find f(x) if f(x+y)=f(x)+f(y) such that f(x) is continuous at 0.
49f'(x)=lim f(x+h)-f(x)h=limf(h)h=m(const!)
h→0 h→0
so, f(x)=mx+c!! [1][1][1]
1absolutely wrong!!!!!!
surprised???
we are just given f(x) is continuous at 0.had we been given tht f(x) is differentiable at all x then u might have been lucky!
1Hi Kunl
See if I am correct.
If f(x) is continuous the n f(x+0)=f(0)
f(x+y)=f(x)+f(y)
Putting y=0
f(x+0)=f(x)+f(0)
So f(x)=0
Am I getting the right answer?
341We have f(0) = 0 by setting x=y=0.
Now, we prove that if f is continuous at x=0, it is continuous everywhere.
We have given any ε>0 a δ>0 such that |x|<δ implies |f(x)|<ε.
Choose δ'=δ/2
Now for any real number a, |f(a+δ')-f(a)| = |f(a) +f(δ')-f(a)| = |f(δ')|<ε.
Hence, f(x) is continuous at every real number.
So, we can breathe a sigh of relief that this is the Cauchy equation we are familiar with and state that f(x) = cx is the family of solutions.
1[1]nice solution sir[infact i guess proving continuity is only possible solution right???or is there some alternate solution too?]..coz i did the same..it took me a lot of time to figure it out!
341i just had a hunch that this could be true.
The proving part is easy.