Let f be a function such that f(x+y)=x+f(y) for any two real numbers x and y. If f(0)=2, then what is f(2012)? Thank you!
To find the value of f(2012), we can start with the given information:
f(x + y) = x + f(y)
We are also given that f(0) = 2. Let's use this information to find some pattern that will help us find f(2012).
Setting x = 0 and y = 0, we have:
f(0 + 0) = 0 + f(0)
f(0) = 0 + f(0)
2 = 0 + 2
This confirms that the equation holds true.
Now, setting y = 0, we have:
f(x + 0) = x + f(0)
f(x) = x + 2
So, for any x, f(x) = x + 2.
To find f(2012), substitute x = 2012 into the equation:
f(2012) = 2012 + 2
f(2012) = 2014
Therefore, f(2012) is equal to 2014.
In summary, using the given information f(x + y) = x + f(y) and f(0) = 2, we can find that f(2012) = 2014 by substituting x = 2012 into the equation f(x) = x + 2.