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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- October 13th 2007, 08:08 PMGodfatherFind a Function G and F[x]
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 - October 13th 2007, 08:30 PMJhevon
**Hint:**note that if we factored this, it would be the difference of two squares, with f being one factor

Quote:

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

simply solve for

Quote:

[c] [f/g][x]=1

**Hint:**any number divided by itself is 1

Quote:

[d] f[g[x]]=9(x^4)+1

solve for

Quote:

[e] g[f[x]]=9(x^4)+1

- October 13th 2007, 08:36 PMGodfather
**Hint:**any number divided by itself is 1

we want

solve for

So it would be X^2 +1 but what is x?

i can't really think of a nice hint to give you here...

[/quote]

How did u get that?

What are all of the x's after [f/g]**[x]** - October 13th 2007, 08:51 PMJhevon
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 answer

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(to the f(g(x)) question: So it would be X^2 +1?

Quote:

but what is x?

Quote:

How did u get that?

we started with

now,

...........note that i added and subtracted one from the x^2 factor, this is like adding zero, so i'm not changing anything.

...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

so now, i just let g(x) be the function that has f(x) plugged into it

Quote:

What are all of the x's after [f/g]**[x]**