Question

Can anyone help me with number 6 & 8?

Answer 1

$\bf \begin{cases} f(x)=x-3\\ g(x)=x^2+1\\ h(x)=-4x+1\\ -------&--------\\ (h\circ g\circ f)(x)=&(h\circ g(~f(x)~)\\ &h(~~g(~~f(x)~~)~~) \end{cases} \\\\\\ g(~~f(x)~~)=[f(x)]^2+1\implies g(~~f(x)~~)=[x-3]^2+1 \\\\\\ g(~~f(x)~~)=[x^2-6x+9]+1\implies \boxed{g(~~f(x)~~)=x^2-6x+10} \\\\\\ h(~~g(~~f(x)~~)~~)=-4[g(~f(x)~)]+1 \\\\\\ h(~~g(~~f(x)~~)~~)=-4[x^2-6x+10]+1 \\\\\\ h(~~g(~~f(x)~~)~~)=-4x^2+24x-40+1 \\\\\\ h(~~g(~~f(x)~~)~~)=-4x^2+24x-39$

Answer 2

6.

So for this, since the circle is open, you are going to be replacing x in g(x) with -4x + 1. It will be solved as such:

$g(h(x))=(-4x+1)^2+1\\ g(h(x))=16x^2-8x+1+1\\ g(h(x))=16x^2-8x+2$

8.

So firstly, we will need to solve g(f(x)) before we can do h(g(f(x))). So replace x in g(x) with x - 3:

$g(f(x))=(x-3)^2+1\\ g(f(x))=x^2-6x+9+1\\ g(f(x))=x^2-6x+10$

So now that we know what g(f(x)) is, we can solve for h(g(f(x))). To do this, replace x in h(x) with x^2 - 6x + 10:

$h(g(f(x)))=-4(x^2-6x+10)+1\\ h(g(f(x)))=-4x^2+24x-40+1\\ h(g(f(x)))=-4x^2+24x-39$

In short, h(g(f(x))) = -4x^2 + 24x - 39.