small doubts

lukaswelker wrote: Now in order to resolve it I replace a=1, b=2. giving me;

f(3)= f(1)+f(2)

Which I would then try for every possibility.

Correct way to do it? or pure luck?

Perfect approach, Lukas!

Below is my approach. It tests a = 1 and b = 1.

For which of the following functions f(a+b) = f(a) + f(b) for all positive numbers a and b?

f(x)= x²
f(x)= x+1
f(x)= √x
f(x)= 2/x
f(x)= -3x

One approach is to plug in numbers. Let's let a = 1 and b = 1

So, the question becomes, "Which of the following functions are such that f(1+1) = f(1) + f(1)?"
In other words, for which function does f(2) = f(1) + f(1)?

A) If f(x)=x², does f(2) = f(1) + f(1)?
Plug in to get: 2² = 1² + 1²? (No, doesn't work)
So, it is not the case that f(2) = f(1) + f(1), when f(x)=x²

B) If f(x)=x+1, does f(2) = f(1) + f(1)?
Plug in to get: 2+1 = 1+1 + 1+1? (No, doesn't work)
So, it is not the case that f(2) = f(1) + f(1)
.
.
.
A, B, C and D do not work.
So, at this point, we can conclude that E must be the correct answer.
Let's check E anyway (for "fun")

E) If f(x)=-3x, does f(2) = f(1) + f(1)?
Plugging in 2 and 1 we get: (-3)(2) = (-3)(1) + (-3)(1)
Yes, it works

The correct answer is E

Cheers,
Brent