We are given that f(x+y)=f(x)+f(y) and that f(1)=3.

We assume that f(x) is continuous. (A non-continuous image is possible, but requires real analysis which is beyond a high school class.)

Then f is of the form f(x)=kx (or y=kx) for some `kinRR` .(f(0+0)=f(0)+f(0)==>f(0)=2f(0) so f(0)=0. Also f(rx)=f(x)+f(x)+ ... +f(x) (r...

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We are given that f(x+y)=f(x)+f(y) and that f(1)=3.

We assume that f(x) is continuous. (A non-continuous image is possible, but requires real analysis which is beyond a high school class.)

Then f is of the form f(x)=kx (or y=kx) for some `kinRR` .(f(0+0)=f(0)+f(0)==>f(0)=2f(0) so f(0)=0. Also f(rx)=f(x)+f(x)+ ... +f(x) (r times)==>f(rx)=rf(x).

So plugging in the known values we get `3=k*1 ==>k=3`

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**The solution is f(x)=3x**

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Checking: f(1+3)=f(4)=12=3+9=f(1)+f(3) as required.

The problem provides the information that `f(1) = 3` , hence, putting `x = 1,y = 0` and f(y) = 0 yields:

`f(1+0) = f(1) + f(0) =gt f(1) = 3`

Putting `x = y` yields:

`f(x + x) = f(x) + f(x) =gt f(2x) = 2f(x)`

Putting `x=y=1` yields:

`f(1+1) = f(1)+f(1) =gt f(2) = 2f(1) =gt f(2) = 2*3 = 6`

Putting `x = 2, y=1` yields:

`f(2+1) = f(2) + f(1) =gt f(3) = 6 + 3 = 9`

`f(k+1) = f(k) + 3 =gt f(k) = f(k+1)-3`

**Hence, evaluating f(x), considering different values for y, yields: for `y = 0, x=1=gt f(1)=3` , for `x=y =gt f(x) = f(2x)/2` ; for `y =1` and `x =1,2,...,x` , then `f(x) = f(x+1)-3` .**