Find a Function G and F[x]

**Godfather** · October 13th 2007, 08:08 PM

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

**Jhevon** · October 13th 2007, 08:30 PM

Originally Posted by Godfather

Let f[x]=(x^2)+1. Find a function g so that
[a] [fg][x]=(x^4)-1

Hint: note that if we factored this, it would be the difference of two squares, with f being one factor

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

we want $x^2 + 1 + g(x) = 3x^2$

simply solve for $g(x)$

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

Hint: any number divided by itself is 1

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

we want $(g(x))^2 + 1 = 9x^4 + 1$

solve for $g(x)$

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

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

$g(x) = (3x - 3)^2 + 1$

**Godfather** · October 13th 2007, 08:36 PM

Hint: any number divided by itself is 1

we want $(g(x))^2 + 1 = 9x^4 + 1$

solve for $g(x)$
So it would be X^2 +1 but what is x?

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

$g(x) = (3x - 3)^2 + 1$ [/quote]
How did u get that?

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

**Jhevon** · October 13th 2007, 08:51 PM

your post makes no sense. separate my quotes from your questions to make sure i don't miss anything.

Originally Posted by Godfather

Hint: any number divided by itself is 1

you said nothing after you quoted this, so i guess you got the answer

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

no. and you can check this. if $g(x) = x^2 + 1$ and $f(x) = x^2 + 1$ , then $f(g(x)) = \left( x^2 + 1 \right)^2 + 1 = x^4 + 2x^2 + 2$ which is not $9x^4 + 1$

but what is x?

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.

How did u get that?

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)

we started with $g(x) = 9x^4 + 1$

now, $9x^4 + 1 = \left( 3x^2 \right)^2 + 1$

$= \left[ 3 \left( x^2 {\color{blue} + 1 - 1} \right) \right]^2 + 1$ ...........note that i added and subtracted one from the x^2 factor, this is like adding zero, so i'm not changing anything.

$= \left[ 3 { \color{red} \left( x^2 + 1 \right) } - 3 \right]^2 + 1$ ...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

$= \left[ 3 { \color{red} \left( f(x) \right) } - 3 \right]^2 + 1$

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

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

i answered this already

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