Question

What are the correct answers and why? Simple and concise explanation please!!!

Answer 1

Answer:

$\large\boxed{-\bigg[(x-3)^2+2x\bigg]+1}\\\\\boxed{-x^2+4x-8}$

Step-by-step explanation:

$(g\ \circ\ f)(x)=g\bigg(f(x)\bigg)$

$f(x)=(x-3)^2+2x\\\\g(x)=-x+1\\\\g\ \circ\ f\to\text{put f(x) instead of x in the function g(x)}:\\\\(g\ \circ\ f)(x)=-\bigg[\underbrace{(x-3)^2+2x}_{x}\bigg]+1$

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$-\bigg[(x-3)^2+2x\bigg]+1=-(x-3)^2-2x+1\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=-(x^2-(2)(x)(3)+3^2)-2x+1=-(x^2-6x+9)-2x+1\\\\=-x^2-(-6x)-9-2x+1=-x^2+6x-9-2x+1\\\\\text{combine like terms}\\\\=-x^2+(6x-2x)+(-9+1)=-x^2+4x-8$

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$-x^2+4x-8=-x^2+4x-2^2+2^2-8=-(x^2-4x+2^2)+4-8\\\\=-(\underbrace{x^2-(2)(x)(2)+2^2}_{(a-b)^2=a^2-2ab+b^2})-4=-(x-2)^2-4=-(x+1-3)^2-4$

$In\ (-x+1-3)^2+2x,\ x^2\ will\ be\ positive.\\In\ (-x+1-3)^2+2(-x+1),\ x^2\ will\ be\ positive.$

Answer 2

Answer:

1st 4th

Step-by-step explanation:

Remark

g ° f has the meaning of

"look at g(x) where ever there is an x in g(x) put all of f(x) where that x is."

g(f(x)) = - f(x) + 1

Choice One

g(f(x)) = - [(x - 3)^2 + 2x] + 1 which is exactly what g(f(x)) comes down to.

Choice 4

g(f(x)) = - [(x - 3)^2 + 2x] + 1

g(f(x)) = - [x^2 - 6x + 9] + 1

g(f(x)) = - x^2 + 6x - 9 + 1

g(f(x)) = - x^2 + 6x - 8

The others.

Some of them are

f(g(x))

Number 2 is half that way.

f(g(x)) = (g(x) - 3)^2 + 2(g(x)) is what it should be if f(g(x)) was what you were asked for the 2x has no substitution made. In any event, choice 2 is incorrect.

I'm not sure how three was obtained. It is not correct if one and four are.

Same comment applies to five. I'm not sure how it was obtained.

Number six is f(g(x)) You have it backwards.