SOLUTION: Hello, I have a few problems, and this should be easy but I'm over thinking it. f9x0 = x/x+t g(x)= x^3 A) Find f(x) + g(x) I found a common denominator, (x+1) but I got confu

Algebra ->  Expressions-with-variables -> SOLUTION: Hello, I have a few problems, and this should be easy but I'm over thinking it. f9x0 = x/x+t g(x)= x^3 A) Find f(x) + g(x) I found a common denominator, (x+1) but I got confu      Log On


   

Question 127634: Hello, I have a few problems, and this should be easy but I'm over thinking it.
f9x0 = x/x+t g(x)= x^3
A) Find f(x) + g(x)
I found a common denominator, (x+1) but I got confused after that...
x^3 is really X^3/1 so to get the common denominator, would you multiply by x since the one is already in the denominator or just by x? and what about the denominator?
Also, I have a function:
f(x)= cube root of (x-1) g(x)= x^3 + 1
c) f(f(x)=?
All you have to do is substitute for this problem, correct?
So... f(f(x))= {cube root[cube root of (x-1)] -1} Would be correct?

If anyone could set me straight with these, I'd appreciate it!
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+x%2F%28x%2Bt%29 g%28x%29=+x%5E3

Is the denominator x%2Bt or x%2B1? I'm going to assume a typo in the original function statement since you go on about x%2B1.

The two denominators are x%2B1 and 1. Since they have no factors in common, the common denominator is the product of the two: %28x%2B1%29%281%29=x%2B1,

So:
%28x%2F%28x%2B1%29%29%2B%28x%5E3%29%2F1


%28x%2B%28x%5E3%28x%2B1%29%29%29%2F%28x%2B1%29

%28x%5E4%2Bx%5E3%2Bx%29%2F%28x%2B1%29



f%28f%28x%29%29=+root%283%2Croot%283%2Cx-1%29%29+-1 is your answer...but you missed it by THAT much.

f%28f%28x%29%29=+root%283%2Croot%283%2Cx-1%29-1%29. See the difference? You just had a misplaced bracket.