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Several questions about functions

by colt
Tags: functions
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colt
#1
Apr4-12, 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 fg = gf? Answer: g(x) = g(0) + x

15.6.1)Be f,g and h functions. Show a proof or a counter-proof for the following identities:
a)(f+g)h = fh + gh
b)h(f+g) = hf + hg
c)1/(fg) = (1/f)g
d)1/(fg) = 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: fg = f(g(x)) = f(x+c) = x+c+1. gf = 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.
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Office_Shredder
#2
Apr4-12, 01:17 PM
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Your first two questions don't make sense as worded.
Proof that there isn't functions fg such that f(x)g(y)=x+y
What does the composition of f and g have to do with the equation f(x)g(y)=x+y?
colt
#3
Apr4-12, 10:09 PM
P: 20
My mistake:

Proof that there isn't functions f and g such that f(x)and g(y) = xy for all values of x and y.
Proof that there isn't functions f and g such that f(x)g(y)=x+y

These are the correct versions


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