SOLUTION: Let f be a function such that f(x+y) = x + f(y) + f(y^2) - y^2 for any two real numbers x and y. If f(0) = -5, then what is f(1)?

Algebra ->  Functions -> SOLUTION: Let f be a function such that f(x+y) = x + f(y) + f(y^2) - y^2 for any two real numbers x and y. If f(0) = -5, then what is f(1)?      Log On


   

Question 1209317: Let f be a function such that
f(x+y) = x + f(y) + f(y^2) - y^2
for any two real numbers x and y. If f(0) = -5, then what is f(1)?
Found 2 solutions by Edwin McCravy, mccravyedwin:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

Something is wrong because:

      f(x+y) = x + f(y) + f(y^2) - y^2

Let x = y = 0

      f(0+0) = 0 + f(0) + f(0^2) - 0^2

        f(0) = 0 + f(0) + f(0) - 0

        f(0) = 2f(0)
 
f(0) - 2f(0) = 0
       
       -f(0) = 0

        f(0) = 0

So f(0) cannot equal -5

Edwin

Answer by mccravyedwin(407) About Me  (Show Source):
You can put this solution on YOUR website!
Now if you want to change the problem so that f(0) = 0, not -5,
then:


       f(x+y) = x + f(y) + f(y^2) - y^2

Let x = -1 and y = 1

      f(-1+1) = -1 + f(1) + f(1^2) - 1^2

         f(0) = -1 + f(1) + f(1) - 1

            0 = 2f(1) - 2

            2 = 2f(1) 

         f(1) = 1

Edwin