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Ch. 1 - Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 1, Problem 39a

Composition of Functions


In Exercises 39 and 40, find


a. (ƒ ○ g) (-1).


ƒ(x) = 1/x , g(x) = 1/√ x + 2

Verified step by step guidance
1
Understand the composition of functions: The composition (ƒ ○ g)(x) means applying g first and then applying ƒ to the result of g(x).
Substitute x = -1 into the function g(x): g(-1) = 1/√(-1 + 2). Calculate the expression inside the square root first.
Simplify the expression inside g(-1): √(-1 + 2) = √1 = 1. Therefore, g(-1) = 1/1 = 1.
Now apply the function ƒ to the result of g(-1): ƒ(g(-1)) = ƒ(1). Substitute 1 into ƒ(x) = 1/x.
Simplify ƒ(1): ƒ(1) = 1/1 = 1. Therefore, (ƒ ○ g)(-1) = 1.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Function Composition

Function composition involves combining two functions to create a new function. The notation (ƒ ○ g)(x) means to apply function g first and then apply function f to the result of g. This is essential for solving the problem, as it requires evaluating g at a specific input and then using that output as the input for f.
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Evaluating Functions

Evaluating a function means substituting a specific value into the function's formula to find the output. In this case, we need to evaluate g at -1, which requires substituting -1 into the function g(x) = 1/√(x + 2). Understanding how to correctly substitute values is crucial for finding the correct outputs in function composition.
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Domain of Functions

The domain of a function is the set of all possible input values for which the function is defined. For the given functions, we must consider the domains of both f(x) = 1/x and g(x) = 1/√(x + 2) to ensure that the inputs do not lead to undefined expressions, such as division by zero or taking the square root of a negative number.
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