Can anyone help me with number 6 & 8?

Question

asked Jul 16, 2019 200k views

2 Answers

Umika · Answer 1 · 2019-07-17T19:13:45+0000

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.