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

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Answer 1 · 2015-12-03T13:37:48

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

$f (g (x))$ $f(g(x))$ basically means that $g (x)$ $g(x)$ is being taken as an "input" for $f$ $f$ instead of the usual "input" $x$ $x$ .

So, to compute $f (g (x))$ $f(g(x))$ , you need to plug $g (x)$ $g(x)$ for every occurance of $x$ $x$ in $f (x)$ $f(x)$ :

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

If f(x)= 1-x^3f(x)=1−x3f(x)= 1-x^3 and g(x)= 1/xg(x)=1xg(x)= 1/x how do you find f(g(x))?