Let f[x]=(x^2)+1. Find a function g so that
[a] [fg][x]=(x^4)-1
[b] [f+g][x]=3(x^2)
[c] [f/g][x]=1
[d] f[g[x]]=9(x^4)+1
[e] g[f[x]]=9(x^4)+1
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Let f[x]=(x^2)+1. Find a function g so that
[a] [fg][x]=(x^4)-1
[b] [f+g][x]=3(x^2)
[c] [f/g][x]=1
[d] f[g[x]]=9(x^4)+1
[e] g[f[x]]=9(x^4)+1
Hint: note that if we factored this, it would be the difference of two squares, with f being one factorQuote:
Originally Posted by Godfather
we want $\displaystyle x^2 + 1 + g(x) = 3x^2$Quote:
[b] [f+g][x]=3(x^2)
simply solve for $\displaystyle g(x)$
Hint: any number divided by itself is 1Quote:
[c] [f/g][x]=1
we want $\displaystyle (g(x))^2 + 1 = 9x^4 + 1$Quote:
[d] f[g[x]]=9(x^4)+1
solve for $\displaystyle g(x)$
i can't really think of a nice hint to give you here...Quote:
[e] g[f[x]]=9(x^4)+1
$\displaystyle g(x) = (3x - 3)^2 + 1$
Hint: any number divided by itself is 1
we want $\displaystyle (g(x))^2 + 1 = 9x^4 + 1$
solve for $\displaystyle g(x)$
So it would be X^2 +1 but what is x?
i can't really think of a nice hint to give you here...
$\displaystyle g(x) = (3x - 3)^2 + 1$[/quote]
How did u get that?
What are all of the x's after [f/g][x]
your post makes no sense. separate my quotes from your questions to make sure i don't miss anything.
you said nothing after you quoted this, so i guess you got the answerQuote:
Originally Posted by Godfather
no. and you can check this. if $\displaystyle g(x) = x^2 + 1$ and $\displaystyle f(x) = x^2 + 1$, then $\displaystyle f(g(x)) = \left( x^2 + 1 \right)^2 + 1 = x^4 + 2x^2 + 2$ which is not $\displaystyle 9x^4 + 1$Quote:
(to the f(g(x)) question: So it would be X^2 +1?
do you not know function notation? the x's here just denote that x is the variable. so saying f(x) means that we have a function of x, that is, a function in which the variable is x.Quote:
but what is x?
like i said, i don't know a nice hint to tell you here. but i essentially tried to get f(x) out of g(x). that is, i rewrote g(x) with the purpose of somehow giving rise to the formula for f(x)Quote:
How did u get that?
we started with $\displaystyle g(x) = 9x^4 + 1$
now, $\displaystyle 9x^4 + 1 = \left( 3x^2 \right)^2 + 1$
$\displaystyle = \left[ 3 \left( x^2 {\color{blue} + 1 - 1} \right) \right]^2 + 1$ ...........note that i added and subtracted one from the x^2 factor, this is like adding zero, so i'm not changing anything.
$\displaystyle = \left[ 3 { \color{red} \left( x^2 + 1 \right) } - 3 \right]^2 + 1$ ...now do you see why i added and subtracted 1? when i multiplied out the -1, i am left with f(x) is the brackets. this is what i was after
$\displaystyle = \left[ 3 { \color{red} \left( f(x) \right) } - 3 \right]^2 + 1$
so now, i just let g(x) be the function that has f(x) plugged into it
i answered this alreadyQuote:
What are all of the x's after [f/g][x]
