Real numbers x and y, f(x+y)=f(x)+f(y)+1. If f(1)=2, what is f(3)?

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The function f satisfies the equation f(x+y) = f(x) + f(y) + 1 for all real numbers x and y, with the known value f(1) = 2. To find f(3), first calculate f(2) using f(2) = f(1+1), which leads to f(2) = f(1) + f(1) + 1, resulting in f(2) = 5. Subsequently, f(3) can be determined using f(3) = f(2+1) = f(2) + f(1) + 1, yielding f(3) = 8. The problem illustrates the application of functional equations in determining specific function values.
Xasuke
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Ok, I'm sure this is an easy problem and all, but it's pissing me off. I'm probably just not understanding it.

The function f has the property that for any real numbers x and y, f(x+y)=f(x)+f(y)+1. If f(1)=2, what is f(3)?

help.
 
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Just use the information provided to find f(2) from which f(3) = 8 follows.
 
wow... I feel completely lost.
How do I find f(2)?
 
f(2)=f(1+1)

If you are given that f(x+y)=f(x)+f(y)+1

then it would be logical to conclude that

f(1+1)=f(1)+f(1)+1 and I think you can take it from there
 
Oh my.. Thanks. I'm such an idiot =)
 
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