If $f (x) = 3 x + 4$ $f(x)=3x+4$ and $g (x) = x^{2} - 1$ $g(x)=x^2-1$ , how do you find $g (g (x))$ $g(g(x))$ ?

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If $f (x) = 3 x + 4$ $f(x)=3x+4$ and $g (x) = x^{2} - 1$ $g(x)=x^2-1$ , how do you find $g (g (x))$ $g(g(x))$ ?

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Answer 1 · 2017-01-29T11:57:56

$x^{4} - 2 x^{2}$ $x^4-2x^2$

Explanation:

To find g(g(x)), substitute $x = x^{2} - 1$ $x=x^2-1$ into g(x)

$g (g (x)) = g (x^{2} - 1) = {(x^{2} - 1)}^{2} - 1$ $g(g(x))=g(color(red)(x^2-1))=(color(red)(x^2-1))^2-1$

$\Rightarrow g (g (x)) = x^{4} - 2 x^{2} + 1 - 1 = x^{4} - 2 x^{2}$ $rArrg(g(x))=x^4-2x^2+1-1=x^4-2x^2$

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If $f (x) = 3 x + 4$ $f(x)=3x+4$ and $g (x) = x^{2} - 1$ $g(x)=x^2-1$ , how do you find $g (g (x))$ $g(g(x))$ ?

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If f(x)=3x+4f(x)=3x+4f(x)=3x+4 and g(x)=x^2-1g(x)=x2−1g(x)=x^2-1, how do you find g(g(x))g(g(x))g(g(x))?

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If $f (x) = 3 x + 4$ $f(x)=3x+4$ and $g (x) = x^{2} - 1$ $g(x)=x^2-1$ , how do you find $g (g (x))$ $g(g(x))$ ?