How do you find (g o f)(x) when $f (x) = 2 x$ $f(x)=2x$ and $g (x) = x^{3} + x^{2} + 1$ $g(x)=x^3+x^2+1$ ?

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Answer 1 · 2018-04-16T15:34:47

$8 x^{3} + 4 x^{2} + 1$ $color(blue)(8x^3+4x^2+1)$

$(g \circ f) (x) = g (f (x))$ $(g @ f)(x)=g(f(x))$

To find the result we substitute $x = f (x)$ $x=f(x)$ in $g (x)$ $g(x)$ :

$g (x) = x^{3} + x^{2} + 1$ $g(x)=x^3+x^2+1$

$g (f (x)) = {(f (x))}^{3} + {(f (x))}^{2} + 1 = {(2 x)}^{3} + {(2 x)}^{2} + 1$ $g(f(x))=(f(x))^3+(f(x))^2+1=(2x)^3+(2x)^2+1$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =8x^3+4x^2+1$

How do you find (g o f)(x) when f(x)=2xf(x)=2xf(x)=2x and g(x)=x^3+x^2+1g(x)=x3+x2+1g(x)=x^3+x^2+1?