anonymous
  • anonymous
Let f(x) = x^2 + x + 2 and g(x) = 2x^2 + 5. Find f(g(x)). Show each step of your work.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@Vocaloid I just need to learn how to do the problem, not the answer.
Nnesha
  • Nnesha
replace x in f(x) function by g(x)
anonymous
  • anonymous
I did that, I got: (2x^2+5)^2+2x^2+2

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Nnesha
  • Nnesha
you forgot something \[ \huge\rm (2x^2+5)^2+\color{blue}{x}+2\] x = 2x^2 + 5
Nnesha
  • Nnesha
you forgot 5
anonymous
  • anonymous
did I? Sorry.
anonymous
  • anonymous
So to solve, is it 2x^4 or 4x^4
Nnesha
  • Nnesha
\[ \huge\rm (2x^2+5)^2+\color{blue}{2x^2 +5}+2\]
Nnesha
  • Nnesha
\[ \huge\rm \color{reD}{x^2}+\color{blue}{x}+2\] \[ \huge\rm \color{ReD}{(2x^2+5)^2}+\color{blue}{2x^2 +5}+2\] x is 2x^2 + 5
anonymous
  • anonymous
Yes, so now we have to square x.
Nnesha
  • Nnesha
yes( 2x^2 +5)^2 is same as (2x^2 +5) (2x^2 +5) so foil it
anonymous
  • anonymous
\[4x^4+20x^2+25\]
anonymous
  • anonymous
right?
Nnesha
  • Nnesha
yes that's right
anonymous
  • anonymous
So now we combine like terms?
Nnesha
  • Nnesha
yesp
anonymous
  • anonymous
\[4x^4+20x^2+25+2x^2+5+2\]
anonymous
  • anonymous
turns into
anonymous
  • anonymous
\[4x^4+22x^2+32\]
anonymous
  • anonymous
Right?
Nnesha
  • Nnesha
yep
anonymous
  • anonymous
So now I need to factor?
Nnesha
  • Nnesha
nope that's ur answer
anonymous
  • anonymous
Oh, is it? Good, my problem was I tried factoring that and had no idea how to do it.
anonymous
  • anonymous
Thanks! :D
Nnesha
  • Nnesha
my pleasure and yes that's it they wants us to find f(g(x)) not factoring
anonymous
  • anonymous
I fanned, so if I need anything, I'll make sure to tag you.
Nnesha
  • Nnesha
thanks i'll try my best :=)
anonymous
  • anonymous
@Nnesha I have one more problem that seems pretty easy.
anonymous
  • anonymous
I think I may know how to solve it but graphing and radicals are my weakpoint.
anonymous
  • anonymous
Describe how to transform \[(\sqrt[6]{x ^{5}})^{7}\] into an expression with a rational exponent
Nnesha
  • Nnesha
alright remember this exponent rule \[\huge\rm \sqrt[m]{x^n} = x^\frac{ n }{ m }\] you can convert root to exponent form
anonymous
  • anonymous
Yes I know this, it would be \[(x^\frac{ 5 }{ 6 })^7\]
anonymous
  • anonymous
then you multiply 7 and 5?
Nnesha
  • Nnesha
yep! right you already know
Nnesha
  • Nnesha
\[\huge\rm (x^m)^n = x^{ m \times n}\]
anonymous
  • anonymous
\[x^\frac{ 35 }{ 6 }\]
anonymous
  • anonymous
thats a 35/6 btw
Nnesha
  • Nnesha
yep
anonymous
  • anonymous
And that is my answer?
Nnesha
  • Nnesha
yep right
anonymous
  • anonymous
K, thanks. Just wanted to check my answer.
Nnesha
  • Nnesha
your answer is correct you're a mathematician :=)

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