Let $f (x) = \frac{1}{1 - 3 x}$ $f(x) = (1) / (1-3x)$ and $g (x) = \frac{1}{x^{2}}$ $g(x) = (1) / (x^2)$ how do you find f(g(x)?

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Let $f (x) = \frac{1}{1 - 3 x}$ $f(x) = (1) / (1-3x)$ and $g (x) = \frac{1}{x^{2}}$ $g(x) = (1) / (x^2)$ how do you find f(g(x)?

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Answer 1 · 2015-10-23T22:56:48

Put $g (x)$ $g(x)$ in place of $x$ $x$ in the formula for $f (x)$ $f(x)$ and simplify to find:

$f (g (x)) = 1 + \frac{3}{x^{2} - 3}$ $f(g(x)) = 1+3/(x^2-3)$

Explanation:

$f (g (x))$ $f(g(x))$

$= \frac{1}{1 - 3 g (x)}$ $= 1/(1-3g(x))$

$= \frac{1}{1 - 3 (\frac{1}{x^{2}})}$ $= 1/(1-3(1/(x^2)))$

$= \frac{x^{2}}{x^{2} - 3}$ $= x^2/(x^2-3)$

$= \frac{x^{2} - 3 + 3}{x^{2} - 3}$ $= (x^2-3+3)/(x^2-3)$

$= 1 + \frac{3}{x^{2} - 3}$ $= 1+3/(x^2-3)$

with restriction $x \neq 0$ $x != 0$

The restriction is necessary because $g (x)$ $g(x)$ is undefined for $x = 0$ $x = 0$ , but $1 + \frac{1}{x^{2} - 3}$ $1+1/(x^2-3)$ is normally defined when $x = 0$ $x = 0$ .

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Let $f (x) = \frac{1}{1 - 3 x}$ $f(x) = (1) / (1-3x)$ and $g (x) = \frac{1}{x^{2}}$ $g(x) = (1) / (x^2)$ how do you find f(g(x)?

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Explanation:

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Let $f (x) = \frac{1}{1 - 3 x}$ $f(x) = (1) / (1-3x)$ and $g (x) = \frac{1}{x^{2}}$ $g(x) = (1) / (x^2)$ how do you find f(g(x)?