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Ch. 1 - Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
1장, 문제 41

In Exercises 41 and 42, (a) write formulas for ƒ ○ g and g ○ ƒ and find the (b) domain and (c) range of each.


ƒ(x) = 2 - x², g(x) = √ x + 2

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1
To find the composition of functions ƒ ○ g, substitute g(x) into ƒ(x). This means replacing every instance of x in ƒ(x) with g(x). So, ƒ ○ g(x) = ƒ(g(x)) = 2 - (√(x + 2))².
Simplify the expression for ƒ ○ g(x). Since (√(x + 2))² simplifies to x + 2, the expression becomes ƒ ○ g(x) = 2 - (x + 2).
To find the composition of functions g ○ ƒ, substitute ƒ(x) into g(x). This means replacing every instance of x in g(x) with ƒ(x). So, g ○ ƒ(x) = g(ƒ(x)) = √(2 - x² + 2).
Determine the domain of ƒ ○ g. The domain of g(x) is x ≥ -2, and for ƒ ○ g(x) to be defined, the expression inside the square root must be non-negative. Therefore, solve x + 2 ≥ 0 to find the domain of ƒ ○ g.
Determine the range of ƒ ○ g. Since ƒ ○ g(x) = 2 - (x + 2), analyze the behavior of this linear function over its domain to find the range. Similarly, analyze the domain and range of g ○ ƒ by considering the restrictions imposed by the square root and the quadratic expression.

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주요 개념

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