
#1
Apr412, 12:47 PM

P: 20

1. The problem statement, all variables and given/known data
I have these problems: 15.6.2)Proof that there isn't functions f and g such that f(x)and g(y) = xy for all values of x and y. Sugestion: Considering x=0 proof that g must be a constant function. After doing that, consider now y=0 e sees what happens with the function. 15.6.3)Proof that there isn't functions f and g such that f(x)g(y)=x+y for alll values of x and y. Sugestion: Considering x=0 proof that f(0) != 0 and determines an expression for g(y). Doing that, consider then y=0 and sees what happens with x. 15.6.4)Be f(x)= x+1.There is such functions f°g = g°f? Answer: g(x) = g(0) + x 15.6.1)Be f,g and h functions. Show a proof or a counterproof for the following identities: a)(f+g)°h = f°h + g°h b)h°(f+g) = h°f + h°g c)1/(f°g) = (1/f)°g d)1/(f°g) = f°(1/g) 3. The attempt at a solution What I tried to do: 15.6.2) I am not even sure if I understand the question.F(x)and G(y) = xy means that both functions are equal. I think they can't. How can a single variable function produces two variables as a answer? 15.6.3)Same problem as above. 15.6.4)I don't understand the answer. Why this unusual answer based on its on value? Why can't g(x) = x or g(x) = x+c ? In my hypothesis: f°g = f(g(x)) = f(x+c) = x+c+1. g°f = g(f(x)) = g(x+1) = x+1+c 15.6.1)a) and b) I know that (f + g) (x) = f(x) + g(x). So (f+g)°h = (f() + g())°h. Not sure if this is right and even if it is, no idea how to show of that the ° operator is distributive (or not). 15.6.1)c and d. No idea in this one. 



#2
Apr412, 01:17 PM

Mentor
P: 4,499

Your first two questions don't make sense as worded.




#3
Apr412, 10:09 PM

P: 20

My mistake:
These are the correct versions 


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