer: f ( g ( x ) ) = x f(g(x))=x f ( g ( x )) = x for x > 0 x>0 x > 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 ) = x f(x)=\sqrt{x} f ( x ) = x and g ( x ) = x 2 g(x)=x^2 g ( x ) = x 2 with domain x ≥ 0 x\ge 0 x ≥ 0 for g g g ? Answer: f ( g ( x ) ) = x f(g(x))=x f ( g ( x )) = x for x ≥ 0 x\ge 0 x ≥ 0 . Substituting: f ( x 2 ) = x 2 = x f(x^2)=\sqrt{x^2}=x f ( x 2 ) = x 2 = x when x ≥ 0 x\ge 0 x ≥ 0 .
Flashcard 28: What is g ( f ( x ) ) g(f(x)) g ( f ( x )) if f ( x ) = x f(x)=\sqrt{x} f ( x ) = x and g ( x ) = x 2 g(x)=x^2 g ( x ) = x 2 with domain x ≥ 0 x\ge 0 x ≥ 0 for f f f ? Answer: g ( f ( x ) ) = x g(f(x))=x g ( f ( x )) = x for x ≥ 0 x\ge 0 x ≥ 0 . Substituting: g ( x ) = ( x ) 2 = x g(\sqrt{x})=(\sqrt{x})^2=x g ( x ) = ( x ) 2 = x when x ≥ 0 x\ge 0 x ≥ 0 .
Flashcard 29: What is f ( g ( x ) ) f(g(x)) f ( g ( x )) if f ( x ) = x 2 f(x)=x^2 f ( x ) = x 2 and g ( x ) = x g(x)=\sqrt{x} g ( x ) = x with domain restriction x ≥ 0 x\ge 0 x ≥ 0 for f f f ? Answer: f ( g ( x ) ) = x f(g(x))=x f ( g ( x )) = x for x ≥ 0 x\ge 0 x ≥ 0 . Substituting: f ( x ) = ( x ) 2 = x f(\sqrt{x})=(\sqrt{x})^2=x f ( x ) = ( x ) 2 = x when x ≥ 0 x\ge 0 x ≥ 0 .
Flashcard 30: What is g ( f ( x ) ) g(f(x)) g ( f ( x )) if f ( x ) = x 2 f(x)=x^2 f ( x ) = x 2 and g ( x ) = x g(x)=\sqrt{x} g ( x ) = x with domain restriction x ≥ 0 x\ge 0 x ≥ 0 for f f f ? Answer: g ( f ( x ) ) = x g(f(x))=x g ( f ( x )) = x for x ≥ 0 x\ge 0 x ≥ 0 . Substituting: g ( x 2 ) = x 2 = x g(x^2)=\sqrt{x^2}=x g ( x 2 ) = x 2 = x when x ≥ 0 x\ge 0 x ≥ 0 .