Finding Functions that Satisfy a Specific Relationship: A Math Olympiad Problem

Click For Summary
The discussion revolves around finding functions f that satisfy the equation f(x² + f(y)) = y - x² for all real x and y. The initial reasoning involves setting x to 0, leading to f(y) = f⁻¹(y), and exploring cases for x greater than and less than zero. For x > 0, the relationship f(k² + f(0)) = -k² + f(0) is derived, while for x < 0, it leads to f(0) = f(k² + f(-k²)) = f(-k²) - k². Participants are encouraged to verify the correctness of these steps and refer to a previously discussed thread for additional context. The inquiry seeks validation of the reasoning presented.
mtayab1994
Messages
584
Reaction score
0

Homework Statement



f\left(x^{2}+f(y)\right)=y-x^{2}


Homework Equations



Find all functions f that satisfy the relationship for every real x and y.

The Attempt at a Solution



is this correct reasoning?

for x=0: f(y)=f^{-1}(y)

for x>0: \existsxεℝ: x=k^{2}

f(k^{2}+f(0))=-k^{2}+f(0)

for x<0 \existsxεℝ: x=-k^{2}

f(0)=f(k^{2}+f(-k^{2})) = f(-k^{2})-k^{2} which entails:

f(-k^{2})=f(0)+k^{2} =-(-k^{2})+f(0)
 
Physics news on Phys.org
mtayab1994 said:

Homework Statement

f\left(x^{2}+f(y)\right)=y-x^{2}...

for x=0: f(y)=f^{-1}(y)
for x>0: \exists xεℝ: x=k^{2}
f(k^{2}+f(0))=-k^{2}+f(0)
for x<0 \exists xεℝ: x=-k^{2}
f(0)=f(k^{2}+f(-k^{2})) = f(-k^{2})-k^{2}...
This question has been previously discussed.

http:
//www.physicsforums.com/showthread.php?t=556487


Is there anything new in what you're posting this time?
 
Please post in the old thead.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
890
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K