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anonymous
 one year ago
Let f(x) = 4x2 + x + 1 and g(x) = x2 – 2. Find g(f(x)). Show each step of your work.
Help please
anonymous
 one year ago
Let f(x) = 4x2 + x + 1 and g(x) = x2 – 2. Find g(f(x)). Show each step of your work. Help please

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@marihelenh can you help me with this one too?

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1You would replace what you have for f(x) into the x position of g(x). \[\left( 4x ^{2}+x+1 \right)^{2}2\] Does that make sense?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but where does the x^22 go

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 2 is there but the x^2 i only see the ^2 added but not the x

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1The f(x) becomes a function of g(x), so you just replace what f was equal to into wherever the x is in the g equation.

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1Ok great! It is a little difficult trying to explain on here sometimes.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.04x^4+x^31 would this be the final answer?

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1Just one sec, let me check.

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1No, the answer is actually even longer.

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1To find the answer you would have to do, \[(4x ^{2}+x+1)(4x ^{2}+x+1) 2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is so hard, i thought i understood but now i don't.

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1Do you have to simplify it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It says Find g(f(x)) so I think I have to but maybe not

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1Well, if you don't have to, you already have your answer, but I will help you simplify it. Just hang in with me. It will seem confusing, but we will get through it and to the answer. OK?

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1The first thing you should do, is multiply the first term of the first set by the second set. dw:1436743172709:dw

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1It would be 16x^4 + 4x^3 + 4x^2

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1Then the same thing with the second term dw:1436743368652:dw 4x^3 + x^2 + x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.04 times 4 = 16 then 16^4 times x = 16x^5 or the x doesnt add ^1?

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1No, it doesn't. You add the current exponents, so it would be 2 + 2 = 4.

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1Then you would do it with the third term. dw:1436743540161:dw 4x^2 + x + 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and what happens to the 2?

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1That will get added on in the end when we put all of the terms together.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so final answer is this 16x^4 + 4x^3 + 4x^2  2 ?

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1That is how you multiply the two trinomials out, so then add them all together. \[16x ^{4}+4x ^{3}+4x ^{2}+4x ^{3}+x ^{2}+x+4x ^{2}+x+12\] (I added in the 2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0^that last part is so confusing

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1Add like terms \[16x ^{4}+8x ^{3}+9x ^{2}+2x1\]

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1Do you understand how I got the three different terms on the pictures by multiplying them out?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes that part i understand

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The f(x) becomes a function of g(x) so you just replace what f was equal to into wherever the x is in the g equation. Then you get: (4x^2+x+1)^2(4x^2+x+1)^22 then multiply 4x^2 by all terms of second equation to get 16x^4 now multiply x by all terms of second equation to get 4x^3 repeat now with last term to get 4x^2 you get 16x^4 + 4x^3 + 4x^2 now to add the 2 you add like terms to get 16x^4 + 8x^3 + 9x^2 + 2x  1 would this be good for the answer ?

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1I just put the nine different terms together and then added the 2.

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1Do you just have to explain how to get the answer?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and get the answer too

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1Ok, well then let me just fix what you have, you were pretty close.

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1then multiply 4x^2 by all terms of second equation to get 16x^4+4x^3+4x^2 now multiply x by all terms of second equation to get 4x^3+x^2+x repeat now with last term to get 4x^2+x+1 you get 16x^4+4x^3+4x^2+4x^3+x^2+x+4x^2+x+1 now to add the 2 and add like terms to get 16x^4+8x^3+9x^2+2x1

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1The first three lines were good. You just had to add a little extra into the next ones, but it was all looking good.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i missed a couple of things, thank you so much for your time :D

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.1It's not a problem. I am soooo sorry for how long it took and how confusing it was.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't care as long as I get it, thank you.
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