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Ch. 2 - Functions and Graphs
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 69

Find a. (fog) (x) b. the domain of f o g. f(x) = x/(x+1), g(x) = 4/x

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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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