33No it is really clear..
Diff wrt x we get...f'(x+y)=f'(x)-8y
now wrt y we get f'(x+y)=f'(y)-8x
so f'(x)+8x=f'(y)+8y
So f'(x)+8x= constant...
Now its all clear...
let y=f(x) and y=g(x) be two continuos functions satisfying f(x+y) =f(x) +f(y) - 8xy for all x,y ε R and g(x+y)=g(x) + g(y) +3xy(x+y) for all x,y εR also f'(0)=8 and g'(0)=-4 then find f(x) and g(x)
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19 Answers
3333Actually Nishant Bhaiya..
My method works as long as given eqaution is symmetric in x and y :P
33Nishant bhaiya ka answer galat hai :P in that g(x)
My "Priyam's diff method" gave right answer without calculation mistake...
:P
1bhaiya, in ur second post (9.12 a.m.) ... 3rd line..
y the sign b4 3x(x+y) term became minus?
33But above both f(x) and g(x) have terms like that of xy and g(x) also have x2y and it worked :P
62priyam your method will work as long as the functions dont have terms of the kind xy or xy2
look back at all those questions u solved.. ;)
33No considering y as a constant... :P
and answer is correct
it is my own method of solving these and had worked for all such problems not a single one has escaped..
I don't know if method is right or wring but it gives answer :p
62where is dy/dx???
if you are taking partial derivative then also there is a mistake ;)
33And this method requires no fitting...
:P
Only it should be differentiable and should contain x and y... i.e. 2 variables
62f(x+y) =f(x) +f(y) - 8xy
substitute x=y=0, we get f(0)=0
f(x+y)-f(x) =f(y)/y - 8x
y
f(x+y)-f(x) ={f(y)-f(0)}/y - 8x
y
take limit y-> 0
f'(x) = f'(0) - 8x
f'(x) = 8-8x
f(x) = 8x - 4x2 + c
f(0)= 0 gives c=0
hence f(x) = 8x - 4x2
62I repeat what one of my teachers once said (not to me!)
"What is obvious to you might not be obvious to me"
He is a brilliant Mathematician :)
33Clearly for f(x)
f'(x)+8x=8
as f'(x)+8x=constant
Now solving this we get
f(x)=8x-4x2
62lol.. yeah i "pinked" my posts..
none of u did.. so i thought it wud be a nice idea to :)
11You've given urself two pink backgrounds!
Brilliant, Nishant bhaiyyah! I'm lovin' it
62g(x+y)=g(x) + g(y) +3xy(x+y)
substitute x=y=0, we get g(0)=0
g(x+y)-g(x) =g(y)/y - 3x(x+y)
y
g(x+y)-g(x) ={g(y)-g(0)}/y - 3x(x+y)
y
take limit y-> 0
g'(x) = g'(0) - 3x(x)
g'(x) = -4 - 3x2
g(x) = -4x - x3