Let $f (x) = x^{2} - 4$ $f(x)=x^2-4$ and $g (x) = \sqrt{x} + 3$ $g(x)=sqrtx+3$ , how do you find g(f(x))?

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Answer 1 · 2016-06-28T17:21:35

$g (f (x)) = \sqrt{x^{2} - 4} + 3$ $g(f(x))=sqrt(x^2-4)+3$

$g (f (x)) = g (x^{2} - 4) = \sqrt{x^{2} - 4} + 3$ $g(f(x))=g(x^2-4)=sqrt(x^2-4)+3$

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