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 factor
we want[b] [f+g][x]=3(x^2)
simply solve for
Hint: any number divided by itself is 1[c] [f/g][x]=1
we want[d] f[g[x]]=9(x^4)+1
solve for
i can't really think of a nice hint to give you here...[e] g[f[x]]=9(x^4)+1
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
no. and you can check this. if and , then which is not(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.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)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
i answered this alreadyWhat are all of the x's after [f/g][x]