If $f (x) = 3 x^{2}$ $f(x)=3x^2$ and $g (x) = \frac{x - 9}{x + 1}$ $g(x)=(x-9)/(x+1)$ , and $x \neq - 1$ $x!=-1$ , then what would $f (g (x))$ $f(g(x))$ equal? $g (f (x))$ $g(f(x))$ ? $f^{- 1} (x)$ $f^-1(x)$ ? What would the domain, range and zeroes for $f (x)$ $f(x)$ be? What would the domain, range and zeroes for $g (x)$ $g(x)$ be?

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If $f (x) = 3 x^{2}$ $f(x)=3x^2$ and $g (x) = \frac{x - 9}{x + 1}$ $g(x)=(x-9)/(x+1)$ , and $x \neq - 1$ $x!=-1$ , then what would $f (g (x))$ $f(g(x))$ equal? $g (f (x))$ $g(f(x))$ ? $f^{- 1} (x)$ $f^-1(x)$ ? What would the domain, range and zeroes for $f (x)$ $f(x)$ be? What would the domain, range and zeroes for $g (x)$ $g(x)$ be?

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Answer 1 · 2018-05-21T17:46:53

$f (g (x)) = 3 {(\frac{x - 9}{x + 1})}^{2}$ $f(g(x))=3((x-9)/(x+1))^2$
$g (f (x)) = \frac{3 x^{2} - 9}{3 x^{2} + 1}$ $g(f(x))=(3x^2-9)/(3x^2+1)$
$f^{- 1} (x) = \sqrt[]{\frac{x}{3}}$ $f^-1(x)=root()(x/3)$

$D_f={x in RR}$ , $R_f={f(x) in RR;f(x)>=0}$
$D_g={x in RR;x!=-1}$ , $R_g={g(x) in RR;g(x)!=1}$

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