Let f(x) = 2x - 5 g(x) = 4x - 1 h(x) = x² + x + 2. Evaluate the indicated function without finding an equation for the function. (fog) (0)
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Intro to Functions & Their Graphs
Problem 69
Textbook Question
Find a. (fog) (x) b. the domain of f o g. f(x) = x/(x+1), g(x) = 4/x
Verified step by step guidance1
First, recall that the composition of functions (fog)(x) means f(g(x)). So, we need to substitute g(x) into the function f.
Given f(x) = \frac{x}{x+1} and g(x) = \frac{4}{x}, substitute g(x) into f(x) to get (fog)(x) = f\left(\frac{4}{x}\right) = \frac{\frac{4}{x}}{\frac{4}{x} + 1}.
Simplify the expression for (fog)(x) by combining the terms in the denominator: \frac{4}{x} + 1 = \frac{4}{x} + \frac{x}{x} = \frac{4 + x}{x}. Then rewrite (fog)(x) as \frac{\frac{4}{x}}{\frac{4 + x}{x}}.
To simplify the complex fraction, multiply numerator and denominator by x to eliminate the denominators inside the fraction: (fog)(x) = \frac{4}{4 + x}.
For the domain of (fog), consider the domain restrictions from both f and g. First, g(x) = \frac{4}{x} requires x \neq 0. Next, f(g(x)) requires the denominator of f(g(x)) to be nonzero, so 4 + x \neq 0, which means x \neq -4. Therefore, the domain of (fog) is all real numbers except x = 0 and x = -4.
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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 applying one function to the result of another, denoted as (f ∘ g)(x) = f(g(x)). It requires substituting the entire expression of g(x) into f(x), which helps in combining two functions into a single expression.
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Domain of a Function
The domain of a function is the set of all input values (x) for which the function is defined. When composing functions, the domain of (f ∘ g) includes all x-values in the domain of g such that g(x) is in the domain of f.
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Restrictions from Denominators
Functions with variables in denominators have restrictions where the denominator cannot be zero. Identifying these values is crucial to determine the domain, especially when composing functions, to avoid division by zero.
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