Given $f (x) = \frac{1}{x - 1}$ $f(x)=1/(x-1)$ , and $g (x) = \frac{2}{x}$ $g(x)=2/x$ , how do you find f(g(x))?

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Given $f (x) = \frac{1}{x - 1}$ $f(x)=1/(x-1)$ , and $g (x) = \frac{2}{x}$ $g(x)=2/x$ , how do you find f(g(x))?

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Answer 1 · 2017-06-04T16:37:04

$f (g (x)) = \frac{x}{2 - x}$ $f(g(x))=x/(2-x)$

Explanation:

To find f(g(x)), simply plug g(x) into x in the equation for f(x):
$f (g (x)) = \frac{1}{(\frac{2}{x}) - 1}$ $f(g(x))=1/((2/x)-1)$
Simplify this by finding a common denominator for the denominator:
$\frac{1}{(\frac{2}{x}) - (\frac{x}{x})}$ $1/((2/x)-(x/x)$
Simplify: $\frac{1}{\frac{2 - x}{x}}$ $1/((2-x)/x)$
Clear the fraction by using keep-change-flip: $1 \cdot (\frac{x}{2 - x})$ $1*(x/(2-x))$
Simplify: $f (g (x)) = \frac{x}{2 - x}$ $f(g(x))=x/(2-x)$

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Given $f (x) = \frac{1}{x - 1}$ $f(x)=1/(x-1)$ , and $g (x) = \frac{2}{x}$ $g(x)=2/x$ , how do you find f(g(x))?

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Explanation:

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Given f(x)=1x−1f(x)=1/(x-1), and g(x)=2xg(x)=2/x, how do you find f(g(x))?

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Given $f (x) = \frac{1}{x - 1}$ $f(x)=1/(x-1)$ , and $g (x) = \frac{2}{x}$ $g(x)=2/x$ , how do you find f(g(x))?