If $f (x) = \frac{1}{x}$ $f(x) = 1/x$ , and $g (x) = \sqrt{x - 2}$ $g(x) = sqrt(x-2)$ , how do you find the domain and range of (f o g)(x)?

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If $f (x) = \frac{1}{x}$ $f(x) = 1/x$ , and $g (x) = \sqrt{x - 2}$ $g(x) = sqrt(x-2)$ , how do you find the domain and range of (f o g)(x)?

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Answer 1 · 2016-02-22T07:57:35

D: $(2, + \infty)$ $(2, +oo)$
R: $(0, + \infty)$ $(0, +oo)$

Explanation:

Given : $f (x) = \frac{1}{x}$ $f(x)=1/x$ and $g (x) = \sqrt{x - 2}$ $g(x)=sqrt(x-2)$

$f (g (x)) = \frac{1}{g (x)} = \frac{1}{\sqrt{x - 2}}$ $f (g(x))=1/g(x)=1/sqrt(x-2)$

Range should be $x > 2$ $x>2$ or $(2, \infty)$ $(2, oo)$ . It cannot include 2 because that will make the function undefine because of $\frac{1}{\sqrt{0 - 0}}$ $1/sqrt(0-0)$

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If $f (x) = \frac{1}{x}$ $f(x) = 1/x$ , and $g (x) = \sqrt{x - 2}$ $g(x) = sqrt(x-2)$ , how do you find the domain and range of (f o g)(x)?

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If $f (x) = \frac{1}{x}$ $f(x) = 1/x$ , and $g (x) = \sqrt{x - 2}$ $g(x) = sqrt(x-2)$ , how do you find the domain and range of (f o g)(x)?