Respuesta :
Answer:
[tex]h \circ m \circ n \text{ where } h(x)=\sqrt{x} \text{ and } m(x)=1+\frac{15}{n} \text{ and } n(x)=x^3-9[/tex]
Step-by-step explanation:
Ok so I see a square root is on the whole thing.
I'm going to let the very outside function by [tex]h(x)=sqrt(x)[/tex].
Now I'm can't just let the inside function by one function [tex]g(x)=\frac{x^3+6}{x^3-9}[/tex] because we need three functions.
So I'm going to play with [tex]g(x)=\frac{x^3+6}{x^3-9}[/tex] a little to simplify it.
You could do long division. I'm just going to rewrite the top as
[tex]x^3+6=x^3-9+15[/tex].
[tex]g(x)=\frac{x^3-9+15}{x^3-9}=1+\frac{15}{x^3-9}[/tex].
So I'm going to let the next inside function after h be [tex]m(x)=1 + \frac{15}{x}[/tex].
Now my last function will be [tex]n(x)=x^3-9[/tex].
So my order is h(m(n(x))).
Let's check it:
[tex]h(m(x^3-9))[/tex]
[tex]h(1+\frac{15}{x^3-9})[/tex]
[tex]h(\frac{x^3-9+15}{x^3-9})[/tex]
[tex]h(\frac{x^3+6}{x^3-9})[/tex]
[tex]\sqrt{ \frac{x^3+6}{x^3-9}}[/tex]
