# How to Solve for Function of Two Variables

• May 14th 2011, 11:43 AM
How to Solve for Function of Two Variables
if f((x+y)/3) = (2+f(x) +f(y))/3 for all real x, y and f’ (2) =2 and f if differentiable function
Then can we find f (x) .in question nothing is said about f(x)(no further information is given)
but they are finding the answer as 2x+2
(Rofl)
• May 14th 2011, 12:13 PM
TheEmptySet
Quote:

if f((x+y)/3) = (2+f(x) +f(y))/3 for all real x, y and f’ (2) =2 and f if differentiable function
Then can we find f (x) .in question nothing is said about f(x)(no further information is given)
but they are finding the answer as 2x+2
(Rofl)

Take the partial derivative with respect to y to get

$\frac{1}{3}\frac{\partial f}{\partial y}\left(\frac{x+y}{3} \right)=\frac{1}{3}\frac{\partial f}{\partial y}(y)$

Now if we evaluate this at the point x=3x y=0)we get (I will now use primes to represent the derivative with respect to y)

$f'(x)=f'(0)$

Since the derivative is constant the function must me linear! If we plug in x=2 we get

$f'(2)=f'(0) =2$

So the equation must have the form

$f(x)=2x+b$

Can you finish from here?
• May 14th 2011, 12:19 PM
abhishekkgp
Quote:

if f((x+y)/3) = (2+f(x) +f(y))/3 for all real x, y and f’ (2) =2 and f if differentiable function
Then can we find f (x) .in question nothing is said about f(x)(no further information is given)
but they are finding the answer as 2x+2
(Rofl)

this is not a very rigorous solution but is good to see.
put x=0,y=0 to get f(0)=2.

the given condition becomes
$f(\frac{0+x+y}{3})=\frac{f(0)+f(x)+f(y)}{3}$.....(1)
note that the centroid of the triangle $OAB$ with $O=(0,f(0)),\,A=(x,f(x)),\,B=(y,f(y))$ is $(\frac{0+x+y}{3},\frac{f(0)+f(x)+f(y)}{3})$.
i am assuming that the curve does not change its concavity(this is where it becomes non-rigorous but this may be established by putting y=x and then using second derivative test).
The centroid of the triangle is inside the triangle but from (1) the centroid is on the curve Y=f(X) and hence outside ot on the triangle .... to see this draw a curve which is concave downwards and mark the three points OAB(this part is also non-rigorous)

so the curve of 'f' has to be a straight line.
put f(X)=aX+b and use f'(2)=2.
• May 14th 2011, 09:48 PM
Quote:

Originally Posted by abhishekkgp
this is not a very rigorous solution but is good to see.
put x=0,y=0 to get f(0)=2.

the given condition becomes
$f(\frac{0+x+y}{3})=\frac{f(0)+f(x)+f(y)}{3}$.....(1)
note that the centroid of the triangle $OAB$ with $O=(0,f(0)),\,A=(x,f(x)),\,B=(y,f(y))$ is $(\frac{0+x+y}{3},\frac{f(0)+f(x)+f(y)}{3})$.
i am assuming that the curve does not change its concavity(this is where it becomes non-rigorous but this may be established by putting y=x and then using second derivative test).
The centroid of the triangle is inside the triangle but from (1) the centroid is on the curve Y=f(X) and hence outside ot on the triangle .... to see this draw a curve which is concave downwards and mark the three points OAB(this part is also non-rigorous)

so the curve of 'f' has to be a straight line.
put f(X)=aX+b and use f'(2)=2.

Sir can you draw the figure ?(Nerd)
• May 14th 2011, 10:04 PM
Quote:

Sir can you draw the figure ?(Nerd)

i am assuming that the curve does not change its concavity(this is where it becomes non-rigorous but this may be established by putting y=x and then using second derivative test).
The centroid of the triangle is inside the triangle but from (1) the centroid is on the curve Y=f(X) and hence outside ot on the triangle .... to see this draw a curve which is concave downwards and mark the three points OAB(this part is also non-rigorous
bhaiya i cant understand even a single line .but it looks beautiful. can you explain in a very simple way whole of the above part