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monroe17Best ResponseYou've already chosen the best response.2
fog.. when f(x)=x2 g(x)=x^2+3x+4 =((x^2+3x+4)2) = 2x^26x8
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
f(x)=x2 f(g(x))=g(x)2 f(g(x))=x^2+3x+42 f(g(x))=x^2+3x+2
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
You're not multiplying, you're subtracting in this case
 one year ago

monroe17Best ResponseYou've already chosen the best response.2
oh.. so I don't distribute the 2?
 one year ago

monroe17Best ResponseYou've already chosen the best response.2
so for fof.. would is be.. (x2)2=x4?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
no, you're not distributing since it's originally x2 and not 2x
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
yes, fof is x4
 one year ago

85295jamesBest ResponseYou've already chosen the best response.0
no the roots is X=1 or 2
 one year ago

monroe17Best ResponseYou've already chosen the best response.2
what about ... (x2)^2+3(x2)+4?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
(x2)^2+3(x2)+4 x^24x+4+3(x2)+4 x^24x+4+3x6+4 x^2x+2
 one year ago

85295jamesBest ResponseYou've already chosen the best response.0
the polynomial discriminant: =1
 one year ago

monroe17Best ResponseYou've already chosen the best response.2
how'd you go from (x2)^2+3(x2)+4 to this.. x^24x+4+3(x2)+4 ?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
(x2)^2 (x2)(x2) x(x2)  2(x2) x^2  2x  2x + 4 x^2  4x + 4
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
So (x2)^2 is the same as x^2  4x + 4
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
which makes (x2)^2+3(x2)+4 the same as x^24x+4+3(x2)+4
 one year ago

monroe17Best ResponseYou've already chosen the best response.2
yea, yeah, I wasn't thinking.
 one year ago

monroe17Best ResponseYou've already chosen the best response.2
I'm gonna try to do this one by myself.. and i'll report back what answer I get lol.. (x^2+3x+4)^2+3(x^2+3x+4)+4...
 one year ago

monroe17Best ResponseYou've already chosen the best response.2
is it simplified to.. x^4+3x^2+8+3x^2+9x+12+4? I know that not all the way simplified.. I just wanna know if im sorta on the right track?
 one year ago

monroe17Best ResponseYou've already chosen the best response.2
x^4+3x^2+16+3x^2+9x+12+4..
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
no, you're supposed to have cubed terms in there
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
keep in mind that (x^2+3x+4)^2 (x^2+3x+4)(x^2+3x+4) x^2(x^2+3x+4)+3x(x^2+3x+4)+4(x^2+3x+4)
 one year ago

monroe17Best ResponseYou've already chosen the best response.2
(x^2+3x+4)^2=x^4+6x^3+17x^2+24x+16?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
that is correct, so (x^2+3x+4)^2+3(x^2+3x+4)+4 becomes x^4+6x^3+17x^2+24x+16+3(x^2+3x+4)+4
 one year ago

monroe17Best ResponseYou've already chosen the best response.2
x^4+6x^3+20x^2+33x+24?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
x^4+6x^3+20x^2+33x+32 is the correct final answer
 one year ago

monroe17Best ResponseYou've already chosen the best response.2
my goodness! finally lol! Now, one final question... How do I find the domain of those... :/
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
The domain of any polynomial is the set of all real numbers.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
since you can plug in any real number into a polynomial function and get a real number out.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
If f(x) and g(x) are both polynomial functions, then the following are also polynomial functions f(g(x)), g(f(x)), f(f(x)), and g(g(x)) or any combination of composition you can think of are all polynomial functions
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
So the domain of each composite function is the set of all real numbers.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
Notice nowhere do we have to worry about a) dividing by zero or b) taking the square root of negative numbers. So there are no restrictions on x.
 one year ago

monroe17Best ResponseYou've already chosen the best response.2
so the domain for all four polynomials is (inf,inf)?
 one year ago

monroe17Best ResponseYou've already chosen the best response.2
phew.. I thought it was gonna b longer then that. lol! Thank you so much. so for any polynomial.. the domain will be (inf,inf) because there are no restrictions?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
Yes, any number will do for the input of a polynomial. So the domain in interval notation for any polynomial is (inf,inf)
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
You'll only run into domain restrictions if you deal with division of variables or having variables in square roots (or logs)
 one year ago

monroe17Best ResponseYou've already chosen the best response.2
okay, I gotta remember that :) thank you!
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
you're welcome
 one year ago
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