1ok
im stuck at this one [41]
Find f(x) if
f(x + y) = x. f(y) + y. f(x) and f '(0) = 1
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19 Answers
1put x=y
f(2x)=2xf(x)
differentiate and put x = 0,we get
2f'(2x)=2[xf'(x) + f(x)]
=> 2f'(0)=2f(0)
=> f(0) = 1 ( as f'(0) = 1) given
but by putting x=y=0 we get f(0) = 0.. so ......................
1Well is it mentioned that the function is differentiable at other points than x=0?
f(x+y)=xf(y)+yf(x)
put y=-x
f(0)=xf(-x)-xf(x)
put x=y=0
f(0)=0
thus we get
f(-x)=f(x)
put y=-2x
f(-x)=xf(-2x)-2xf(x)
but f(-x)=f(x) and f(-2x)=f(2x)=2xf(x)
so we get
f(x)=2x2f(x)-2xf(x)
which gives f(x)=0
But then f'(0)=0 opposing to the condition given in question
62the second part has 0/0 form ..so we can apply l'hospital
so f'(x) =f(x) + x f'(h)
statement is slightly wrong..
but the next step follows from limit h tending to zero
btw see dimension's explanation to show the fallacy with this question..
1f' (x)=f(x+h)-f(x)/h where h tends to 0
{using first principle}
=xf(h)+hf(x)-f(x)/h
=f(x)+ [xf(h)-f(x)]/h
but f(x)=xf(0) using given condition
so we get
f'(x)=f(x) + [x{f(h)-f(0)}]/h where h tends to 0
the second part has 0/0 form
.. so we can apply l'hospital so
f'(x) =f(x) + x f'(h) putting h=0
we get f'(X)=f(x)+xf'(0) or
f'(x)=f(x)+x as f'(0)=1
1put x=y
f(2x)=2xf(x)
differentiate
f'(2x)=xf'(x)+f(x)
put x=0
we get f(0)=1
this is in contradicton when we put x=y=0 in org. equn,by which we get f(0)=0
so,i think, no function satisfies the given condition !!!
62f(x + y) = x. f(y) + y. f(x) and f '(0) = 1
substitute y=0
f(x) = x f(0) + 0
thus, f(x) = x. f(0)
now we just need to find f(0)
substitute x=y=0 ..
f(0) = 0.
Thus f(x) = x is the solution..
Have i missed something! ?