Given $f (x) = \frac{6}{x^{2}}$ $f(x)=6/x^2$ and $g (x) = x - 3$ $g(x)=x-3$ , how do you find g(f(x))?

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Answer 1 · 2016-01-30T17:02:03

To make a composition of a function you must plug one function into another, or on this case $f$ $f$ into g.

$f (x)$ $f(x)$ = $\frac{6}{x^{2}}$ $6/x^2$

Plug in $f$ $f$ for x in g, since

g( $\frac{6}{x^{2}}$ $6/x^2$ ) = $\frac{6}{x^{2}}$ $6/x^2$ - 3

So, $g (f (x))$ $g(f(x))$ = $\frac{6}{x^{2}}$ $6/x^2$ - 3

Practice exercises:

Knowing that $f (x)$ $f(x)$ = $2 x^{2}$ $2x^2$ + 15x + 22 and that $g (x)$ $g(x)$ = $- \frac{5}{x}$ $-5/x$ ,
find:

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

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

c) $f (g (- 2))$ $f(g(-2))$

d). $g (f (7))$ $g(f(7))$

Hopefully this helps.

Given f(x)=6/x^2f(x)=6x2f(x)=6/x^2 and g(x)=x-3g(x)=x−3g(x)=x-3, how do you find g(f(x))?