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let f(x+y/2)= f(x)+ f(y) / 2 for all real x and y . if f'(0) exists and equals -1 and f(0)=1,find f(2) ans is -1

let f(x+y/2)= f(x)+ f(y) / 2 for all real x and y . if f'(0) exists and equals -1 and f(0)=1,find f(2) 
ans is -1

Grade:11

2 Answers

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
9 years ago
Hi
f(x/2) = (1/2)f(x) + (1/2)f(0) = (1/2)f(x) + 1/2
f(x) = (1/2)f(2x) + 1/2
f ' (0) = lim(x->0) (f(x) - f(0))/x
f ' (0) = lim(x->0) (f(x) - f(0))/x
f ' (0) = lim(x->0) (f(2x) - 1)/(2x) = (1/2)lim(x->0) (f(2x) -1)/x
f ' (x) = lim(y->0) ( f(x+y) - f(x) ) /y
f(x+y) = f( (2x)/2 + (2y)/2).
f(x+y) = (1/2)f(2x) + (1/2)f(2y)
f '(x) = lim(y->0) [ (1/2)f(2x) + (1/2)f(2y) - f(x)]/y
(1/2)f(2x) - f(x) = -1/2
f ' (x) = lim(y->0) [ (1/2)f(2y) - 1/2] /y = (1/2) lim(y->0) [f(2y) - 1]/y
(1/2) lim(y->0) [f(2y) - 1]/y = f'(0)
f '(x) = -1, so f(x) = -x + c
1 = f(0) = c so f(x) = 1 - x
f(2) = 1 - 2 = -1
Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty
Ayush Singh
28 Points
2 years ago
Well,
If we see carefully then f(x+y/2) is simply fx+fy/2  so if we try to take fx = x+c then f(0)=1 condition is satisfied with c=1.
But since f'(0) = -1 so if we try to substitute
f(x)=(-)x+1 then all conditions given are satisfied with f(x+y/2)= -(x+y/2) +1
Which is basically [(-x+1) + (-y+1)]/2 and which is fx+fy/2 
So now we can directly see that f(2)=-2+1 =(-1). ;--)

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