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izat

#1 h Jan 28, 2016, 4:41 AM • 1 Y

Y by Adventure10

FInd all $f : \mathbb{R}\to\mathbb{R}$ such that $f(x^2+y+f(y))=2y+f(x)^2$ . For all $x, y$

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9045 posts

#2 h Jan 28, 2016, 8:03 PM • 1 Y

Y by Adventure10

This is an old problem. But since I just can't find it I will present the standard solution:
Solution

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KMI

#3 h Mar 5, 2016, 5:00 AM • 1 Y

Y by Adventure10

Tintarn wrote:

we obtain $a^2=b^2$ . So $a^2=b^2 \Leftrightarrow f(a)^2=f(b)^2$ .
Since $f$ is surjective this implies $f(-x)=-f(x)$ for all $x$ .

for surjective function ? not injective?
typo or error?

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#4 h Mar 5, 2016, 5:25 AM • 2 Y

Y by Adventure10, Mango247

Another link of problem http://www.artofproblemsolving.com/community/c6t298f6h1207079_functional_eq

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#5 h Mar 5, 2016, 6:50 AM • 2 Y

Y by Tintarn, Adventure10

KMI wrote:

Tintarn wrote:

we obtain $a^2=b^2$ . So $a^2=b^2 \Leftrightarrow f(a)^2=f(b)^2$ .
Since $f$ is surjective this implies $f(-x)=-f(x)$ for all $x$ .

for surjective function ? not injective?
typo or error?

Maybe Tintarn meant that, for all $x$ positive $f(x)$ is positive. $f$ is surjective, so we have that there is $y$ such that $f(y)=-f(x)$ . We know that $y$ could be only $-x$ or $x$ . The latter is wrong, so $y=-x$ .

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